Bihar Board Class 12 Physics Solutions Chapter 3 Current Electricity Textbook Questions and Answers, Additional Important Questions, Notes.

## BSEB Bihar Board Class 12 Physics Solutions Chapter 3 Current Electricity

**Bihar Board Class 12 Physics **Current Electricity Textbook Questions and Answers

Question 1.

The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

Answer:

Here, E = e.m.f of storage battery = 12 V

r = internal resistance = 0.4 Ω.

I = maximum current that can be drawn from the battery = ?

The maximum current is drawn from the battery when the external resistance in the circuit is zero i.e., R = 0.

Thus using the relation,

Question 2.

A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Answer:

Here, E = 10 V

r = 3 Ω

I = current in the circuit = 0.5 A.

R = resistance of the resistor = ?

V = terminal voltage of battery when circuit is closed = ?

Using the relestion I = \(\frac {E}{R + r}\) , we get

R = \(\frac {E}{I}\) – r = \(\frac {10}{0.5}\) – 3 = 10 x 2 – 3 = 20 – 3

or R = 17 Ω

Now using V = IR, we get

V = 0.5 x 17 = 8.5 A.

Question 3.

(a) Three resistors lfi, 1Ω,2Ω Rand 3Ω are combined in series. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Answer:

Here,

R_{1} = 1Ω

R_{2} = 2Ω

R_{3} = 3Ω

Where R_{1}, R_{2}and R_{3} are the three registor connected in series.

(a) Let Rs = total resistance of the series combination = ?

In series, the total resistance is given by

R_{s} = R_{1} + R_{2} + R_{3} = 1 + 2 + 3 = 6 Ω.

(b) Here, E = e.m.f. of battery connected across the combination = 12 v:

r = 0. Let V_{1}, V_{2} and V_{3} be the potential drop across each resistor = ?

Also Let I = current flowing through the combination.

.’. V_{1} = Potential drop across R_{1} = IR_{1} = 2 x 1 = 2V

V_{2} = Potential drop across R_{2} = IR_{2} = 2 x 2 = 4V

V_{3}= Potential drop across R_{3} = IR_{3} = 2 x 3 = 6V

Question 4.

(a) (a) Three resistors 2Ω, 4Ω and 5Ω are combined in parallel. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

Answer:

Let R_{1}, R_{2} and R_{3} be the three resistances connected in parallel.

Here,

R_{1} = 2 Ω

R_{2} = 4 Ω

R_{3} = 5 Ω

(a) If R_{p} be the resistance of the parallel grouping = ?

(b) Here, E = 20 V,r = 0

Let I = total current in the circuit = ?

and I_{1}, I_{2},I_{1} be the current through R_{1}R_{2} and R_{3}, respectively?

∴ I_{1} = current through R_{1} = \(\frac{\mathrm{E}}{\mathrm{R}_{1}}\) = \(\frac {20}{4}\) = 1OA

I_{2} = current through R_{2} = \(\frac{\mathrm{E}}{\mathrm{R}_{2}}\) = \(\frac {20}{4}\) = 5 A

I_{3} = current through R_{3} = \(\frac{\mathrm{E}}{\mathrm{R}_{3}}\) = \(\frac {20}{5}\) = 4 A

Question 5.

At room temperature (27.0 °C) the resistance of a heating element is 100 12. What is the temperature of the element if the resistance is found to be 11712, given that the temperature coefficient of the material of the resistor is 1.70 x 10^{-4} °C^{-1}.

Answer:

Let t°C = temperature of the element = ?

Here, R_{t} = Resistance at t°C = 117 Ω.

R_{27} = resistance at 27°C = 100 Ω.

α = temperature coefficient of resistance = 1.70 x 10^{-4} °C^{-1}

Using the relation,

Question 6.

A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 x 10^{-7} m^{2}, and its resistance is measured to be 5.01 Ω . What is the resistivity of the material at the temperature of the experiment?

Answer:

Here, l =. length of wire = 15 m.

A = area of cross-section of the wire = 6.0 x 10^{-7} m^{2}.

R = resistance of wire = 5.012.

ρ = resistivity of the wire = ?

Using the relation, R = \(\frac{\rho l}{\mathrm{~A}}\), we get

Question 7.

A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 oc. Determine the temperature coefficient of resistivity of silver.

Answer:

Here, R_{0} = R_{27.5} resistance of 27.5°C = 2.1 Ω

R_{t} = R_{100} = resistance at 100°C = 2.7 Ω

∝ = temperature coefficient of resistivity =?

∆t = 100 – 27.5 = 72.5°C.

Using the relation,

a = ,weget

Question 8.

A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds

to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C Temperature coefficient

of resistance of nichrome averaged over the temperature range involved is 1.70 x 10^{-4} °C^{-1}

Answer:

Here, V = supply voltage = 230 V.

I_{1} = initial current 3.2 A

θ_{1} = initial temperature 27°C = room temperature

I_{2} = steady current = 2.8 A

θ_{2} = steady temperature =?

∝ = temperature coeïficient of resistance = 1.7

If R_{1} and R_{2}, be the resistance of wire atθ_{1} and θ_{2} respectively, then

Question 9.

Determine the current in each branch of network shown in figure.

Answer:

Let I be the total current in the circuit.

I_{1} = Current flowing through AB.

.’. I – I_{1} = Current flowing through AD.

I_{2} – Current flowing throug BD.

.’. I_{1} – I_{2} = current flowing through BC. and I_{1} – I_{2} + I_{2} = Current flowing through DC.

Applying loop law to ABDA, we get 10 I_{1} + 5 I_{2} – 5 (I – I_{1}) = 10

or 3I_{1} + I_{2} – I = 0 ……….(i)

Again applying loop law to BCDB, we get

5 (I_{1} -I_{2}) -10 (I – I_{1} + I_{2}) – 5I_{2} = 0

or 15I_{1} – 2I_{2} – 10I = 0

or 3I_{1} – 4I_{2} – 2I = 0 …….(ii)

Applying loop law to ABCEFA, we get

10I + 10I_{1} + 5 (I_{1} – I_{2}) = 10

or 3F_{1} – I_{2} + 2I = 2 …………(iii)

(ii) + (iii) gives, 6I_{1} – 5I_{2} = 2 …….(iv)

Multiplying (i) by 2 and then adding to (iv) we get

9I_{1} + I_{2} = 2 ……..(v)

(vi) + 5 x (v) gives,

6I_{1} – 5I_{2} + 45I_{1} + 5I_{2} = 2 + 10

or 51 I_{1} = 12

or I_{1} = \(\frac {4}{17}\) A ……..(vi)

∴ Current in branch AB, I_{1} = \(\frac {4}{17}\) A

∴ From (v) and (vi), we get

I_{2} = 2 – 9 x \(\frac {9}{17}\) = – \(\frac {2}{17}\) A

– ve sign shows that I2 is actually from D to B. Now from (i), we get

Question 10.

(a) In a metre bridge [Fig.], the balance point is found to be at 39.5 cm from the end A, when the resistor Vis of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

(b) Determine the balance point of the bridge above if X and V are interchanged.

(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

Answer:

Here, lx – 39.5 cm; R – X = ? S = Y = 12.5 Ω

we know that

Thick copper strips are used as they have negligible resistance, thus their use minimizes the resistance of connections so that the value of resistors of wheat stone/metre bridge are not changed.

(b) When X and Y are inter changed, then R = Y = 12.5 Ω, S = X = 8.16 0,l = ?

(c) Now A and C will be at the same potential so no current will flow through galvanometer. When at balance point galvanometer and cell are interchanged. No, it will not show any current.

Question 11.

A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Answer:

Here, r = internal resistance of battery = 0.5 Ω

R = series register = 15.5 Ω

Ej = e.m.f. of storage battery = 8.0 V

E2 = Voltage of D.C. Supply= 120 V

As the storage battery of e.m.f. E_{1} is charged with a d.c. supply of e.m.f. E_{2}, SO the net e.m.f. ‘E’ in the circuit is given by

E = E_{2} – E_{1} = 120 – 8 = 112 V

Total resistance in the circuit, R’ = R + r = 15.5 + 0.5 = 16.0 Ω.

Let V = terminal voltage of the battery during charging = ?

If I = current in the circuit during charging, then

I = \(\frac{E}{R+r}\) = \(\frac{112}{16}\) = 7A.

During charging, the voltage of the D.C. supply in a circuit must be equal. to the sum of the voltage drop across R and terminal voltage of the battery.

i.e., E_{2} = V + IR

or 120 = V + 7 x 15.5

or V = 120 – 108.5 = 11.5 V

V can also be calculated as :

V = sum of e.m.f. of battery + P.D. across r = E + Ir = 8 + 7 x 0.5 = 11.5 V.

The purpose of having a series register in the charging circuit is to limit the current drawn from the external source of d.c. supply which is dangerously high in its absence.

Question 12.

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

Answer:

Here,

E_{1} = e.m.f of first cell = 125 V

l_{1} = balancing length for first cell 35 cm

E_{2} = e.m.f of 2nd cell = ?

l_{2} = balancing length for second cell = 63 cm

Using the relation,

\(\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{l_{1}}{l_{2}}\)

E_{2} = \(\frac{l_{2}}{l_{1}} \times \mathrm{E}_{1}=\frac{63}{35}\) x 1.25

= 9 x 0.25 = 2.25 V.

Question 13.

The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 x 10^{28} m^{– 3}. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 x 10^{-6} m2 and it is carrying a current of 3.0 A.

n = 8.5 x 10^{28} m^{-3}

Answer:

Here, n = number density of free electrons = 8.5 x 10^{28} m^{-3}

l = length of wire = 3 m

A = Area of cross-section of wire = 2.0 x 10^{-6} m^{2}

I = current in the wire = 3.0 A

e = 1.6 x 10^{-19} C

Let t = time taken by electron to drift from one end to another of the wire = ?

Using the relation, I = neA υ_{d} , we get

v_{d} = I/neA

Question 14.

The earth’s surface has a negative surface charge density of 10^{-9} C m^{-2}. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 x 10^{6} m).

Answer:

Here, a = surface charge density of earth = 10^{-9} cm^{2}

R = radius of earth = 6.37 x 10^{6} m

I = Current over entire globe = 1800 A

V = P.D. between the top of atmosphere and surface of earth = 4000 KV.

Let t = time required to neutralise earth’s surface = ?

∴ Area of whole globe, A = 4πR^{2}

= 4π x (6.37 x 10^{6})^{2} = 509.64 x 10^{12} m^{2}.

Also we know that

σ = \(\frac {q}{A}\) = \(\frac {It}{A}\)

t = \(\frac {σ A}{I}\) = \(\frac{10^{-9} \times 509.64 \times 10^{12}}{1800}\)

= 0.283 x 10^{3} S

= 283 s = 4 min 43 s.

Question 15.

(a) Six lead-acid type of secondary cells each of tmf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?

(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

Answer:

Here, e.m.f. of each secondary cell, E = 2.0 V

n = no. of cells = 6

r = internal resistance of each cell = 0.015 Ω

R = external resistance = 8.5 W

Total e.m.f = nE = 6 x 2 = 12 V R + nr = 8.5 + 6 x 0.015 = 8.59 Ω.

(i) If I be the total current, then

\(\frac{\mathrm{nE}}{\mathrm{R}+\mathrm{n} \mathrm{r}}=\frac{12}{8.59}\) = 1.4 A

(ii) Let V = terminal voltage = IR = 1.4 x 8.50 = 11.9 V

V can also be calculated as :

V = nE – Im = 12-1.4 x 0.015 x 6 = 11.876 = 11.9 V

(b) Here, E = e.m.f. of cell = 1.9 V.

r = internal resistance of cell = 380 Ω.

I _{max}= maximum current that can be drawn from the cell = ?

We know that I = \(\frac {E}{R + r}\)

For I = I_{max} , R must be zero.

I_{max} = \(\frac {E}{R}\) = \(\frac {1.9}{380}\) = 0.005 A = 5 x 10^{-3} A = 5 mA.

No, the cell cannot be used to drive the starting motor of a car as initially it needs a current of nearly 100 A.

Question 16.

Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (ρ_{al} = 2.63 x 10^{-8} m, ρ_{Cu} = 1.72 x 10^{-8} Ω m, Relative density of Al = 2.7, of Cu = 8.9).

Answer:

Let A_{1} A_{2} be the area of cross-section of aluminium and copper wires respectively.

l = lenth of each wire

Also let R_{1}, R_{2} be the resistance of aluminium and copper wires respectively. .

R_{1} = R_{2} (given) ……(i)

For Aluminium Wire ρ_{2} = resistivity of Al = 2.63 x 10^{8} Ωm.

relative density of Al = 2.7

∴ Its density, d_{1} = 2.7 x 10^{3} kg m^{-3}.

Thus using the relation,

R = ρ \(\frac {l}{A}\) we Set R_{1} = p_{1}\(\frac{l}{\mathrm{~A}_{1}}\)

or R = 2.63 x 10^{-8} x \(\frac{l}{\mathrm{~A}_{1}}\) ………..(ii)

if m_{1} = mass of al wire then

m_{1} = density x volume

= d_{1} x A_{1}l_{1} = 2.7 x 10^{3} x A_{1} l ………(iii)

p =resistivity of Cu = 1.72 x 10^{-8}Ωm.

relative densitv of Cu = 8.9

∴ Its density, d_{2} = 8.9 x 10^{3} kg m^{-3}.

∴ R_{2} = \(\frac{\rho_{2} \cdot l_{2}}{\mathrm{~A}_{2}}\) = \(\frac{1.72 \times 10^{-8} l}{\mathrm{~A}_{2}}\) ……….(iv)

m_{2} = mass of Cu wire = A_{2}d_{2},P_{2}

= A_{2} x 8.9 x 10^{3} x I ………..(v)

From (i), (ii) and (iv), we get

It follows from (vii) that the Aluminium wire of given length is lighter than copper wire of same length since for the same values of resistance and length, the aluminium wire has lesser mass than copper wire, so aluminium wire is preferred for overhead power cables. A heavy cable may sag down owing to its own weight.

Question 17.

What conclusion can you draw from following observations on a resistor made of alloy manganin:

Answer:

Ohm’s law is valid to a high accuracy.

We know that R = \(\frac {V}{I}\)

From the observations given here, for all currents from 0.2 A to 8.0 A, the resistance of the registor is almost same and is equal to 19.7 Q. As the current increases, temperature also increases but resistance has no effect. Also we know that the resistance of alloys i.e., manganin here does not change with temperature and their temperature coefficient of resistance is negligibly small. Thus the resistance as well as resistivity of the alloy manganin is nearly independent of temperature.

Question 18.

Answer the following questions:

(a) A steady current flows in a metallic conductor of non-uniform cross- section. Which of these quantities is constant along the conductor current, current density, electric field, drift speed?

(b) Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.

(c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?

(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why? ‘

Answer:

(a) We know that current density, electric field and drift speed all are inversely proportional to the area of cross-section of the conductor. As the steady current flows in the metallic conductor of non-uniform area of cross-section, so only current remains constant, when it flows through a conductor of non-uniform area of cross-section.

(b) No, Ohm’s law is not universally applicable for all conducting elements. It is valid only for metallic conductors and that also when the physical conditions like temperature, stress etc. of the conductor remains the same.

The examples of elements which don’t obey Ohm’s law are vacuum diode semi conductor, diodes, thermistors, thyristor SCR etc.

(c) The maximum current that can be drawn from a voltage supply is given by

I_{max} = \(\frac {E}{r}\)

Where E is the e.m.f. of the source and r is the internal resistance of the source. Clearly for I_{max} to be large, r must be small.

(d) If tire internal resistance (r) of the HT supply is low, large amount of current will be drawn from the supply during the short circuiting which will cross the safety limits and damage the HT source. But if r is large, then current in the circuit will not exceed the safe limit. Hence a high voltage supply must have a very large internal resistance.

Question 19.

Choose the correct alternative:

(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.

(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.

(c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature^

(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (1022/1023).

Answer:

(a) greater

(b) lower

(c) nearly independent of

(d) 10^{22}

Question 20.

(a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?

(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) \(\frac {11}{3}\) Ω (ii) \(\frac {11}{5}\) Ω, (iii) 6 Ω, (iv) 6 Ω ?

(c) Determine the equivalent resistance of networks shown in Figure.

Answer:

(a) (i) The resistance will be maximum when the resistor are connected in series. Thus if R_{max} be the maximum resistance, then

R_{max} = R + R + R + …… + n times = nR.

(ii) The resistance will be minimum when the resistors are connected in parallel thus if R_{max} be the minimum resistance, then

(b) (i) When 3 Ω resistance is connected in series with the parallel combination of 1 and 2 Ω, we will get the required resistance.

Here R_{1} = 1Ω, R_{2}= 2Ω, R_{3} = 3Ω.

If R_{eff}be the net resistance of the combination, then

(ii) When 1 Ω resistance is connected in series with the parallel combination of 2 Ω and 3 Ω, we will get the required resistance.

if R_{eff} be the effictive resistance, then

(iii) If all the resistances are connected in series, then

R_{s} = R_{1} + R_{2} + R_{3} = 1 + 2 + 3 = 6 Ω

(iv) When all the resistors are connected in parallel, then

Answer:

(c) (a) The given network is a series combination of four equal units. Each unit has 4 resistances in which two resistances (1Ω. each being in series) are in parallel with 2 resistances (2 Ω each in series) If R be the net resistance of one unit, then

\(\frac{1}{R_{p}}\) = \(\frac{1}{2}+\frac{1}{4}=\frac{2+1}{4}=\frac{3}{4}\)

\(\frac{1}{R_{p}}\) = \(\frac{4}{3}\) Ω

If R be the total resistance of the network, then

R = 4 x R_{p} = 4 x \(\frac{4}{3}\) = \(\frac{16}{3}\) =5.33 Ω

(b) Let us connect a battery across a battery across the points A and B. We see that the same current flows through all the five resistances as all the 5 resistances are connected in series.

If R’ be the total resistance of the network between the point A and B, then

R’ = 5R.

Question 21.

Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by infinite network shown in figure. Each resistor has resistance of 1Ω.

Answer:

Let X be the equivalent resistance of the network. Since network is infinite adding one more set of three resistances each of value R = 1 Ω across the terminals will not affect the total resistance i.e., it should still remain equal to X. Thus this network can be represented as

Let R_{eq} be the equivalent resistance of this network, then

R_{eq} = R + equivalent resistance of parallel combination of X and R) + R

Addition of 3 resistances to resistance X of infinite network should not alter the total resistance of the infinite network. Thus

= R + \(\frac{XR}{X + R}\) + R

= 2R + \(\frac{XR}{X + R}\)

Addition of 3 resistances to resistance X of infinite network should not alter the total resistance of the infinite network. Thus

R_{eq} = X

But the value of resistance cannot be – ve

∴ X = 2.732 W

E = e.m.f. of supply = 12 V, r = 0.5 Ω

Let I = be the current drawn from the supply = ?

Question 22.

Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 k Ω is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf e and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is the value el

(b) What purpose does the high resistance of 600 k have?

(c) Is the balance point affects by this high resistance?

(d) Is the balance point affected by the internal resistance of the driver cell?

(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?

(f) Would the circuit work well for determing an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

Answer:

Here, E_{1} = e.m.f. of standard cell = 1.02 V. ”

E_{2} = e.m.f. cell connected = e = ?

l_{1} = balancing length for E_{1} = 67.3 cm

l_{2} = balancing length for E_{2} = 82.3 cm.

E = e.m.f. of driver cell = 2.0 V, r = its internal resistance = 0.40 Ω.

(a) As = \(\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{l_{1}}{l_{2}}\) or E_{2} = E_{1} x \(\frac{l_{2}}{l_{1}}\)

∴ ε = E_{1} x \(\frac{l_{2}}{l_{1}}\) = 1.02 x \(\frac{82.3}{67.3}\) = 1.247 V = 1.25 V

(b) Here R = 600 K Ω connected in series with cell. The purpose of using this high resistance is to allow very small current through the galvanometer when the movable contact is far from the balance point.

(c) No, the balance point is not affected by the presence of this high resistance.

(d) No, the balance point is not affected by the internal resistance of the driver cell.

(e) No, the method will not work as the balance point will not be obtained on the potentiometer wire if the e.m.f. of the driver cell is less than the e.m.f. of the standard cell.

(f) No, the circuit will not work for measuring extremely small e.m.f. of the order of milli vilt because the balance point will be just close to the end A and the percentage error in measurement will be very large. The circuit is modified by putting a suitable resistance R in series with the wire AB so that potential drop across AB is only slightly greater than the emf. to be measured. Then, the balance point will be at larger length of the wire and the percentage error will be much smaller.

Question 23.

Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε?

Answer:

Here, R = 10 Ω, X = ?

l_{1} = balancing length with standard resistance R = 58.3 cm

l_{2} = balancing length with standard resistance X = 68.5 cm

For a potential we know that

\(\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{l_{2}}{l_{1}}\)

Also we know that

In case we don’t get balance point on potentiometer wire, e.m.f. of driver cell is less than the e.m.f to be measured. Then we need driver cell of higher e.m.f. When we don’t get balance point, it also means that the potential drop across R or X is greater than the potential drop across wire AB. A suitable\series resistor can be put in the external circuit to reduce the current in the outside circuit and hence potential drop across Ror X.

Question 24.

Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Answer:

Here, e.m.f. of driver cell, E = 2.0 V

e.m.f. standard cell, E_{1} = 1.5 V

r = internal resistance of cell = ?

l_{1}= balancing length when cell is in open circuit = 76.3 cm

l_{2} = balancing length when cell in closed circuit = 64.8 cm.

R = Resistance used = 9.5 Ω

We know that

**Bihar Board Class 12 Physics **Chapter 3 Current Electricity Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.

Name the carriers of electric current in

(a) A bar made of Ag.

(b) A hydrogen discharge tube.

(c) A voltaic cell.

(d) Lead Accumulator being charged by external supply.

(e) A germanium semiconductor.

(f) A wire made of the alloy nichrome.

(g) A super conductor.

Answer:

(a) Electrons.

(b) Electrons and Positive ions of Hydrogen.

(c) + ve ions of H^{+} and – ve ions of SO_{4}.

(d) – do –

(e) Holes and electrons.

(f) Electrons

(g) Electrons.

Question 2.

Name the current carriers in solid conductors, liquids and gases.

Answer:

Current carriers in solid conductors, liquids and gases are free electrons, + ve/ – ve ions and + ve/ – ve ions respectively.

Question 3.

Define steady current.

Answer:

It is defined as the current whose magnitude does not change with time.

Question 4.

Define varying current.

Answer:

It is defined as the current whose magnitude changes with time.

Question 5.

Define current density. Tell whether it is a scalar or a vector quantity. Also give its S.I. unit.

Answer:

It is defined as the current flowing per unit area of the conductor. It is a vector quantity and its SI unit is Ampere-metre”2 (Am”2). .

Question 6.

How does the relaxation time of electrons in the good conductor varies with temperature?

Answer:

The relaxation time of electrons in the good conductors is inversely proportional to the temperature. As the temperature decreases, relaxation

time increases and vice-versa.

Question 7.

What is the direction of current density?

Answer:

It flows in the same direction as that of the flow of current.

Question 8.

Define resistance of a conductor. What is a resistor?

Answer:

It is defined as the opposition offered by the conductor to the flow of electrons. A resistor is defined as a conducting substance which offers resistance to the flow of current in the circuit.

Question 9.

Define the resistivity/specific resistance of the material of a conductor?

Answer:

It is defined to be numerically equal to the resistance of a conductor of unit length and unit area of cross-section.

Question 10.

How the resistivity of a material changes when its length and area of cross-section are trippled?

Answer:

We know that the resistivity of the material a conductor is independent of the dimensions of the conductor, hence it will remain same.

Question 11.

Define the conductivity of the material of a conductor? What is its S.I. unit?

Answer:

It is defined as the reciprocal of the resistivity. It is denoted by o. S.I. unit of a is mho m”1 or siemen m_1 (Snr1).

Question 12.

How does the conductivity of a semi conductor vary with temperature?

Answer:

The conductivity of a semiconductor increases with the increase in its temperature.

Question 13.

State the condition in which terminal voltage across a secondary cell is equal to its e.m,f.

Answer:

When the cell is in the open circuit i.e., when no current is drawn from the cell, then terminal voltage is equal to its e.m.f.

Question 14.

On what factors does (he internal resistance of a cell depend?

Answer:

- It depends upon the following factors:
- nature of the electrolyte of the cell.
- nature of the electrodes of the cell.
- distance between the electrodes of the cell.
- Area of the electrodes inside the electrolyte.

Question 15.

Why manganin is used for construction of standard resistance coil.

Answer:

It is used due to the following reasons:

(i) The resistance of manganin is very high. It is amount 40 times than that of copper wire of same dimensions, hence small length of it is sufficient for making standard resistance coil.

(ii) Manganin has very small temperature coefficient of resistance ^ r as compared to copper wire, so the variation of resistance due to heat is i negligible.

Question 16.

Is a wire carrying current charged? Why?

Answer:

No. At any instant, the number of protons is equal to the f number of electrons, hence wire is uncharged. Simply there is a continuous flow of electrons.

Question 17.

Two wire A and B of the same metal have the same area of cross-section and have their lengths in the ratio 3:2. What will be the raitio of currents flowing through them when the same potential difference is applied across length of each of them?

Answer:

We know that R where l is the length of the conductor. Now as the lengths of the two wires are in the ratio 3 :2, so the ratio of their resistances is also 3:2. Also we know that I ∝ \(\frac {1}{R}\), so the currents in the two wires will be in the ratio 2:3 when the same potential difference is applied across the two wires.

Question 18.

A piece of metal and another of silicon are cooled from room temperature to 100 K. What will happen to their conductivities? Explain.

Answer:

We know that the temperature coefficient of resistivity of metals is positive while that of semiconductor is negative. Th&s when their temperatures are lowered from room termperature to 100 K, the resistivity of metal decreases and that of silicon increases, hence the conductivity of metal increases and that of the silicon decreases.

Question 19.

Can the potential difference across a battery be greater than its e.m.f.?

Answer:

Normally not. But during the charging when the battery is connected to the external source, the terminal potential difference V is greater than E i.e.,

V_{applied} = E + 1R.

Question 20.

Is there any net field inside the cell when the circuit is closed and a steady current passes through?

Answer:

Yes, there is net field inside the cell opposite to the outside electrostatic field i.e., when circuit is closed, current flows in the direction of outside electrostatic field but it is opposite to the electrostatic field inside the cell as is shown in the figure here.

Question 21.

Can Kirchhoff s laws be applied both to D. C (direct current) and A.C. Circuits?

Answer:

Yes, Kirchhoff’s laws can be applied both to D.C. and A.C. circuits. .

Question 22.

Define a wheat stone bridge.

Answer:

It is defined as an arrangement of four resistances used to measure one of them in terms of other three. ”

Question 23.

Define a slide wire bridge.

Answer:

It is defined as a device used to measure the unknown resistance.

Question 24.

Define a potentiometer.

Answer:

It is defined as a device used to measure the internal difference across the terminals of a cell.

Question 25.

State the principle of potentiometer?

Answer:

It states that when a constant current is passed through a wire of uniform area of cross-section, the potential drop across any part of the wire is directly proportional to its length, i.e., V ∝ l.

Question 26.

Why is a potentiometer named so?

Answer:

It is named so because it is used to measure the potential difference across the terminals of a cell.

Question 27.

Define electric energy.

Answer:

It is defined as the heat energy produced in a conductor of resistance R on passing a current I through it for a given time t.

i.e„ U = I^{2}Rt = \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\)t = VIt.

Question 28.

Define KWh. Derive the relation between 1 KWh and 1J.

Answer:

Electric energy consumed or dissipated in an electrical circuit is said to be 1 KWh if a device of 1 KW power is used for one hour.

Relation between 1 KWh 1 J –

We know that

1J = IV x 1A x 1s

Also we know that

1 KWh = 1 KW x 1 hour

= 10^{3} watt x 1600 s

= 10^{3} x 1 V x 1 A x 3.6 x 10^{3} s

or 1 KWh = 3.6 x 10^{6} J.

Question 29.

Why does a potentiometer act like an ideal voltmeter?

Answer:

An ideal voltmeter has infinite resistance. Potentiometer draws no current from the circuit and hence its resistance is infinite as R = \(\frac {V}{I}\) = \(\frac {V}{O}\) = ∞ So it acts like an ideal voltmeter.

Question 30.

If a wire is stretched to four times its original length without the loss of mass, how will the resistivity of the wire be influenced?

Answer:

As the resistivity depends upon the nature of the conductor and not upon its dimensions, so it remain unchanged.

Question 31.

Arrange the following materials in order, starting with the best conductor and ending with the best insulator. mercury, nichrome, platinum, copper, silverm, tungsten, glass, iron, carbon, silicon, hard rubber.

Answer:

Silver, Copper, tungsten, iron, platinum, mercury nichrome, carbon, silicon, glass, hard rubber.

Question 32.

Define temperature coefficient of resistance of a given material.

Answer:

It is defined as the change in resistance per unit resistance per degree rise or fall in the temperature of a given material.

Question 33.

Comment on the direction of conventional and electronic currents.

Answer:

The direction of conventional current is in the direction of motion of positive charges while the direction of electronic current is opposite to the

conventional current and is in the direction of motion of electrons.

Question 34.

What is the effect of increase in temperature on the drift speed of free electrons in a metallic conductor?

Answer:

As the temperature of conductors is increased, average relation time decreases due to the increase in the collision frequency of electrons and thus their resistance increases. We know that V_{d} ∝ I∝ \(\frac {I}{R}\), thus drift velocity of electrons will decrease in a metallic conductor.

Question 35.

What is the effect of temperature on the conductivity of electrolytes?

Answer:

Th£ conductivity of electrolytes will increase because of the large number of ions and greater mobility of ions.

Question 36.

What happens to the drift velocity of the charge carriers when the potential difference across a given copper wire is increased?

Answer:

We know that the drift velocity of charge carriers is given by

Now as V_{d} is directly proportional to the P.D. applied, so an increase in V will lead to an increase in Vd.

Question 37.

Bends in a rubber pipe reduce the flow of water through it. How would the bends in wire affect its electric resistance?

Answer:

The bends in a wire will not affect its resistance. This is because the electrons are extremely small in size as compared to the bends. Thus electrons can easily change their direction of motion. Hence the resistance will remain unchanged.

Question 38.

Why effective resistance increases when the resistances are connected in series?

Answer:

When resistances are connected in series, then there is an increase in the effective length. Since resistance is directly proportional to the length, so the effective resistance is increased.

Question 39.

When the resistances are connected in parallel, the effective resistance decreases. Why?

Answer:

When the resistances are connected in parallel, then there is an effective increase in the area of cross-section. As the resistance is inversely proportional to the area of cross-section, hence the effective resistance decreases due to the increase in area.

Question 40.

Why is it not advisable to use copper wire in a potentio¬meter?

Answer;

It is not advisable to use copper wire in a potentiometer because the copper has small resistivity and large temperature coefficient of resistance.

Question 41.

Can the internal resistance of a cell may be called a defect?

Answer:

Yes, this is because a part of the electrical energy is used up in overcoming the internal resistance of the cell.

Question 42.

How are the coils wound in a resistance box?

Answer:

The resistance coil is doubly wound on a bobbin to avoid electromagnetic induction. The two ends of the resistance coil are connected to the two brass studs having a small gap between them. A plug is fitted into this gap. When this plug is taken out of the gap, the resistance is included in the circuit. .

Question 43.

Why is the terminal voltage across a battery more than the e.m.f. during recharging?

Answer:

During recharging of the battery, the voltage across the internal resistance (= Ir) and the e.m.f. are in the same direction, hence they get added up to give terminal voltage. So the terminal voltage is greater than thee.m.f. ‘

Question 44.

Why do we prefer a potentiometer of longer bridge wire?

Answer:

We prefer a potentiometer of longer bridge wire so as to increase its sensitivity. When the bridge wire is longer, the potential gradient is smaller. Smaller the potential gradient, more is the sensitivity of the potentiometer wire.

Question 45.

On what factors does the e.m.f. of a cell depend?

Answer:

The e.m.f. of a cell depends upon the following factors :

- Nature of the electrolyte.
- Temperature of the electrolyte.
- Concentration of the electrolyte.
- Nature of the electrodes.

Question 46.

Why the light of a motor car becomes slightly dim when it is started?

Answer:

The starter initially draw a larger current from the battery that causes a large voltage drop across the internal resistance of the battery. Hence the potential difference across the terminals of the battery is reduced leading the light to get dim.

Question 47

Does the e.m.f. represents force? Does e.m.f. has an electrostatic origin?

Answer:

No, the e.m.f. is not a force but it is the work done to move a unit charge from one terminal to the another terminal of the cell. The e.m.f. does not have the simple electrostatic origin.

Short Answer Type Questions

Question 1.

Prove that ρ = \(\frac{m}{n e^{2} \tau}\) where the symbols have usual meaning.

Answer:

We know that

I = neAυ_{d} ……(1)

Where A = area of cross-section of the conductor.

n = number of electrons per unit volume in the conductor. It is also Called carrier density of electrons in the given conductor.

υ_{d} = drift veLocity and is given by

υ_{d} = \(\frac{\mathrm{eE}}{\mathrm{m}}\) τ = \(\frac{\mathrm{e}}{\mathrm{m}}\).\(\frac{\mathrm{V}}{\mathrm{l}}\) τ ……….(2)

where = \(\frac{\mathrm{V}}{\mathrm{l}}\) = electric field across-the conductor.

m = mass of electrons.

τ = average relaxation time of electrons in the conductor.

∴ From (1) and (2), we get

I = neA. \(\frac{\mathrm{e}}{\mathrm{m}}\) (\(\frac{\mathrm{V}}{\mathrm{l}}\))τ

= \(\frac{\mathrm{ne}^{2}}{\mathrm{~m}} \tau\left(\frac{\mathrm{V}}{l}\right)\) A

or V = \(\frac{\mathrm{m}}{\mathrm{ne}^{2} \tau} \cdot \frac{l}{\mathrm{~A}} \cdot \mathrm{I}\) ……..(3)

Also according to Ohm’s we know that,

V = RI ……..(4)

∴ From (3) and (4), we get

R = \(\frac{m}{n e^{2} \tau} \cdot \frac{l}{A}\) ……….(5)

Also we know that

R = ρ . \(\frac {l}{A}\) …….(6)

From (5) and (6), we get

ρ = \(\frac{\mathrm{m}}{\mathrm{ne}^{2} \tau}\)

Hence proved

Question 2.

Explain colour code of resistance.

Answer:

The value of resistances used in electric and electronic circuits vary over a large range. Such resistances are marked on them according to colour code. These high resistances are generally carbon resistances which have four bands of different colours named as A, B, C, D. The first two bands A and B represent first two significant figures of the value of resistance (in Ω). The third band C represents the decimal multiplier and fourth band D gives the tolerance of the resistance known as the limit of accuracy or the permitted variation in the value of resistance. The colour code can be remembered with follow sentence:

Question 3.

Give the table that gives the colour code for carbon resistance.

Answer:

Question 4.

Derive the expression for equivalent resistance of series grouping.

Answer:

Two or more resistances are said to be connected in series if they are connected end to end and same current flows through each of them on applying some potential difference across the combination.

Let R_{1} R_{2} and R_{3} be the three resistances connected in series.

V = P.D. applied across the combination

I = current flowing through each of them.

If V_{1} V_{2} and V_{3} be the potential differences across R_{1} R_{2} and R_{3} respectively, then

V_{1} =IR_{1},V_{2} = IR_{2},V_{3} = IR_{3}

and V = V_{1} + V_{2} + V_{3}

Also let R_{s} be the equivalent resistance of the series grouping.

∴ V = IR_{s} ……(3)

From (1), (2) and (3), we get

IR_{s} = IR_{1} + IR_{2} + IR_{3}

or R_{s} = R_{1} + R_{2} + R_{3}

Question 5.

Derive the expression for equivalent resistance of the parallel grouping.

Answer:

Two or more resistances are said to be connected in parallel if they are connected between two common points and potential difference across each of them is same as the applied potential difference. Let R_{1} R_{2} and R_{3} be the three resistances connected in parallel.

V = Potential difference applied across the combination = Pot. diff. across each of them.

Let I = total current in the circuit.

If I_{1}, I_{2} and I_{3} be the currents flowing through R_{1}, R_{2} and R_{3} respectively, then

V = I_{1}R_{1} or I_{1} = \(\frac{\mathrm{V}}{\mathrm{R}_{1}}\)

Similarly

I_{2} = \(\frac{\mathrm{V}}{\mathrm{R}_{2}}\)

and I_{3} = \(\frac{\mathrm{V}}{\mathrm{R}_{2}}\) ……..(1)

If R_{p} be the equivalent resistance of the parallel grouping, then

V = IR_{p} or I = \(\frac{\mathrm{V}}{\mathrm{R}_{p}}\) ……..(2)

Also I = I_{1} + I_{2} + I_{3}

Question 6.

Define drift velocity and prove that \(\overrightarrow{\mathrm{V}_{\mathrm{d}}}\) = – \(\frac{\mathrm{e} \overrightarrow{\mathrm{E}}}{\mathrm{m}} \tau\) where the symbols have their usual meaning.

Answer:

Definition:

It is defined as the average velocity acquired by free electrons in a direction opposite to the direction of applied electric

field. It is denoted by \(\overrightarrow{\mathrm{V}_{\mathrm{d}}}\)

To prove \(\overrightarrow{\mathrm{V}_{\mathrm{d}}}\) = – \(\frac{\mathrm{e} \overrightarrow{\mathrm{E}}}{\mathrm{m}} \tau\) We know that a conductor has plenty of free electrons that move randomly in all possible directions with equal thermal velocities in the absence of an external electric field as shown in figure 1.

Thus their average thermal velocity is zero, i.e., \(\overrightarrow{v_{\mathrm{av}}}\) = 0 ……(i)

When an electric field is applied across the conductor, the free electrons get accelerated in a direction opposite to the direction of the applied field. The electrons gain extra velocity for a short time due to this acceleration as they collide with other free electrons or ions in the conductor. Let V = P.D. applied across the conductor PQ having area of cross-section A and length l.

If \(\overrightarrow{\mathrm{E}}\) electric field set from P to Q, then

\(\overrightarrow{\mathrm{E}}\) = \(\frac {V}{l}\) ………(ii)

Let \(\overrightarrow{\mathrm{F}}\) = Force experienced by the electron in the field \(\overrightarrow{\mathrm{E}}\) , then

\(\overrightarrow{\mathrm{E}}\) = -e \(\overrightarrow{\mathrm{E}}\) …(iii)

Where-ve sign shows that \(\overrightarrow{\mathrm{F}}\) and \(\overrightarrow{\mathrm{E}}\) are in opposite direction.

Let \(\overrightarrow{\mathrm{a}}\) be the acceleration of the electron between two successive collisions

\(\overrightarrow{\mathrm{a}}\) = \(\frac{\overrightarrow{\mathrm{F}}}{\mathrm{m}}=-\frac{\mathrm{e} \overrightarrow{\mathrm{E}}}{\mathrm{m}}\) ……..(iv)

where m = mass of an electron.

Let τ be the average relaxation time between two successive collisions.

The drift velocity is given by

which is the required expression for V_{d.}

Question 7.

Derive the relation between electric current and drift velocity.

Or

Prove that I = neAV_{d}.

Answer:

Let l, A be the length and area of cross-section of the conductor PQ. V = P.D. applied across it.

If \(\overrightarrow{\mathrm{E}}\) be the electric field produced from P to Q, then

E = \(\frac {V}{l}\) ………(i)

Also let n = number of free electrons per unit volume of the conductor.

∴ Volume of conductor = Al

Thus if N be total no. of free electrons in the conductor, then

N = n x Al ……(ii)

Let q be the charge flowing in the conductor from end Q to P, then q = Ne = nAle …..(iii)

Where e is the magnitude of the charge on an electron.

Let t be the time taken by the free electrons to flow from Q to P, then

t = \(\frac{l}{v_{\mathrm{d}}}\) ….(v)

where υ_{d} = drift speed of electrons.

If I be the electric current in the conductor, then by def.

Hence proved.

Question 8.

Derive the relation between current density (J) and υ_{d}.

Answer:

We know that J = \(\frac {1}{A}\) = current per unit area …….(1)

Also we know that I = neAυ_{d}

∴ From (1) and (2), we get

J = \(\frac{\text { neA } v_{\mathrm{d}}}{\mathrm{A}}\)

or J = neυ_{d}

or J ∝ υ_{d}

Question 9.

On what factors does the resistance of a conductor depend? Give the corresponding relation.

Answer:

Let R be the resistance of a conductor of length Z and area of cross-section A.

The R depends upon the following two factors :

(i) It is directly proportional to the length of the conductor: i.e.,

R ∝ Z …..(i)

(ii) It is inversely proportional to the area of cross-section of the conductor.

R ∝ \(\frac {1}{A}\) …..(ii)

Combining (i) and (ii) we get,

R ∝ \(\frac {1}{A}\)

R ∝ ρ . \(\frac {1}{A}\) …….(iii)

Where p is the proportionality constant known as resistivity or specific resistance of the conductor, eqn. (iii) is the required relation.

Question 10.

Prove that J = σE.

Answer:

We know that

I = neAυ_{d} …….(i)

Also we know that

υ_{d} = \(\frac{\mathrm{eE}}{\mathrm{m}} \tau\) ……..(ii)

∴ From (i) and (ii), we get

Now \(\frac {1}{A}\) = J = current density and

∴ From (iii) and (iv), we get

J = σE.

Question 11.

On what factors does the resistance and resistivity of a conductor depend’

Answer:

The resistance of a conductor depends upon the following factors :

(i) It depends upon the nature of the conductor.

(ii) It depends upon the dimensions (i.e., length and area of cross-section) of the conductor.

(iii) It depends upon the physical conditions (temperature, mechanical strain etc) of the conductor.

But R is independent of the values of V and I.

Resistivity of a conductor depends upon the following factors :

(i) It depends upon the nature of conductor i.e., ρ ∝ \(\frac {1}{n}\)

(ii) It depends upon the physical conditions of the conductor i.e.,

ρ ∝ \(\frac {1}{τ}\) But ρ is independent of the dimensions of the conductor.

Question 12.

Explain the temperature dependence of the resistivity of a conductor.

Answer:

We know that the resistivity of a conductor is given by

ρ = \(\frac{\mathrm{m}}{\mathrm{ne}^{2} \tau}\)

or ρ = \(\frac {1}{τ}\) ……..(i)

In other words, the resistivity of a conductor is inversely proportional to the average relaxation time of free electrons in the conductor. When the temperature of the conductor is increased, the amplitude of oscillations of the positive ions in the conductor also increases, thus collision frequency of electrons with the oscillating positive ions also increases, hence x decreases, thus ρ will increase according to eqn. (i) as shown in the graph given here. In other words, conductors become insulators at very high temperatures. .

Thus mathematically, The variation of ρ with temperature T (K) can be expressed as p = p0 (1 + ∝∆T).

Where ρ_{0} = resistivity at 0°C or 273 K.

Question 13.

Explain the temperature dependence of p of an alloy and a semiconductor insulator.

Answer:

The resistivity of an alloy like nichrome is very large, thus it is nearly independent of the variation of temperature. In other words, p depends weakly on temperature for an alloy as shown in figure.

The resistivity of a semiconductor and an insulator has an exponential dependence on the temperature expressed as

ρ = \(\rho_{0} e^{E_{g} / k_{B} T}\)

where E_{g} = band gap between conduction and valence band. K_{B} is Boltzmann’s constant.

Now as the temperature is increased, ρ decreases exponentially as shown in the figure here. At 0 K, both the semiconductor and the insulators have infinite value of resistivity.

Question 14.

Define thermistors. How it differs from an ordinary resistance. Give their some important applications.

Answer:

It is defined as a highly heat sensitive resistor generally made of a semiconducting material. .

Thermistor differs from an ordinary resistance in the following ways:

(i) The temperature coefficient of a thermistor is very high.

(ii) Their temperature coefficient of resistivity can be both positive and negative.

(iii) The resistance of a thermistor changes very rapidly with the change of temperature.

Applications:

(1) They are used for voltage stabilisation, temperature control and remote sensing.

(2) A thermistor with negative temperature coefficient of resistivity is used in resistance thermometers to measure very low temperature of the order of 10 K.

(3) They are used in the protective circuits of electrical equipments like transformers, motors and generators.

(4) They are used in temperature control units in an industry.

(5) They are used to safe guard the heater of the picture tube of a TV set against the variation in current.

Question 15.

What are the differences between e.m.f. and terminal potential difference?

Answer:

e.m.f.

(1) It is defined as the potential difference between the two terminals of a cell when it is in the open circuit.

(2) It is independent of the resistance of the external circuit.

(3) It is a cause of electric current.

(4) e.m.f. of a cell is greater than the potential difference between the two terminals of the cell.

(5) It is used for source.

Terminal Potential difference :

(1) It is defined as the potential difference between the two terminals of a cell when it is in the closed circuit.

(2) Potential difference between any two points of a circuit is proportional to the resistane between these two points.

(3) It is an effect of electric current.

(4) It is lesser than the e.m.f. of the cell.

(5) It can be measured between any two points of the circuit.

Question 16.

Derive the relation between the e.m.f. and terminal potential difference of a cell.

Answer:

Consider a cell of e.m.f. E having internal resistance r connected to an external resistance R through a key K. When K is open, no current is drawn from the cell. So the voltmeter connected across the cell gives the value of e.m.f. (E).

Let I be the current drawn from the cell when K is closed.

As R and r are in series, so

total resistance = R + r

Thus according to Ohm’s law

I = \(\frac{E}{R+r}\)

or E = I(R + r) = IR + Ir. …..(1)

Since the external resistance R is connected parallel to the electrodes of the cell, so the terminal potential difference of the cell is equal to the potential difference across R.

i.e., V = IR …..(2)

∴ From (1) and (2), we get

E = V + Ir

or V = E – Ir

which is the required relation between E and V.

Question 17.

Derive the expression for the internal resistance of a cell.

Answer:

We know that

I = \(\frac{E}{R+r}\) ………(i)

Also we know that

which is the required expression for internal resistance of a cell.

Question 18.

Derive the relation for variation of resistivity of insulators and semiconductors with temperature.

Answer:

We know that ρ = \(\frac{\mathrm{m}}{\mathrm{ne}^{2} \tau}\) ……..(i)

Also from relation of thermodynamic equilibrium.

We know that

n = \(n_{0} e^{-E_{g} / k_{B} T}\) …….(ii)

∴ From (i) and (ii), we get

where ρ = \(\frac{m}{n_{0} e^{2} \tau}\) = resistivity at ok

Question 19.

State Kirchhoff’s laws of electrical circuits. What are their sign conventions?

Answer:

The following are the two Kirchoff’s laws of electrical circuits.

Kirchhoff’s first law :

It states that the algebraic sum of all the currents meeting at a point in an electrical circuit is always zero. It is also called Kirchhoff’s current law (KCL) or junction rule.

Sign Conventions :

(i) The currents flowing towards the junction are taken as positive.

(ii) The currents flowing away from the junction are taken as negative.

Kirchhoff’s Second Law or Kirchhoff’s Voltage Law (KVL) :

It states that in any closed part of an electrical circuit, the algebraic sum of the e.m.f.s is always equal to the algebraic sum of the products of resistances and the currents flowing through them. It is also called loop rule.

i.e., Mathematically,

Sign Conventions :

(i) The product of I and R is taken as + ve if we move in the direction of flow of current and – ve if we move opposite to the direction of flow of current in the closed loop.

(ii) The e.m.f. is taken as + ve if we move through the electrolyte from – ve to + ve terminal of the cell and it will be taken as – ve if we move from + ve to – ve terminal of the cell.

Question 20.

Derive the condition of balanced wheatstone bridge circuit using Kirchhoff’s laws.

Answer:

The connections of wheatstone bridge are shown in the figure.

Let I be the main current in the circuit.

Let I_{1} and I_{2} be the currents flowing through P and R respectively.

I_{g}= Current flowing through the galvanometer on pressing Key K_{1}.

Thus currents flowing through Q and X are I_{1} – I_{g} and I_{2} + I_{g} respectively.

Applying KCl at junction A, we get

I = I_{1} + I_{2} …..(1)

Applying loop rule to closed circuits ABDA and BCDB, we get

I_{1}P + I_{g}G – I_{2}R = 0 …..(2)

and(I_{1} – I_{g})Q – (I_{2} + I_{g})X – I_{g}G = 0 …(3)

Now if the resistance R is adjusted in such away that point B and D are at the same potential, then the bridge is said to be balanced and the galvanometer shows no deflection in this condition.

ie., I_{g} = 0 …(4)

∴ From (2) and (3), we get

I_{1}P – I_{2}R = 0 or II_{1}p = I_{2}R …(5)

and I_{1}Q – I_{2}X = 0 or I_{1}Q = I_{2}X ….(6)

Question 21.

How is potentiometer used to compare e.m.f.s of two given cells?

Answer:

Circuit diagram is shown here.

Let I = Constant current flowing in the circuit.

First insert the plug between terminals 1 and 3 and slide the jockey to find the balancing length for cell of e.m.f. E_{1} Let it be l_{1}

x = resistance per unit length of the wire.

∴ E_{1} = (xl_{1}) I …….(1)

Now take out the plug from 1 and 3 and insert it between 2 and 3. The cell of e.m.f. E_{2} will be in the circuit. If l_{2} be the balancing length, then

E_{2} = (xl_{2})I …..(2)

Question 22.

How to measure the internal resistance of a cell using potentiometer?

Answer:

Let I be the constant current flowing through the circuit. Keep K2 out and Jockey is moved over the potentiometer wire so as to balance the e.m.f. of the cell having resistance r.

If l_{1} be the balancing length, then

E = (xl_{1},)I …..(i)

Now insert the plug in key K_{2} and introduce a resistance S, find the balancing length l_{2} (say), then the terminal potential difference is given by,

V = (xl_{2})I …..(ii)

Question 23.

How to find out unknown resistance using meter bridge?

Answer:

The circuit diagram is shown here.

from blanced wheat stone bridge, we know that

\(\frac {P}{Q}\) = \(\frac {P}{X}\) …….(1)

Let l = balancing length from end A for a resistance R from resistance box.

∴ P ∝ l,Q ∝ (100 – 1)

∴ \(\frac {P}{Q}\) = \(\frac{l}{100-l}\) …….(2)

or \(\frac {P}{Q}\) = \(\frac{l}{100-l}\)

or X = \(\frac{100-l}{l}\) R …….(3)

Thus knowing l and R, X can be calculated from equation (3).

Question 24.

Define and derive the expression for eIectri energy. Give its S.l. unit.

Answer:

It is defined as the amount of work done by a source to maintain a current in an electrical circuit.

Expression :

Let Rbe the resistance of an electrical circuit or device.

I = currentflowingthroughitfrompointAtoBinatimetifqbethe charge flowing from B to A.

Then

q = It

Let V = P.D. between the two points A and B.

If W be the work done to carry the charge from B to A, then

W =Vq = Vit.

This work done is equal to the electric energy consumed in the circuit (= U) ie.,

U = VIt

= (IR) It = I^{2}Rt (∴V = IR)

= V. \(\frac {V}{R}\) t = \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\) t

U = Vit = \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\)t = I^{2}Rt.

SI unit of electric energy is Joule (J).

1J = 1V x 1A x 1S.

Question 25.

Define electric power. Give its expressions and define 1 watt.

Answer:

It is defined as the rate at which work is done by the source of e.m.f. in maintaining the electric current in a circuit. It is denoted by P

P = \(\frac {U}{t}\) = VI = \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\) = I^{2}R

Si. unit of P is watt (W).

1W x 1V x 1A.

The electric power of a device is said to be I watt if one ampere of current flows through it when a constant potential difference of one volt

is applied across it.

Question 26.

Under what conditions will the same amount of current flow through the same number of cells connected in series and then in parallel?

Answer:

Let n number of cells each of e.m.f. E and internal resistance r be connected in series and then in parallel.

Also let I_{1} and I_{2} be the currents through the circuit in series and parallel groupings respectively.

∴ I_{1} = \(\frac {nE}{R + nr}\) ……(1)

and I_{2} = \(\frac{E}{R+\frac{n}{r}}\) …….(2)

where R is the external resistance.

i.e., the currents in two combinations will be same when external resistance is equal to the internal resistance of individual cell.

Question 27.

Clarify your elementary notions about current in a metallic conductor by answering the following questions:

(a) The electron drift speed is estimated to be only a few mm / second, for currents in the range of a few amperes. How then is current established almost the instant a circuit is closed?

(b) The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. But force should cause acceleration. Why then do the electrons acquire a steady average drift speed?

(c) If the electron drift speed is so small, and the electron’s charge is small, how can we still obtain large amounts of current in a conductor?

(d) When electrons drift in a metal from lower to higher potential, does it mean that all the ‘free’ electrons of the metal are moving in the same direction?

(e) Are the paths of electrons straight lines between successive collisions (with the positive ions of the metal) in the (i) absence of electric field, (ii) presence of electric field?

Answer:

(a) Electric field is established throughout the circuit almost instantly, with the speed of light. This causes a ‘local electron drift’ at every point in the conductor. The establishment of a current does not have to wait for electrons from one end of the conductor to travel to the other end. For reaching the steady state, it takes a little while.

(b) Each free electron does accelerate due to the force acting on it due to electric fields and increases the drift speed until it collides with a positive ion of the metal. It loses its drift speed after collision but starts to accelerate and increases it drift again only to suffer a collision again and so on. Thus the motion of electrons in the conductor under electric field is basically the stop and go motion. In the absence of continuous acceleration, the electrons appear to drift with a certain steady velocity.

(c) The number density of free electrons i.e., number of free electrons per unit volume is very large (-10^{29} m^{3}. So these large number of electrons cause-large current in the conductor.

(d) No. All the free electrons are not moving in the same direction. This is because the drift velocity is superposed over the random thermal velocities of electrons. Hence the electrons may still be moving at random.

(e) (i) Yes, the paths of free electrons in the absence of field are straight lines.

(ii) No, the paths of electrons in the presence of field are not straight lines but are generally curved due to the acceleration produced by the electric field.

Long Answer Type Questions

Question 1.

Derive the expression for the current in a circuit having series, parallel and mixed grouping of cells.

Answer:

(1) Cells in series:

Let E be the e.m.f. of each of the n-cells connected in series in a circuit having external resistance R.

r = internal resistance of each cell.

∴ total e.m.f. of the cells = E + E +…+ n times = nE.

Total internal resistance of the cells = r + r + ….+ n times = nr.

∴ Total resistance of the circuit = R + nr.

If I be the current flowing in the circuit,

Then

Following cases arise :

(a) When R> > nr, then

(b) When R < < nr, then

Thus in order to get a large amount of current from the cells ; connected in series, the external resistance should be very large as compared to the total internal resistance of the cells.

(2) Cells in parallel :

Let m cells each of e.m.f. E and internal resistance r be connected in parallel so as to send a current I through the external resistance R.

∴ total e.m.f. of parallel combination of cells = e.m.f. of a single cell = E.

If r_{eff} be the equivalent internal resistance of m cells connected in parallel.

∴ \(\frac{1}{r_{\text {eff }}}\) = \(\frac{1}{r}+\frac{1}{r}+\frac{1}{r}\) + …….. + m times

= \(\frac {m}{r}\)

or r_{eff} = \(\frac {r}{m}\)

∴ total resistance of the circuit = R + r_{eff}

Following cases arise:

(a) When R > > \(\frac {r}{m}\), then I = \(\frac {E}{R}\) = Current due to a single cell,

(b) When R < < \(\frac {r}{m}\), then m

I = \(\frac{E}{\left(\frac{r}{m}\right)}\) = m(\(\frac {E}{R}\))

= m times the current due to a single cell.

Thus in order to get a large current in a circuit, cells may be connected in parallel when the total internal resistance of the cells is very large as compared to the external resistance in the circuit.

(3) Mixed grouping of Cells :

Let rm be the number of cells connected to an external resistance R such that n cells are there in a row arranged in m rows.

∴ total e.m.f. of the cells = total e.m.f. of n cells in a row = nE.

Total internal resistance of the cells in each row = nr.

Since there are m rows each having internal resistance nr, so the equivalent resistance of the combination is given by

imh

∴ Total resistance of the circuit = R + \(\frac {nr}{m}\)

If I be the current flowing in the circuit, then

The current in the circuit will be maximum if mR + nr is minimum.

∴ For mR + nr be minimum,

\((\sqrt{m R}-\sqrt{n r})^{2}\) = 0

or \(\sqrt{\mathrm{mR}}-\sqrt{\mathrm{nr}}\) = 0

or mR = nr

or R = \(\frac {nr}{m}\)

Thus when external resistance is equal to the total internal resistance of the mixed grouping of cells, the current is maximum in the circuit.

Numerical Problems

Question 1.

A solution of sodium chloride discharges 6.1 x 1016 Na+ ions and 4.6 x 1016 Cl” ions in 2s. Find the current passing through the solution.

Answer:

Here, n^{+} = Na^{+} = 6.1 x 10^{16}

n^{–} = no. of Cl^{–} = 4.6 x 10^{16}

time = 2s.

e = 1.6 x 10^{-19} C.

If q_{1} and q_{2} be the total charge on Na^{+} and Cl^{–} ions discharged,

Then

q_{1} = n^{+} e = 6.1 x 10^{16} x 1.6 x 10^{-19} C.

and q_{2} = n^{–}e = 4 6 x 10^{16} x 16 x 10^{-19} C

If q be the total charge discharged, then

q = q_{1} + q_{2} .

= (6.1 x 10^{16}+ 4.6 x 10^{16}) x 1.6 X 10^{-19} C.

If I be the current passing through the solution, then

Question 2.

The amount of charge passing through the cross-section of a wire in time t is given by q = – at^{2} + bt^{3} + C.

(a) What are the dimensional formulae of constants a, b and c ?

(b) If the values of constants a, b, c are 3,5 and 2 in S.I. units, find the value of current at t = 3 s.

(c) Find the initial current.

(d) The time after which the value of current reaches a maximum value.

(e) Maximum or minimum value of current.

Answer:

Here, q = – at^{2} + bt^{3} + c …….(1)

(a) As the L.H.S. represents charge, so each term on R.H.S. must represent charge i.e.,

Putting a = 3, b = 5, c = 2, t = 3 s, in equation (2), we get

∴ I = -2 x 3 x 3 + 3 x 5 x 3^{2}

= 18 + 135 = 117 A.

(c) Let I = I_{0} be the initial current i.e., at t = 0.

∴ From (2), at t = 0,

I_{0} = 0

(d) For I to be maximum,

(e) Putting this value of t = 0.2 s in (2), we get

I = – 2 x 3 x (0.2) + 3 x 5 (0.2)

= -1.2+ 3 = 1.8 A

As the value of I is more than that at t = 0, so it must be maximum I_{max} = 1.8 A.

Question 3.

A current of 1A flows through a wire of length 0.24 m and area of cross-section 1.2 mm2, when it is connected to a battery of 3 V. Find the number density of free electrons in the wire, if the electron mobility is 4.8 x 10^{-6} m^{2}V^{-1} s^{-1}.

Answer:

Here, V = 3 V, l = 0.24 m, I = 1 A.

A = 1.2 mm^{2} = 1.2 x (10^{-3} m)^{2} = 1.2 x 10^{-6} m^{2}

μ = mobility of electron = 4.8 x 10^{-6} m^{2} V^{-1}S^{-1}

n = no. density of free electrons = ?

∴ E = \(\frac {V}{l}\) = \(\frac {2}{0.24}\) = 12.5 Vm^{-1}

If υ_{d} be the drift velocity of electrons, then

υ_{d} = μ E = 4.8 x 10^{-6} x 12.5 = 60 x 10^{-6} ms^{-1}.

Using the relation, I = neA V_{d}, we get

Question 4.

A wire of resistance 5 Ω is drawn out so that its length is increased by twice its original length. Calculate its new resistance.

Answer:

Here, R = initial resistance of wire = 5 Ω

Let l, A be its initial dimensions i.e. length and area of cross-section respectively.

If ρ be the resistivity of the wire, then

R = ρ \(\frac {1}{2}\) …….(1)

Now let R’ be its new resistance V, A’ be its new dimentsions, then

As the volume of the wire remains the same, so

Al = A’l’

Question 5.

A copper wire is stretched to make it 0.2% longer. What is the percentage change in resistance?

Answer:

Let l, A be the length and area of cross-section of the wire before stretching.

V, A’ = length and area of cross-section of the wire after strecthing.

R,R’ be its resistance before and after stretching.

If ρ be the resistivity of the wire, then

R = \(\frac {ρl}{A}\) ……….(1)

And R’ = ρ\(\frac {l}{A}\) ……..(2)

Now l’ = l + 0.2% l = l + \(\frac {0.2}{100}\) l

= 1.002 l.

Al = A’l’ (∴ Volume remains constant)

From A’ = \(\frac{\mathrm{A} l}{l^{\prime}}=\frac{\mathrm{A} l}{1.002 l}=\frac{\mathrm{A}}{1.002}\) …….(4)

Percentage increase in R is given by

Question 6.

Six equal resistances each of 4 Ω are connected to form a network as shown below. What” is the equivalent resistance between points A and B.

Answer:

Here, R = 4 Ω.

Let R_{AB} be the equivalent resistance of the network between A and B = ?

The 5 resistances of arms AO, AC, BO, BC and CO form a balanced wheat stone bridge, so no current will flow in the arm CO. Thus the given network reduces to the following :

Now the resistances in arms AC and CB are in series.

If Rj be the equivalent resistance, then

R_{1} = R + R = 4 + 4 = 8Ω.

Also the resistances of arms AO and OB are also in series.

If R_{2}, be the equivalent resistance, then

R_{2} = R + R = 2R = 2 x 4 = 8Ω.

Now R_{1}, R_{2} and R are in parallel between the points A and B.

Question 7.

Two cells of e.m.f. 1.5 V and 2.0 V having internal resistances 2 Ω and 1 respectively have their negative terminals joined by a wire! of 6 Ω and positive terminals by another wire of 4 Ω. A third wire of 8 Ω connects the mid-point of these two wires. Find the current through; and the potential difference at the ends of the third wire.

Answer:

Let E_{1} and E_{2} be the e.m.fs. of 1st and 2nd cell having internal resistances r_{1} and r_{2} respectively connected by resistances; R_{1} and R_{2} at – ve and + ve terminals respectively. j

∴ E_{1} = 1.5 V, r_{1} = 2 Ω

E_{2}= 2.0 V, r_{2} = 1 Ω

R_{1} = 6 Ω, R_{2} = 4 Ω

Let R_{3}= 8 Ω be the third resistance connecting R_{1} and R_{2} at their mid-points.

Let I_{1} and I_{2} be the currents produced by cell 1 and 2 respectively.

Applying KVL to closed loops ABFGA and CBFDC, we get

2I_{1} + 2I_{1} + 8(I_{1} + I_{2}) + 3 I_{1} = 1.5.

or 15I_{1} + 8 I_{2} = 1.5 ……..(1)

and 1 x I_{2} + 2I_{2} + (I_{1} + I_{2}) x 8 +3I_{2} = 2

or 8I_{1} + 14I_{2} = 2

or 4I_{1} + 7I_{2} = 1 …….(ii)

15 (ii) – (i) x 4 gives

60 I_{1} + 105I_{2} – 60I_{1} – 320I_{2} = 15 – 6

or 105 I_{2} – 32 I_{2} = 9

or 73 I_{2} = 9

∴ I_{2} = \(\frac {9}{73}\) A.

∴ From (ii) 4I_{1} + 7 x \(\frac {9}{73}\) = 1

or 4I_{1} = 1 – \(\frac {63}{73}\) = \(\frac {10}{73}\)

or I_{1} = \(\frac{10}{73} \times \frac{1}{4}=\frac{5}{146}\) A

∴ Current through 8 Ω wire = I_{1} + I_{2}

= \(\frac{5}{146}+\frac{9}{73}\)

= \(\frac{5+18}{146}=\frac{23}{146}\) A

If V be the potential difference across R_{3}, then

V = (I_{1} + I_{2})R_{3}

= \(\frac{23}{146} \times 8=\frac{184}{146}\) = 1.26 V.

Question 8.

A potential difference of 2 V is applied between the point A and B of the network given here. Calculate:

(a) equivalent resistance between points A and B.

(b) the magnitude of the currents flowing in the arms AFCEB and AFDEB.

Answer:

(a) The given network can be converted into Wheatstone bridge by turning points A and B on other side of the plane as shown here :

Here \(\frac {P}{Q}\) = \(\frac {R}{S}\)

or \(\frac {2}{2}\) = \(\frac {2}{2}\)

so the bridge is balanced hence no current flows through the resistance of 20 in the arm CD, so it can be neglected. Now the circuit reduces to the following equivalent circuit. Let RAB be the equivalent resistance of the network.

The resistances each of 20 in arm FC and CE are in series. Also resistances in arms FD and DE each of 2 Ω are in series, so R_{AB} is given by,

(b) The arms AFCEB and AFDEB are in parallel each having net resistance = 2 + 2 = 4 Ω.

Let I = total current through the circuit.

If I_{1} and I_{2} be the currents through arms AFCEB and AFDEB, then

I = I_{1} + I_{2}

Also 4I_{1} = 4I_{2} or I_{1} = I_{2}. (∴ P.D. across FCE = P.D. across FDE)

∴ I = 2I_{1} or I_{1} = \(\frac {I}{2}\)

Similarly Also

I_{1} = \(\frac {I}{2}\)

∴ Also R_{AB} = R_{AB} x I

or 2 = 2 x I

or 1 = 1A

∴ I_{1} = I_{2} = \(\frac {I}{2}\) = \(\frac {1}{2}\) = 0.5 A.

Question 9.

(a) Calculate the currents in each of the branches of the circuit shown in the network here.

Calculate the potential difference between points P and Q.

Answer:

(a) The various currents are shown in the figure here.

At Point P, I_{1} + I_{2} = I_{3} ………(i)

Applying KVL to iooj,s land ‘respective.ly we get,

3I_{2} – I_{1} – 4I_{1} = 30 – 5

or 3I_{2} – 5I_{1} = 25 ……(ii)

and 2I_{3} – 3I_{2} – 5I_{2} = 25 – 30

or 2I_{3} + 8 I_{2} = 5 ……(iii)

or 2 (I_{1} + I_{2}) + 8I_{2} = 5 (by using (ii))

or 2I_{1} + 10I_{2} = 5 ……(iv)

(ii) x 2 + 5 (iv) gives,

6I_{2} – 10 I_{1} + 10 I_{1} + 5O I_{2} = ,25 x 2 +5 x 5

or 56I_{2} = 50 + 25 = 75.

II_{2}, = \(\frac {75}{56}\) =1.34A.

∴ From (iv),

2I_{1}, + 10 x 1.34 = 5

or 2I_{1} = 5 – 13.4 = -8.4

∴I_{1} = -4.2 A.

∴ I_{3} = -4.2 + 1.34 = -2.86 A.

(b) Let V = P.D. between P and Q

V = I_{2}R = 1.34 x 3 = 4.02 = 4.0 V.

Question 10.

Find the potential difference between the points A and B in the figure given below.

Answer:

The two parts on either side of the cell are in parallel.

Here P.D. between C and D,V = 2V.

The equivalent circuit is given below :

The 3 resistances in path.CBD are in series, so their net resistance is 5 + 5 + 5 = 15 Q. Similarly the three resistances in path C’D’ are in series, so net resistance is 5 + 5 + 5 = 15 Q. Both these are now in parallel. If R be the equivalent resistance of the network, between C and D.

Then

Let I = total current in the circuit, then

As the resistances in CD and C’D’ paths are same, so the current will divide equally in these two arms or paths and is given by

I’ = \(\frac {I}{2}\) = \(\frac {1}{2}\) x \(\frac {4}{15}\) = \(\frac {2}{15}\) A.

Let V_{1}, V_{2} be the P.D. between C’ or C and A, C and B, then

If V’ be the potential difference between A and B, then

Question 11.

A potential difference of 220 V is maintained across a 12,000 Ω rheostat as shown in the figure here. The voltmeter V has a resistance of 6000, D and the point C is at one fourth of the distance from A to B. What is the reading of voltmeter?

Answer:

Let R_{AB} be the resistance of the rheostat

∴ R_{AB} = 12,000 Ω

If R_{AC}be the resistance of AC part, then

R_{AC} = \(\frac {1}{4}\) R_{AB} = \(\frac {1}{4}\) x 12,000 = 3,000 Ω

Resistance of the part BC is given by

R_{BC} = 12000 – 3,000 = 9000 Ω

R_{V} = resistance of voltmeter

= 6000 Ω (given)

Now R_{V} and R_{AC} are in parallel, so if R_{1} be their equivalent resistance, then

Now R_{BC} and R_{1} are in series, If R_{CD} be the equivalent resistance of the circuit between points C and D, then

R_{CD} = R_{BC} + R_{1} = 9000 + 2000 = 11,000 Ω .

Let I be the current in the circuit, then

I = \(\frac {1}{2}\) = \(\frac {1}{2}\) = 20 x 10^{-3} A.

If V’ be the voltmeter reading, then

V’ = Potential difference between A and C

= P.D. across R_{1}

= R_{1} x I = 2000 x 20 x 10^{-3} = 40 V.

Question 12.

Two cells each of same e.m.f. E but having internal resistances r_{1} and r_{2} are connected in series through an external resistance R. Find the value of R in terms of r_{1} and r_{2} for which first cell will have zero potential difference across it.

Answer:

Let E_{1} and E_{2} be the e.m.f .s of two cells. r_{1} r_{2} are their internal resistances respectively.

∴ E_{1} = E_{2} = E (given)

Let R = external resistance connecting the two cells in series

R = ?

Let V_{1} = P.D. across the first cell,

.’. V_{1} = 0 (given).

Let R’ be the equivalent resistance of the circuit.

Here r_{1} r_{2} and R are in series.

∴ R’ = r_{1} + r_{2} + R

Let E’ = total e.m.f. = E + E = 2E.

If I be the current flowing through the circuit, then

I = \(\frac{E^{\prime}}{R^{\prime}}=\frac{2 E}{r_{1}+r_{2}+R}\)

Now potential drop across of first cell is

I r_{1} = \(\frac{2 \mathrm{Er}_{1}}{\mathrm{r}_{1}+\mathrm{r}_{2}+\mathrm{R}}\)

We know that if the potential drop across the internal resistance of a cell is equal to the e.m.f. of the given cell, then the terminal potential difference of the cell will be zero.

Thus-for V_{1} to be zero for first cell.

or 1r_{1} = E

or \(\frac{2 \mathrm{Er}_{1}}{\mathrm{r}_{1}+\mathrm{r}_{2}+\mathrm{R}}\) = E

or 2r_{1} = rj_{1} + r_{2} + R

or R = 2r_{1} – r_{1} – r_{2}

or R = r_{1} – r_{2}.

Question 13.

Find the equivalent resistance between the points X and Y of the network of resistors given below if each resistor has a resistence r.

Answer:

Let E be the e.m.f. of a cell connected between X and Y.

Let I = current entering the network at A and leaving it at B.

The currents in various branches are shown in the figure.

As the current in the branch DE and EC is same, so it will remain same even when the contact at point E between the branches DEC and AEB is broken. Thus equivalent circuit becomes.

Let R_{1} and R_{2} be the net resistances of the branches ADCB and AB respectively.

Let R’ be the equivalent resistance of the Cricuit, Then

R’ = Parallel Grouping R_{1} and R_{2}

Aliter:

Applying KVL to the loops DECO and ADEA, we get

Let R’ be the equivalent resistance of the network,

∴ V_{A} – V_{B} = IR’ …….(4)

Where V_{A} – V_{B} is the potential difference between the points A and B.

Also V_{A} – V_{B} = I_{1}r + I_{1}r = 2I_{1}r.

= 2. \(\frac {4}{7}\) Ir = \(\frac {8}{7}\) Ir …….(5)

From (4) and (5), we get

IR’ = \(\frac {8}{7}\) Ir 7

or R’ = \(\frac {8}{7}\) r.

Question 14.

A part of the circuit in steady state along with the currents flowing in the branches, the values of resistances etc. is shown in the network here. Find the energy stored in the capacitor.

Answer:

We know that the capacitor offers infinite resistance to the flow of d.c., so d.c. cannot flow through arm ab, hence the circuit between a and b is open.

Applying KCl at junction ‘a’ and ‘b’ respectively, we get

I_{1}= 2 + 1 = 3A .

and I_{2} + 1 = 2 or I_{2} = 2 – 1 = 1 A.

Let V_{a} and V_{b} be the potential at points a and b respectively.

Applying KVL to the loop aBFba, we get

5I_{1} + 1 x I_{1} + 2I_{2} = V_{a} – V_{b}

or V_{a} – V_{b} = 6I_{1} + 2I_{2}

= 6 x 3 + 2 x 1 = 20V.

= P.D. across the capacitor

C = 4µF = 4 x 10^{-6} F

U = energy stored across the capacitor = ?

V = P.D. across C = V_{a} – V_{b} = 20 V

Using the relation,

U = \(\frac {1}{2}\) CV^{2},we get

U = \(\frac {1}{2}\) x 4 x 10^{-6} x (20)^{2}

= 2 x 10^{-6} x 400 = 8 x 10^{-4} J.

Question 15.

Theire CD of a slide wire bridge is 400 cm long in the figure here. Where should the free end of the galvanometer be connected to CD so that it shows no deflections.

Answer:

Here, P = 16 Ω, Q = 24 Ω, CD = 400 cm.

Let J be the point where free end of the galvanometer is connected such that it shows no deflection.

Let CJ = l cm.

∴ DJ = (400 – l) cm.

Now as the bridge is balanced, so

Question 16.

Find the equivalent resistance between the point A and B of the networks given here :

Answer:

(a) The given figure is equivalent to two balanced wheat stone bridges EFGO and HOJI connected at point O, thus no current will flow through arms FO and OI. So the equivalent network becomes.

It can further be redrawn as :

Now Let R_{1}, R_{2}, R_{3} be the net resistances in arms H I J, AOB and EFG respectively.

R_{1}, R_{2}, and R_{3} are in parallel, so if R_{AB} be the equivalent resistance between A and B, then

(b) As the current will divide equally along arm CHB and CHFED, so they may be assumed that at F, there is no contact. Thus equivalent network is given below:

Let R_{1} be the resistance of branch CD.

R_{1} = r + r = 2r …..(1)

Also let R_{2} be the resistance of branch CGFED

R_{2} = ‘resistance of parallel combination of part FGHE + resistance of CG and

If R_{AB} be the equivalent resistance of the network but A and B, then

Question 17.

Three cells are connected in parallel with their like poles connected together with wires of negligible resistance. If the e.m.f .s of the cells are 2,1 and 4v respectively and their internal resistances are 4,3 and 2 Q respectively, then find the current through each cell.

Answer:

Here, E_{1} = e.m.f. of 1^{st} cell = 2 V

E_{2} = e.m.f. of2^{nd} cell = 1 V

E_{3} = e.m.f. of 3^{rd} cell = 4 V

Let r_{1}, r_{2}, r_{3} be their internal resistances respectively.

∴ r_{1} = 4Ω, r_{2} = 3Ω, r_{3} = 2Ω .

Let I_{1}, I_{2} and I_{3} be the currents flowing through the cells of e.m.f. E_{1},, E_{2}, E_{3} respectively.

Applying KCL at point A, we get

I_{1} + I_{2} + I_{3} = 0 …….(1)

Applying KVL to the closed loops ACDBA and ABFEA respectively, we get

∴ From (4) and (6) we , get

∴ From (2) and (7) we , get

Question 18.

(a) The external diameter of a 5 m long hollow tube is 10 cm and thickness of its walls is 5 mm. If the specific resistance of the copper wire be 1.7 x 10^{-8} Ω m, then find its resistance.

(b) Six resistance each of 6 Ω are connected in the manner shown in the figure and a current of 0.5 A flows through the network. What is the potential diffemce between the points P and Q?

Answer:

(a) Let D_{1} and D_{2} be the external and internal diameters of the cylinder of length l.

Here D_{1} = 10 cm.

r_{1} = external radius of the cylinder

= \(\frac{\mathrm{D}_{1}}{2}\) = 5 cm = 5 x 10^{-2} m.

t = thickness of the wall of the cylinder = 5 mm = 0.5 cm

r_{2} = internal radius of cylinder

= r_{1} – t = 5 – 0.5 = 4.5 cm

= 4.5 x 10^{2} m

l = 5 m

ρ = specific resistance of the copper = 1.7 x 10^{-8} Dm.

Let R = resistance of the wire = ?

A = area of cross-section of the cylinder is given by

We know that

(b) Here, the three resistances each of 6 Ω in network (1) are connected in parallel. If R_{p} be their equivalent resistance, then

For network (2), two resistances each of 60 are in series and another of 60 is in parallel with it.

Thus if R’ be the equivalent resistance of this network, then

which is equivalent to

Now R_{p}, and R’ are in series.

If R be tire equivalent resistance between the points P and Q, then

R = R_{p} + R

= 2 + 4 = 6 Ω

I = current flowing between P and Q = 0.5 A.

If V be the potential difference between the points P and Q, then V = IR = 0.5 x 6 = 3 V.

Question 19.

An electric bulb is marked 100 W, 230 V. If the supply voltage drops to 115 V, what is the heat and light energy produced by the bulb in 20 minutes?

Answer:

Here, P = 100 W, V_{1}= 230 V energy produced = ?

Let R be the resistance of the bulb.

Using the relation, P = \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\), we get

R = \(\frac{\mathrm{V}^{2}}{\mathrm{P}}\) = \(\frac{230 \times 230}{100}\) = 529 Ω

V_{2}=115V, t = 20min = 20 x 60 = 1200s.

If U be the energy (heat and light) produced, then

U = I^{2}Rt = \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\) t

= \(\frac{(115)^{2} \times 1200}{529}\) = 30000 J

= 30 KJ.

Question 20.

An electric power station (50 MW) transmits power. to a distant load through long and thick cables. Power is transmitted at (a)

50 x 10 V and (b) 500 V. Calculate the power loss in each case. (c) In which case will it be lesser.

Answer:

Let R be the resistance of cables.

P = Power of station =50 MW =50 x 10^{6} V.

(a) Transmission at 50 x 10^{3} V

Power loss m cables, P1 = I_{2}^{2} R.

where I_{1} = line current

(b) Transmission at 500 V. Let I_{2} be the line current in this case, V = Transmission Voltage = 500 V

I_{2} = \(\frac{\mathrm{P}}{\mathrm{V}}=\frac{50 \times 10^{6}}{500}\) = 10^{5}A

If P_{2} be the power loss in the cables, then

P_{2} = I_{2}^{2}R

= (105)^{2} R

= 10^{10}R Watt.

(c) P_{1}< P_{2}, i.e., transmission at 50 x 10^{3} V wouLd result into less power Loss.

Fill In The Blanks

Question 1.

For ‘n’ identical resistances, the ratio of the total resistances of series combination to the total resistance of parallel combination is ……..

Answer:

\(\frac{\mathrm{R}_{\mathrm{s}}}{\mathrm{R}_{\mathrm{p}}}=\mathrm{n}^{2}\) (Hints : R_{P} = \(\frac {r}{n}\)

Question 2.

The drift velocity of electron is of the order of while the thermal average thermal velocity of the electron at room temperature is …….

Answer:

10^{-1} ms^{-1}, 10^{5} to 106 ms^{-1}.

Question 3.

The temperature coefficient of resistance lies between ……. to ……. °C^{-1}

Answer:

10^{-2}, 10^{-4}

Question 4.

Manganin has a resistance which is ……… ti ……….. as compared to pure copper and its temperature coefficient is °C^{-1}.

Answer:

30 to 40,10^{-5} .

Question 5.

The resistivity of a conductor depends upon the nature of the material of the conductor but is independent of the ………….

Answer:

Physical dimensions i.e., length and area of cross-section of the conductor.

Question 6.

The direction of current inside the cell is from ……. electrode but outside the cell, it is from ………. electrode.

Answer:

negative to positive electrode, positive to negative electrtode.

Question 7.

The terminal potential difference of a cell is zero when the potential drop across its internal resistance is while for the cell to be in open circuit the terminal P.D. is ……..

Answer:

equal to the e.m.f. of the cell, equal to tfie e.m.f. fo the cell. (Hint I = 0 for open circuit and V = E as. E = V + Ir).

Question 8.

In order to get a large amount of current from the cells connected in series, the external resistance should be ……..

Answer:

Very large as compared to the net internal resistance of the cells.

Question 9.

In order to get the maximum current in the circuit, the mixed grouping of cells must be done in such a way that the external resistance is ……

Answer:

equal to the net (effective) internal resistance of all the cells.

Question 10.

Nichrome is an alloy of ………. and …….

Answer:

Ni, Cu, Cr.

Question 11.

Commercial unit of electrical energy is and is ………. times the J.

Answer:

KWh or B.O.T. unit or simply unit, 3.6 x 10^{6}.

Question 12.

For the wheatstone bridge to be balanced, the necessary conditions for the four resistances is …….

Answer:

\(\frac {P}{Q}\) = \(\frac {R}{X}\)

Question 13.

For a potentiometer, the ratio of the e.m.f.s of two cells is ………

Answer:

equal to the ratio of the balancing lengths for the two cells i.e., Ei h mathematically, \(\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{l_{1}}{l_{2}}\)

Question 14.

For insulators, the resistance ………… with the increase in temperature.

Answer:

increases.

Question 15.

The relation ……….. is used when supplied electric power is lost or dissipated as heat.

Answer:

P = I2R.

Question 16.

The equivalent resistance of a network ………… due to parallel grouping of resistances and becomes ………

Answer:

decrease, lesser than the least resistance in the network.