Class 12 Physics Chapter 3 NCERT Solutions - Current Electricity: Questions and Answers PDF

Q.3.1 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?

Ans.3.1 EMF of the battery, E = 12 V

Internal resistance of the battery, r = 0.4 Ω

Let the maximum current drawn = I

According to OHM’s law, E = Ir

So I =  $\frac{E}{r}$ =  $\frac{12}{0.4}$ amp = 30 amp

Therefore, the maximum current can be drawn is 30 ampere.

 

Q.3.2 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?

Ans.3.2 EMF of the battery = 10 V

Internal resistance of the battery, r = 3 Ω

Current in the circuit, I = 0.5 A

Let the resistance of the resistor be R

According to Ohm’s law

I =  $\frac{E}{(R+r)}$

R + r =  $\frac{E}{I}$ or R =  $\frac{E}{I}$ - r =  $\frac{10}{0.5}$ - 3 = 17 Ω

Terminal voltage of the battery when the circuit is closed is given by

V = IR = 0.5  $\times 17=8.5V$

Therefore the resistance is 17 Ω and the terminal voltage is 8.5 V

 

Q.3.3 (a) Three resistors 1 Ω, 2 Ω, and 3 Ω 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.

Ans.3.3 The equivalent resistance of the resistor in series is given by

R = 1 + 2 + 3 = 6 Ω

From Ohm’s law, I =  $\frac{V}{R}$ we get I =  $\frac{12}{6}$ = 2 A.

Potential drop across 1 Ω resistor = I  $\times R$ = 2  $\times 1$ = 2 V

Potential drop across 2 Ω resistor = I  $\times R$ = 2  $\times 2$ = 4 V

Potential drop across 3 Ω resistor = I  $\times R$ = 2  $\times 3$ = 6 V

 

Q.3.4 (a) Three resistors 2 Ω, 4 Ω and 5 Ω 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.

Ans.3.4 Let  $R_1$ = 2 Ω,  $R_2$ = 4 Ω,  $R_3$ = 5 Ω

If the equivalent resistance is R, then  $\frac{1}{R}$ =  $\frac{1}{R_1}$ +  $\frac{1}{R_2}$ +  $\frac{1}{R_3}$ =  $\frac{1}{2}$ +  $\frac{1}{4}$ +  $\frac{1}{5}$ =  $\frac{10+5+4}{20}$ =  $\frac{19}{20}$

R =  $\frac{20}{19}$ = 1.05 Ω

The EMF of the battery = 20 V

Current through  $R_{1,}$  $I_1$ =  $\frac{V}{R_{1,}}$ =  $\frac{20}{2}$ = 10 A

Current through  $R_{2,}$  $I_2$ =  $\frac{V}{R_{2,}}$ =  $\frac{20}{4}$ = 5A

Current through  $R_{3,}$  $I_3$ =  $\frac{V}{R_{3,}}$ =  $\frac{20}{5}$ = 4 A

Total current I =  $I_1$ +  $I_2$ +  $I_3$ = 10 + 5 + 4 = 19 A

 

Q.3.5 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10^–4 °C^–1.

Ans.3.5 Let T be the room temperature and R be the resistance at room temperature and  $T_1$ be the required temperature and  $R_1$ be the resistance at that temperature and α be the coefficient of resistor.

We have:

T = 27  $℃,$ R = 100 Ω,  $T_1$ = ?,  $R_1$ = 117 Ω, α = 1.70  $\times {10}^{-4}$ °  ${\mathit{C}}^{-1}$

We know the relation of α can be given as

α =  $\frac{R_1-R}{R(T_1-T)}$ =  $\frac{117-100}{100(T_1-27)}$ = 1.70  $\times {10}^{-4}$

or (  $T_1$ - 27) =  $\frac{117-100}{100\times 1.70\mathrm{}\times {10}^{-4}}$ = 1000

$T_1$ = 1000 +27 = 1027  $℃$

 

Q.3.6 A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10^–7 m², and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

Ans.3.6 Given, length of the wire, l = 15 m

Uniform cross section, A = 6.0  $\times {10}^{-7}$  $m^2$

Resistance measured, R = 5.0 Ω

From the relation of

R =  $\rho \frac{l}{A}$ , where  $\rho$ is the resistivity of the material, we get

$\rho =\frac{AR}{l}$ =  $\frac{6.0\mathrm{}\times {10}^{-7}\times 5}{15}$ = 2  $\times {10}^{-7}$ Ωm.

 

Q.3.7 A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

Ans.3.7 Let us assume R = 2.1 Ω, T = 27.5 °C,  $R_1$ = 2.7 Ω,  $T_1$ = 100  $℃$ , α = resistivity of silver

We know the relation of α can be given as

α =  $\frac{R_1-R}{R(T_1-T)}$ =  $\frac{2.7-2.1}{2.1(100-27.5)}$ = 3.94  $\times {10}^{-3}$  ${℃}^{-1}$

 

Q.3.8 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 × 10^–4 °C^–1.

Ans.3.8 Given

Supply voltage, V = 230 V

Initial current drawn, I= 3.2 A

Final current drawn,  $I_1$ = 2.8 A

Room temperature, T = 27.0 °C

Steady temperature,  $T_1$ = ?

From Ohm’s law, we get initial resistance,  $R$ =  $\frac{V}{I}$ =  $\frac{230}{3.2}$ Ω = 71.875 Ω

Final resistance,  $R_1$ =  $\frac{V}{I_1}$ =  $\frac{230}{2.8}$ Ω = 82.143 Ω

From the relation of α =  $\frac{R_1-R}{R(T_1-T)}$ , where α is the temperature coefficient of resistance, we get

1.7  $\times {10}^{-4}$ =  $\frac{82.143-71.875}{71.875(T_1-27)}$

$T_1-27=$ 840.34

$T_1=867.34℃$

Therefore the steady temperature of heating element required is  $867.34℃$

 

Q.3.9 Determine the current in each branch of the network shown in Fig. 3.30:

 

Ans.3.9

 

Let us assume

$I_1$ = Current flowing through the outer circuit

$I_2$ = Current flowing through the branch AB

$I_3$ = Current flowing through the branch AD

$I_4$ = Current flowing through the branch BD

(  $I_2$ -  $I_4)$ = Current flowing through the branch BC

(  $I_3$ +  $I_4)$ = Current flowing through the branch DC

For the closed circuit ABDA, potential is zero, i.e.

10  $I_2$ + 5  $I_4$ - 5  $I_3$ = 0

$I_3=I_4+2I_2$ …………(1)

For the close circuit BCDB, potential is zero, i.e.

5(  $I_2$ -  $I_4)$ - 10(  $I_3$ +  $I_4)$ - 5  $I_4$ = 0

5  $I_2$ - 5  $I_4$ - 10  $I_3$ - 10  $I_4$ - 5  $I_4$ =0

5  $I_2$ - 10  $I_3$ - 20  $I_4$ = 0

$I_2-2I_3$ - 4  $I_4$ = 0

$I_2=2I_3$ +4  $I_4$ …………….(2)

For the close circuit ABCFEA, potential is zero, i.e.

-10 + 10  $I_1+$ 10  $I_2$ + 5(  $I_2$ -  $I_4)$ = 0

10  $I_1$ + 15  $I_2$ - 5  $I_4$ = 10

2  $I_1$ + 3  $I_2$ -  $I_4$ = 2 …………(3)

Solving equations (1) and (2) we get

$I_2=-2I_4$

$I_3=-3I_4$

Substituting these values in equation (3), we get

$I_4=-\frac{2}{17}$ A

$I_2=\frac{4}{17}$ A

$I_3=\frac{6}{17}$ A

$I_1$ =  $I_2+I_3$ =  $\frac{10}{17}$ A

 

Q.3.10 (a) In a meter bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor S is of 12.5 Ω. Determine the resistance of R. 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 R and S 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?

 

Ans.3.10 Balance point from end A,  $l_1$ = 39.5 cm

Resistance of the resistor, S = 12.5 Ω

Condition for the balance is given as,

$\frac{R}{S}$ =  $\frac{l_1}{{100-l}_1}$

$R=S\times \frac{l_1}{100-l_1}$ = 12.5  $\times \frac{39.5}{100-39.5}$ = 8.16 Ω

$\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{b}\mathrm{e}\mathrm{t}\mathrm{w}\mathrm{e}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{s}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{a}\mathrm{}\mathrm{W}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{t}\mathrm{s}\mathrm{t}\mathrm{o}\mathrm{n}\mathrm{e}$ bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula.

If R & S are interchanged, then  $l_1$ and (100 -  $l_1$ ) will also get interchanged. The balance point will be (100-  $l_1)$ from A. So the new balance point is 100 – 39.5 = 60.5 cm from A.

When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection, no current will flow through galvanometer.

 

Q.3.11 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?

Ans.3.11 EMF of the battery, E = 8 V

Internal resistance of the battery, r = 0.5 Ω

DC supply voltage, V = 120 V

Resistance of the resistor, R = 15.5 Ω

Let the effective voltage in the circuit be  $V_1$

Then  $V_1$ = V – E = 120 – 8 = 112 V

If the current flown in the circuit be I then from the relation

I =  $\frac{V_1}{R+r}$ we get I =  $\frac{112}{15.5+0.5}$ = 7 A

Voltage across the resistor R = IR = 7  $\times 15.5=$ 108.5 V

Hence, Terminal voltage of the battery

= DC supply voltage – Voltage drop across the resistor

= 120 – 108.5 = 11.5 V

The purpose of having a series resistor in a charging circuit is to limit the current drawn from the external source. Otherwise, the current will be extremely high.

 

Q.3.12 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?

Ans.3.12 EMF of the cell,  $E_1$ = 1.25 V

Let the EMF of the replaced cell be  $E_2$

Existing balance point,  $l_1$ = 35 cm

New balance point,  $l_2$ = 63 cm

From the relation of balance condition, we get

$\frac{E_1}{E_2}$ =  $\frac{l_1}{l_2}$ , we get  $E_2$ =  $\frac{E_1\times l_2}{l_1}$ =  $\frac{1.25\times 63}{35}$ = 2.25 V

Therefore the emf of the another cell is 2.25V

 

Q.3. 13 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10²⁸ 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 × 10^–6 m² and it is carrying a current of 3.0 A.

Ans.3.13 Given, number of free electron in a copper conductor, n = 8.5  $\times {10}^{28}$  $m^{-3}$

Length of the copper wire, l = 3.0 m

The area of cross section, A = 2  $\times {10}^{-6}$  $m^2$

Current carried by the wire, I = 3.0 A

From the relation I = nAe  $V_d$ where

e = Electric charge = 1.6  $\times {10}^{-19}$ C

$V_d=\mathrm{D}\mathrm{r}\mathrm{i}\mathrm{f}\mathrm{t}\mathrm{}\mathrm{v}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{y}$

We get  $V_d$ =  $\frac{I}{nAe}$

Again, Drift velocity (  $V_d)$ =  $\frac{Lengthofthewire\left(l\right)}{Timetakentocoverl\left(t\right)}$

t =  $\frac{l}{V_d}$ =  $\frac{lnAe}{I}$ =  $\frac{3\times 8.5\mathrm{}\times {10}^{28}\times 2\mathrm{}\times {10}^{-6}\times 1.6\mathrm{}\times {10}^{-19}}{3}$ = 27.2  $\times {10}^3$ seconds

 

Q.3.14 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 neutralize 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 × 10⁶ m.)

Ans.3.14 Surface charge density of the earth, $\sigma$ =  ${10}^{-9}$ C  $m^{-2}$

Current over entire Globe, I = 1800 A

Radius of the earth, r = 6.37  $\times {10}^6$ m

Hence, surface area of the earth, A = 4  $\pi r^2$ = 4  $\times \pi \times {(6.37\mathrm{}\times {10}^6)}^2$ = 5.1  $\times {10}^{14}$  $m^2$

Charge on the earth surface, q =  $\sigma$  $\times A$ = 0.51  $\times {10}^6$ C

If t is time taken to neutralize the earth surface, then q = I  $\times t$

t =  $\frac{q}{I}$ =  $\frac{0.51\mathrm{}\times {10}^6}{1800}$ = 283.28 seconds

 

Q.3.15 (a) Six lead-acid type of secondary cells each of emf 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?

Ans.3.15 Number of secondary cells, n = 6

emf of each secondary cell = 2 V

Internal resistance of each secondary cell, r = 0.015 Ω

Resistance of the resistor, R = 8.5 Ω

Current drawn from the supply, I is given as I =  $\frac{Totalvoltage}{R+totalinternalresistance}$

=  $\frac{6\times 2}{8.5+6\times 0.015}$ = 1.396 A

Terminal voltage = I  $\times R$ = 1.396  $\times 8.5$ = 11.87 V

After long use emf = 1.9 V

Internal resistance of the cell, r = 380 Ω

Maximum current can be drawn =  $\frac{1.9}{380}$ = 5  $\times {10}^{-3}$ A

No, the cell cannot drive the starter motor of the car as it requires high current.

 

Q.3.16 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. (  ${\mathit{\rho }}_{\mathit{A}\mathit{l}}$ = 2.63 × 10^–8  Ωm,  ${\mathit{\rho }}_{\mathit{C}\mathit{u}}$ = 1.72 × 10^–8 Ω m, Relative density of Al = 2.7, of Cu = 8.9.)

Ans.3.16 Given, resistivity of aluminium,  ${\rho }_1$ = 2.63  $\times {10}^{-8}$ Ωm

Resistivity of copper,  ${\rho }_2$ = 1.72  $\times {10}^{-8}$ Ωm

Relative density of aluminium = 2.7

Relative density of copper = 8.9

For Aluminium wire: Let

$l_1$ = length,  $m_1$ = mass,  $R_1$ = resistance,  $A_1$ = cross-section area,  ${\rho }_1$ = density

For Copper wire: Let

$l_2$ = length,  $m_2$ = mass,  $R_2$ = resistance,  $A_2$ = cross-section area,  ${\rho }_2$ = density

From the relation R =  $\frac{\rho l}{A}$ , we get

$R_1=\frac{{\rho }_1{l}_1}{A_1}$ ………(1), and

$R_2=\frac{{\rho }_2{l}_2}{A_2}$ ………(2)

It is given  $R_1$ =  $R_2$ and  $l_1$ =  $l_2$ . Hence

$\frac{A_1}{A_2}$ =  $\frac{{\rho }_1}{{\rho }_2}$ =  $\frac{2.63}{1.72}$ (  ${10}^{-8}$ cancel each other)

From the relation mass = volume  $\times \mathrm{r}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{y}$ = length  $\times$ cross sectional area  $\times$ relative density, we get

$m_1=l_1{A}_1{d}_1$ and  $m_2=l_2{A}_2{d}_2$ , where  $d_1$ = relative density of aluminium and  $d_2$ = relative density of copper

$\frac{m_1}{m_2}=\frac{l_1{A}_1{d}_1}{l_2{A}_2{d}_2}$ =  $\frac{A_1{d}_1}{A_2{d}_2}$ =  $\frac{2.63}{1.72}\times \frac{2.7}{8.9}$ = 0.464

It can be inferred from this ratio that  $m_1$ is less than  $m_2$ . Hence, aluminium is preferred as overhead power cables over copper.

 

Q.3.17 What conclusion can you draw from the following observations on a resistor made of alloy manganin?

Current (A)	Voltage (V)	Current (A)	Voltage (V)
0.2	3.94	3.0	59.2
0.4	7.87	4.0	78.8
0.6	11.8	5.0	98.6
0.8	15.7	6.0	118.5
1.0	19.7	7.0	138.2
2.0	39.4	8.0	158.0

Ans.3.17 From Ohm’s law, R =  $\frac{V}{A}$

From the given table we get  $\frac{3.94}{0.2}$ =  $\frac{7.87}{0.4}$ =  $\frac{11.8}{0.6}$ ……………  $\frac{158}{8}$  $\approx 19.7\Omega$ = constant

Hence manganin is an ohmic conductor.

 

Q.3.18 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?

Ans.3.18 When a steady current flows in a metallic conductor of non-uniform cross section, only the current flowing is constant. Current density, Electric field and Drift speed are inversely proportional to the cross section area, hence not constant.

Ohm’s law is not applicable to all conductors, vacuum diode semi-conductor is a non-ohmic conductor.

According to ohm’s law V = IR, voltage is directly proportional to current, hence to draw high current from a low voltage source, internal resistance ®, needs to be low.

To prevent the drawing of extra current, which can cause short circuit, the internal resistance for a high voltage system needs to be high.

 

Q.3.19 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).

Ans.3.19 Alloys of metal usually have greater resistivity than that of their constituent metals.

Alloys usually have much lower temperature coefficients of resistance than pure metals.

The resistivity of the alloy manganin is nearly independent of increase of temperature.

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

 

Q.3.20 (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) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?

(c) Determine the equivalent resistance of networks shown in Fig. 3.31.

 

Ans.3.20 Total number of resistors = n

Resistance of each resistor = R

When the resistors are connected in series, effective resistance  $R_{eff}$ is maximum.  $R_{eff}$ = nR

When n resistors are connected in parallel, the effective resistance,  $R_{eff}$ is minimum.  $R_{eff}$ =  $\frac{R}{n}$ . The ratio of maximum to minimum resistance =  $\frac{nR}{\frac{R}{n}}$ =  $n^2$

Let us assume,  $R_1$ = 1 Ω,  $R_2$ = 2 Ω,  $R_3$ = 3 Ω

Required equivalent resistance, R = $\frac{11}{3}$Ω

 

From the circuit the equivalent resistance is given by

R =  $\frac{2\times 1}{2+1}$ + 3 =  $\frac{2}{3}$ + 3 =  $\frac{11}{3}$ Ω

Required equivalent resistance, R =  $\frac{11}{5}$ Ω

From the circuit, the equivalent resistance is given by

R =  $\frac{2\times 3}{2+3}$ + 1 =  $\frac{6}{5}$ + 1 =  $\frac{11}{5}$ Ω

Required equivalent resistance, R = 6 Ω

From the circuit, the equivalent resistance is given by

R = 1 + 2+ 3 = 6 Ω

 

Required equivalent resistance, R =  $\frac{6}{11}$ Ω

 

From the circuit, the equivalent resistance is given by

R =  $\frac{1\times 2\times 3}{1\times 2+2\times 3+3\times 1}$ =  $\frac{6}{11}$ Ω

(a)

 

It can be observed from the given circuit that in the first small loop, two resistors of resistance 1 Ω each are connected in the series. Hence the equivalent resistance is = 1 + 1= 2  $\Omega$ . At the bottom part 2 resistors of 2 Ω each are connected in a series and thus make equivalent resistance of 2 + 2 = 4 Ω. Thus the circuit can be redrawn as 2 Ω and 4 Ω resistances are connected in parallel. Hence the equivalent resistance of each loop is R =  $\frac{2\times 4}{2+4}$ =  $\frac{4}{3}$ Ω. All these 4 loop resistors are connected in series. Thus the total equivalent resistance is  $\frac{4}{3}$ Ω  $\times 4=\frac{16}{3}$ Ω

Here all 5 resistors are connected in series. So the equivalent resistance is

5  $\times R=5R$

 

Q.3.21 Determine the current drawn from a 12V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.

 

Ans.3.21 Let the equivalent resistance of the given circuit be R’. The equivalent resistance of an infinite network is given by

R’ = 2 +  $\frac{R\text{'}}{(R^{\text{'}}+1)}$ or R’ =  $\frac{2R^{\text{'}}+2+R\text{'}}{(R^{\text{'}}+1)}$ or  ${R\text{'}}^2$ + R’ = 2R’ + 2 + R’

${R\text{'}}^2-2R^{\text{'}}-2=0$

R’ =  $\frac{2\pm \sqrt{4+8}}{2}$ = 1  $\pm \surd 3$

Since R’ cannot be negative, hence R’ = 1+  $\surd 3$ = 2.73 Ω

Internal resistance, r = 0.5Ω

Total resistance = 2.73 + 0.5 = 3.23 Ω

Current drawn from the source =  $\frac{12}{3.23}$ A= 3.72 A

Q. 3.22 Figure 3.33 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 up to 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 kΩ 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 $\mathit{ϵ}$ and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

 

(a) What is the value e ?

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

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

(d) 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?

(e) Would the circuit work well for determining 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?

Ans.3.22 Constant emf of the standard cell,  $E_1$ = 1.02 V

Balance point on the wire,  $l_1$ = 67.3 cm

A cell of unknown emf,  $ϵ$ , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm.

The relation of connected emf and balance point is,

$\frac{E_1}{l_1}$ =  $\frac{ϵ}{l}$

Hence,  $ϵ$ =  $\frac{l}{l_1}\times E_1$ =  $\frac{82.3}{67.3}\times 1.02$ = 1.247 V

The purpose of using high resistance of 600 KΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.

The balance point is not affected by the presence of high resistance.

The point is not affected by the internal resistance of the driver cell.

The method would not work if the emf of the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.

The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.

The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.

 

Q.3.23 Figure 3.34 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.

 

Ans.3.23 Internal resistance of the cell = r

Balance point of the cell in open circuit,  $l_1$ = 76.3 cm

An external resistance ® is connected to the circuit with R = 9.5 Ω

New balance point of the circuit,  $l_2$ = 64.8 cm

Current flowing through circuit = I

The relation connecting resistance and emf is,

r = (  $\frac{l_1-l_2}{l_2}$ )R =  $\frac{76.3-64.8}{64.8}$  $\times 9.5=$ 1.69 Ω

Therefore, the internal resistance is 1.69 Ω.