Current Electricity-Solutions

 CBSE Class 12 Physics

NCERT Solutions
Chapter - 3
Current Electricity

---

1: The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is Ω$\mathrm{\Omega }$ 0.4, what is the maximum current that can be drawn from the battery?
Ans: Emf of the battery, E = 12 V
Internal resistance of the battery, r = 0.4 Ω
Maximum current that can be drawn from the battery, I= E/r= 12/0.4=30 A
So, the maximum current drawn from the given battery is 30 A.

---

2: A battery of emf 10 V and internal resistance 3Ω$\mathrm{\Omega }$ 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: Emf of the battery, E=10 V
Internal resistance of the battery ,r =3Ω$\mathrm{\Omega }$
Current in the circuit, I=0.5 A
Resistance of the resistor = R
As by Ohm’s law
I=ER+r$I=\frac{E}{R+r}$
R+r=EI$R+r=\frac{E}{I}$
=100.5=20Ω$=\frac{10}{0.5}=20\mathrm{\Omega }$
∴R=20−3=17Ω$\therefore R=20-3=17\mathrm{\Omega }$
Terminal voltage of the resistor=V
Using Ohm’s law,
V = IR
=0.5×17$=0.5\times 17$
= 8.5 V
Therefore, the resistance of the resistor is 17Ω$17\mathrm{\Omega }$ and the terminal voltage is 8.5 V.

---

3:

1. Three resistors 1Ω,2Ω$1\mathrm{\Omega },2\mathrm{\Omega }$ and 3Ω$3\mathrm{\Omega }$ are combined in series. What is the total resistance of the combination?
2. If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Ans:

1. Three resistors of resistances 1Ω,2Ω$1\mathrm{\Omega },2\mathrm{\Omega }$and 3Ω$3\mathrm{\Omega }$ are combined in series. In series combination of resistances ,total resistance of the combination is given by the formula as
   R = R1 + R2 + R3
   So total resistance=1 + 2 + 3 = 6Ω$6\mathrm{\Omega }$
2. Current flowing through the circuit = I
   Emf of the battery, E=12 V
   Total resistance of the circuit by Ohm’s law is given as,
   I=ER=126=2A$I=\frac{E}{R}=\frac{12}{6}=2\mathrm{A}$
   Let potential drop across 1Ω$1\mathrm{\Omega }$ resistor = V1
   by Ohm's law
   V1=2×1=2V$V_1=2\times 1=2V$ ........ (i)
   Let potential drop across 2Ω$2\mathrm{\Omega }$ resistor = V2
   by ohm's law
   V2=2×2=4V$V_2=2\times 2=4V$ ...... (ii)
   Let potential drop across 3Ω$3\mathrm{\Omega }$ resistor = V3
   By ohm's law
   V3=2×3=6V$V_3=2\times 3=6V$ ...... (iii)
3. Therefore, the potential drop across the 1Ω,2Ωand3Ω$1\mathrm{\Omega },2\mathrm{\Omega }\mathrm{a}\mathrm{n}\mathrm{d}3\mathrm{\Omega }$ resistors are 2V, 4V, and 6V respectively.

---

4.

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

1. There are three resistors of resistances,
   R1 = 2, R2 = 4, and R3 = 5
   They are connected in parallel. Hence, total resistance(R) of the combination of parallel resistances is given by,
   1R=1R1+1R2+1R3$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$
   =12+14+15=10+5+420=1920$=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{10+5+4}{20}=\frac{19}{20}$
   ∴R=2019Ω$\therefore R=\frac{20}{19}\mathrm{\Omega }$
   Therefore, total resistance of the combination is 2019Ω$\frac{20}{19}\mathrm{\Omega }$ .
2. Emf of the battery, V = 20V
   Let currents I1,I2andI3$I_1,I_{2}and{I}_3$ flow through the resistors R1, R2 and R3 and is given
   I1=VR1=202=10$I_1=\frac{V}{R_1}=\frac{20}{2}=10$
   I2=VR2=204=5A$I_2=\frac{V}{R_2}=\frac{20}{4}=5A$
   I3=VR3=205=4A$I_3=\frac{V}{R_3}=\frac{20}{5}=4\mathrm{A}$
   Total current, I = I1 + I2 + I3 = 10 + 5+ 4 = 19 A
   Therefore, the current through each resistor is 10A, 5A, and 4A, respectively and the total current is 19A.

---

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 1.70×10−40C−1$1.70\times {10}^{-4}{}^0{\text{C}}^{-1}$?
Ans: Room temperature, T=27oC
Resistance of the heating element at temperature T1 is R1=117Ω$R_1=117\mathrm{\Omega }$
Temperature co-efficient of the material of the filament,
α=1.70×10−40C−1$\alpha =1.70\times {10}^{-4}{}^0{\text{C}}^{-1}$
α$\alpha$ is given by the relation,
α=R1−RR(T1−T)$\alpha =\frac{R_1-R}{R\left(T_1-T\right)}$
T1−T=R1−RRα$T_1-T=\frac{R_1-R}{R\alpha }$
T1−27=117−100100(1.7×10−4)$T_1-27=\frac{117-100}{100\left(1.7\times {10}^{-4}\right)}$
T1 - 27 = 1000
Therefore, at 1027∘C${1027}^{\circ }C$ the resistance of the element is 117Ω$117\mathrm{\Omega }$.

---

6: A negligibly small current is passed through a wire of length 15 m and uniform cross section =6.0×10−7m2$=6.0\times {10}^{-7}{\mathrm{m}}^2$ and its resistance is measured to be 5.0Ω$\mathrm{\Omega }$. What is the resistivity of the material at the temperature of the experiment?
Ans: Resistivity of material is given by ρ=Ral$\rho =\frac{Ra}{l}$
Length of the wire, l =15 m
Area of cross-section of the wire,a=6.0×10−7m2$a=6.0\times {10}^{-7}{\mathrm{m}}^2$
Resistance of the material of the wire, R=5.0Ω$R=5.0\mathrm{\Omega }$
Resistivity of the material of the wire= ρ$\rho$
Resistivity of material of wire is given as:
ρ=Ral$\rho =\frac{Ra}{l}$ =5×6×10−715$=\frac{5\times 6\times {10}^{-7}}{15}$
=2×10−7Ωm$=2\times {10}^{-7}\mathrm{\Omega }m$
Therefore, the resistivity of the material is 2×10−7Ωm$2\times {10}^{-7}\mathrm{\Omega }\mathrm{m}$.

---

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: Temperature, T1 = 27.5°C
Resistance of the silver wire at temperature T1 is R1=2.1Ω$R_1=2.1\mathrm{\Omega }$
Temperature, T2 = 1000oC
Resistance of the silver wire at temperature T2 is R2=2.7Ω$R_2=2.7\mathrm{\Omega }$
Let temperature coefficient of resistance of silver =α$\alpha$
It is related with temperature and resistance as
α=R2−R1R1(T2−T1)=2.7−2.12.1(100−27.5)$\alpha =\frac{R_2-R_1}{R_1\left(T_2-T_1\right)}=\frac{2.7-2.1}{2.1(100-27.5)}$
= 0.0039oC-1
Therefore, the temperature coefficient of silver is 0.0039oC-1.

---

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 =1.7×10−4C−1$=1.7\times {10}^{-4}{C}^{-1}$
Ans: Supply voltage, V=230V
Initial current drawn=I1 = 3.2A
Initial resistance= R1 which is given by the relation,
R1=VI=2303.2=71.87Ω$R_1=\frac{V}{I}=\frac{230}{3.2}=71.87\mathrm{\Omega }$
Steady state value of the current, I2 = 2.8A
Resistance at the steady state = R2 which is given as
R2=2302.8=82.14Ω$R_2=\frac{230}{2.8}=82.14\mathrm{\Omega }$
Temperature co-efficient of resistance of nichrome, α=1.70×10−4C−1$\alpha =1.70\times {10}^{-4}{\mathrm{C}}^{-1}$
Initial temperature of nichrome, T1 = 27.0°C
Steady state temperature reached by nichrome= T2
Since,
α=R2−R1R1(T2−T1)$\alpha =\frac{R_2-R_1}{R_1\left(T_2-T_1\right)}$
Therefore,
T2−27∘C=82.41−71.8771.87×1.7×10−4=840.5$T_2-{27}^{\circ }C=\frac{82.41-71.87}{71.87\times 1.7\times {10}^{-4}}=840.5$
T2 = 840.5 + 27 = 867.5° C
Therefore, the steady temperature of the heating element is 867.5oC.

---

9: Determine the current in each branch of the network shown in fig 3.30:
Ans: Current flowing through various branches of the circuit is represented in the given figure.

I1 = Current flowing through the outer circuit
I2 = Current flowing through branch AB
I3 = Current flowing through branch AD
I2 - I4 = Current flowing through branch BC
I3 + I4 = Current flowing through branch CD
I4 = Current flowing through branch BD
Applying Kirchhoff's loop rule to the closed circuit ABDA,
10I2 + 5I4 - 5I3 = 0
2I2 + I4 - I3 = 0
I3 = 2I2 + I4 ..... (1)
Applying Kirchhoff's 2nd law to the closed circuit BCDB,
5(I2 + I4) - 10(I3 + I4) - 5I4 = 0
5I2 + 5I4 - 10I3 - 10I4 - 5I4 = 0
5I2 - 10I3 - 20I4 = 0
I2 = 2I3 + 4I4..........(2)
Applying Kirchhoff's 2nd law to the closed circuit ABCFEA,
-10 + 10(I1) + 10(I2) + 5(I2 - I4) = 0
10 = 15I2 + 10I1 - 5I4
3I2 + 2I1 - I4 = 2..........(3)
From equations (1) and (2), we obtain
I3 = 2(2I3 + 4I4) + I4
I3 = 4I3 + 8I4 + I4
-3I3 = 9I4
-3I4 = +I3 ...... (4)
Putting equation (4) in equation (1), we obtain
I3 = 2I2 + I4
-4I4 = 2I2
I2 = -2I4 ...... (5)
It is evident from the given figure that,
I1 = I3 + I2 ....... (6)
Putting equation (6) in equation (1), we obtain
3I2 + 2(I3 + I2) - I4 = 2
5I2 + 2I3 - I4 = 2...... (7)
Putting equations (4) and (5) in equation (7), we obtain
5(−2I4)+2(−3I4)−I4=2$5\left(-2I_4\right)+2\left(-3I_4\right)-I_4=2$
10I4−6I4−I4=2$10I_4-6I_4-I_4=2$
17I4 = -2
Equation (4) reduces to
I3=−3(I4)$I_3=-3\left(I_4\right)$
=−3(−217)=617A$=-3\left(\frac{-2}{17}\right)=\frac{6}{17}A$
I2=−2(I4)$I_2=-2\left(I_4\right)$
=−2(−217)=417A$=-2\left(\frac{-2}{17}\right)=\frac{4}{17}A$
I2−I4=417−(−217)=617A$I_2-I_4=\frac{4}{17}-\left(\frac{-2}{17}\right)=\frac{6}{17}\mathrm{A}$
I3+I4=617+(−217)=417A$I_3+I_4=\frac{6}{17}+\left(\frac{-2}{17}\right)=\frac{4}{17}\mathrm{A}$
I1=I3+I2$I_1=I_3+I_2$
=617+417=1017A$=\frac{6}{17}+\frac{4}{17}=\frac{10}{17}\mathrm{A}$
Therefore, current in branch AB =417A$=\frac{4}{17}\mathrm{A}$
current in branch BC =617A$=\frac{6}{17}\mathrm{A}$
current in branch CD =−417A$=\frac{-4}{17}\mathrm{A}$
current in branch AD=617A$=\frac{6}{17}\mathrm{A}$
current in branch BD=−217A$=\frac{-2}{17}\mathrm{A}$
Therefore total current = 417+617+−417+617+−217=1017A$\frac{4}{17}+\frac{6}{17}+\frac{-4}{17}+\frac{6}{17}+\frac{-2}{17}=\frac{10}{17}\mathrm{A}$.

---

10:

1. In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end when the resistor Y is of 12.5 Ω.
2. Determine the balance point of the given bridge if X and Y are interchanged.
3. What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

Ans: A metre bridge with resistors X and Y is represented in the given figure.

1. Balance point from end A is l1 =39.5 cm
   Resistance of the resistor Y =12.5Ω$=12.5\mathrm{\Omega }$
   Condition for the balance is given as,
   XY=l1100−l1$\frac{X}{Y}=\frac{l_1}{100-l_1}$
   XY=39.5100−39.5×12.5=8.2Ω$\frac{X}{Y}=\frac{39.5}{100-39.5}\times 12.5=8.2\mathrm{\Omega }$
   Therefore, the resistance of resistor X is 8.2Ω$8.2\mathrm{\Omega }$.
2. If X and Y are interchanged, then, l1 and (100 -l1) get interchanged.
   The balance point of the bridge will be at a distance (100 -l1) from A.
   100 -l1 = 100 − 39.5 = 60.5 cm
   Therefore, the balance point is 60.5 cm from A.
3. When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection so no current would flow through the galvanometer.

---

11: A storage battery of emf 8.0 V and internal resistance 0.5Ω$\mathrm{\Omega }$ is being charged by a 120 V dc supply using a series resistor of 15.5Ω$\mathrm{\Omega }$. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Ans: Emf of the storage battery, E
Internal resistance of the battery, r =0.5Ω$0.5\mathrm{\Omega }$
DC supply voltage, V = 120 V
Resistance of the resistor, R = 15.5Ω$\mathrm{\Omega }$
Effective voltage in the circuit =V1
R is connected to the storage battery in series. Hence, it can be written as
V1 = V - E
V1 = 120 - 8 = 112V
Current flowing in the circuit can be calculated as: I=V1R+r=11215.5+0.5=7A$I=\frac{V_1}{R+r}=\frac{112}{15.5+0.5}=7A$

Voltage across resistor R is given by the product , IR= 7 ×$\times$ 15.5 = 108.5 V
DC supply voltage = Terminal voltage of battery + Voltage drop across R

Terminal voltage of battery = 120 - 108.5 = 11.5 V.

A series resistor in a charging circuit limits the current drawn from the external source and saves the circuit from extremely high current which is very dangerous.

---

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: Emf of the cell, E1 =1.25 V
Balance point of the potentiometer, I1 = 35 cm
The cell is replaced by another cell of emf E2
New balance point of the potentiometer, I2 =63 cm
By the condition of balance of potentiometer
E1E2=l1l2$\frac{E_1}{E_2}=\frac{l_1}{l_2}$
E2=E1×l2l1$E_2=E_1\times \frac{l_2}{l_1}$
=1.25×6335=2.25V$=1.25\times \frac{63}{35}=2.25V$
Therefore, emf of the second cell is 2.25 V.

---

13: The number density of free electrons in a copper conductor estimated in Example is 8.5×1028m−3$8.5\times {10}^{28}{m}^{-3}$. 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 2×1028m−3$2\times {10}^{28}{m}^{-3}$ and it is carrying a current of 3.0 A.
Ans: Number density of free electrons in a copper conductor, n=8.5×1028m−3$n=8.5\times {10}^{28}{m}^{-3}$
Length of the copper wire, l = 3.0 m
Area of cross-section of the wire, A=2.0×10−6m2$A=2.0\times {10}^{-6}{m}^2$
Current carried by the wire, I=A,
as I=neAvd$I=neA{v}_d$
Where,
e = Electric charge= 1.6×10−19C$1.6\times {10}^{-19}C$
vd=lt$v_d=\frac{l}{t}$
I=nAelt$I=n\mathrm{A}e\frac{l}{t}$
t=nAelI$t=\frac{nAel}{I}$
=3×8.5×1028×2×10−6×1.6×10−193$\frac{3\times 8.5\times {10}^{28}\times 2\times {10}^{-6}\times 1.6\times {10}^{-19}}{3}$
=2.7×104s$=2.7\times {10}^{4}s$
Therefore, the time taken by an electron to drift form one end of wire to the other is 2.7×104s$2.7\times {10}^{4}s$

---

14: The earth’s surface has a negative surface charge density of 10-9cm-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 part of the globe [Radius of earth = 6.37×106m$6.37\times {10}^6\mathrm{m}$.
Ans: Surface charge density of the earth, δ=10−9Cm−2$\delta ={10}^{-9}C{m}^{-2}$
Current over the entire globe, I = 1800A
Radius of the earth, r=6.37×106m$r=6.37\times {10}^{6}m$
Surface area of the earth,
A=4πr2$A=4\pi r^2$
=4×3.14×(6.37×106)2$=4\times 3.14\times {\left(6.37\times {10}^6\right)}^2$
=5.09×1014m2$=5.09\times {10}^{14}{\mathrm{m}}^2$
Charge on the earth surface = Surface charge density of the earth x Surface Area of the earth
=10−9×5.09×1014=5.09×105C$={10}^{-9}\times 5.09\times {10}^{14}=5.09\times {10}^{5}C$
Time taken to neutralize the earth’s surface = t
I=qt$I=\frac{q}{t}$
t=qI$t=\frac{q}{I}$
=5.09×1051800=282.77s$=\frac{5.09\times {10}^5}{1800}=282.77s$
Therefore, it takes 282.77 s to neutralize the earth’s surface.

---

15:

1. Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015Ω$\mathrm{\Omega }$ are joined in series to provide a supply to a resistance of 8.5Ω$\mathrm{\Omega }$. What are the current drawn from the supply and its terminal voltage?
2. A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380Ω$\mathrm{\Omega }$. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

Ans:

1. Number of secondary cells, n
   Emf of each secondary cell, E= 2.0V
   Internal resistance of each cell, r= 0.015Ω$\mathrm{\Omega }$
   series resistor is connected to the combination of cells.
   Resistance of the resistor, R = 8.5Ω$\mathrm{\Omega }$
   Current drawn from the supply=I is given by the relation,
   I=nER+nr$I=\frac{nE}{R+nr}$
   =6×28.5+6×0.015=128.59=1.39A$=\frac{6\times 2}{8.5+6\times 0.015}=\frac{12}{8.59}=1.39\mathrm{A}$
   Terminal voltage, V = IR = 1.39 ×$\times$ 8.5 = 11.87 A
   Therefore, the current drawn from the supply is 1.39 A and terminal voltage is 11.87 A.
2. After a long use, emf of the secondary cell, E=1.9V
   Internal resistance of the cell, r= 380 Ω$\mathrm{\Omega }$
   Maximum current
   E=Er=1.9380=0.005A$E=\frac{E}{r}=\frac{1.9}{380}=0.005\mathrm{A}$
   Therefore, the maximum current drawn from the cell is 0.005 A. This much current is not sufficient to start the motor of a car, as it requires very high current..

---

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. (ρAl=2.63×10−8Ωm${\rho }_{\text{Al}}=2.63\times {10}^{-8}\mathrm{\Omega }\text{m}$, ρCu=1.72×10−8Ωm${\rho }_{Cu}=1.72\times {10}^{-8}\mathrm{\Omega }\mathrm{m}$ Relative density of Al = 2.7, of Cu = 8.9.)
Ans: Resistivity of aluminium, ρAl=2.63×10−8Ωm${\rho }_{\text{Al}}=2.63\times {10}^{-8}\mathrm{\Omega }\text{m}$
Relative density of aluminium, d1 = 2.7
Let l1 be the length of aluminium wire and m1 be its mass.
Resistance of the aluminium wire = R1
Area of cross-section of the aluminium wire = A1
Resistivity of copper, ρCu=1.72×10−8Ωm${\rho }_{Cu}=1.72\times {10}^{-8}\mathrm{\Omega }\mathrm{m}$
Relative density of copper, d2 = 8.9
Let l2 be the length of copper wire and m2 be its mass.
Resistance of the copper wire = R2
Area of cross-section of the copper wire = A2
The two relations can be written as
R1=ρ1l1A1$R_1={\rho }_1\frac{l_1}{A_1}$ ...... (1)
R2=ρ2l2A2$R_2={\rho }_2\frac{l_2}{A_2}$ ...... (2)
It is given that,
R1 = R2
∴ρ1l1A1=ρ2l2A2$\therefore {\rho }_1\frac{l_1}{A_1}={\rho }_2\frac{l_2}{A_2}$
And,
l1 = l2
∴ρ1A1=ρ2A2$\therefore \frac{{\rho }_1}{A_1}=\frac{{\rho }_2}{A_2}$
AlA2=ρlρ2$\frac{A_{\mathrm{l}}}{A_2}=\frac{{\rho }_{\mathrm{l}}}{{\rho }_2}$
=2.63×10−81.72×10−8=2.631.72$=\frac{2.63\times {10}^{-8}}{1.72\times {10}^{-8}}=\frac{2.63}{1.72}$
Mass of the aluminium wire,
m1 = Volume ×$\times$Density
= A1l1 ×$\times$ d1 = A1 l1d1 … (3)
Mass of the copper wire,
m2 = Volume ×$\times$ Density
= A2l2 ×$\times$ d2 = A2 l2d2 ….. (4)
Dividing equation (3) by equation (4), we obtain
m1m2=A1l1d1A2l2d2$\frac{m_1}{m_2}=\frac{A_1{l}_1{d}_1}{A_2{l}_2{d}_2}$
For l1 = l2,
m1m2=A1d1A2d2$\frac{m_1}{m_2}=\frac{A_1{d}_1}{A_2{d}_2}$
For A1A2=2.631.72$\frac{A_1}{A_2}=\frac{2.63}{1.72}$
m1m2=2.631.72×2.78.9=0.46$\frac{m_1}{m_2}=\frac{2.63}{1.72}\times \frac{2.7}{8.9}=0.46$
It can be inferred from this ratio that m1 is less than m2. Hence, aluminium is lighter than copper.
Since aluminium is lighter and its cost would be less and is preferred for overhead power cables over copper.

---

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

CURRENT	VOLTAGE	CURRENT	VOLTAGE
0.2	3.94	3	59.2
0.4	7.87	4	78.8
0.6	11.8	5	98.6
0.8	15.7	6	118.5
1.0	19.7	7	138.2
2.0	39.7	8	158.0

Ans: It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm’s law. According to Ohm’s law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistance of manganin is 19.7Ω$\mathrm{\Omega }$.

---

18: Answer the following questions:

1. A steady current flows in a metallic conductor of non uniform cross section of non uniform cross section. Which of these quantities is constant along the conductor,: current ,current density ,electric field, drift speed?
2. Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.
3. A low voltage supply from which one needs high currents must have very low resistance. Why?
4. A high tension (HT) supply of, say, 6 kV must have a very large internal resistance

Ans:

1. When a steady current flows in a metallic conductor of non uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant in a conductor of varying cross section.
2. No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode semi-conductor is a non-ohmic conductor. Ohm’s law is not valid for it.
3. According to Ohm’s law,
   I=VR$I=\frac{V}{R}$
   If V is low, then internal resistance R must be very low, so that high current can be drawn from the source.
4. In order to prohibit the current from exceeding the safety limit, a high tension supply must have a very large internal resistance. If the internal resistance is small, then the current drawn can exceed the safety limits in case of a short circuit which can be very dangerous.

---

19: Choose the correct alternative:

1. Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
2. Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
3. The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.
4. The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (1022/103).

Ans:

1. Alloys of metals usually have greater resistivity than that of their constituent metals.
2. Alloys usually have lower temperature coefficients of resistance than pure metals.
3. The resistivity of the alloy, manganin, is nearly independent of increase of temperature.
4. The resistivity of a typical insulator is greater than that of a metal by a factor of the order of 1022.

---

20:

1. 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?
2. Given the resistances of 1Ω,2Ω,3Ω,$1\mathrm{\Omega },2\mathrm{\Omega },3\mathrm{\Omega },$ how will be combine them to get an equivalent resistance of
   (i) (11/3)Ω$\mathrm{\Omega }$ (ii) (11/5)Ω$\mathrm{\Omega }$ (iii) 6Ω$\mathrm{\Omega }$ (iv) (6/11)Ω$\mathrm{\Omega }$?
3. Determine the equivalent resistance of networks shown in fig.
   
   

Ans:

1. Total number of resistors= n
   Resistance of each resistor= R
   1. When n resistors are connected in series, effective resistance R1 is the maximum, given by the product nR.
      Hence, maximum resistance of the combination,R1 = nR
   2. When n resistors are connected in parallel, the effective resistance (R2) is the minimum, given by the ration Rn$\frac{R}{n}$.
      Hence, minimum resistance of the combination, R2=Rn$R_2=\frac{R}{n}$
   3. The ratio of the maximum to the minimum resistance is,
      R1R2=nRR/n=n2$\frac{R_1}{R_2}=\frac{nR}{R/n}=n^2$
2. The resistance of the given resistors is,R1=1Ω,R2=2Ω,R3=3Ω$R_1=1\mathrm{\Omega },R_2=2\mathrm{\Omega },R_3=3\mathrm{\Omega }$
   1. Equivalent resistance, R′=113Ω$R^{\mathrm{\prime }}=\frac{11}{3}\mathrm{\Omega }$
      Consider the following combination of the resistors.
      Equivalent resistance of the circuit is given by,
      
      R′=2×12+1+3=23+3=113Ω$R^{\mathrm{\prime }}=\frac{2\times 1}{2+1}+3=\frac{2}{3}+3=\frac{11}{3}\mathrm{\Omega }$
   2. Equivalent resistance, R′=115Ω$R^{\mathrm{\prime }}=\frac{11}{5}\mathrm{\Omega }$
      Equivalent resistance of the circuit is given by,
      
      R′=2×32+3+1=65+1=115Ω$R^{\mathrm{\prime }}=\frac{2\times 3}{2+3}+1=\frac{6}{5}+1=\frac{11}{5}\mathrm{\Omega }$
   3. Equivalent resistance, R′=6Ω$R^{\mathrm{\prime }}=6\mathrm{\Omega }$
      Consider the series combination of the resistors, as shown in the given circuit.
      Equivalent resistance of the circuit is given by the sum,
      R′=1+2+3=6Ω$R^{\mathrm{\prime }}=1+2+3=6\mathrm{\Omega }$
      Consider the series combination of the resistors, as shown in the given ciruit.
      Equivalent resistance of the circuit is given by,
      
      R′=1×2×31×+2×3+3×1=611Ω$R^{\mathrm{\prime }}=\frac{1\times 2\times 3}{1\times +2\times 3+3\times 1}=\frac{6}{11}\mathrm{\Omega }$
3. 1. It can be observed form the given circuit that two resistors of resistance 1Ω$1\mathrm{\Omega }$ each are connected in series.
      Hence, their equivalent resistance= (1 + 1) = 2Ω$2\mathrm{\Omega }$
      It can also be observed that two resistors of resistance 2Ω$2\mathrm{\Omega }$ each are connected in series.
      Hence, their equivalent resistance= (2 + 2) = 4Ω$4\mathrm{\Omega }$
      Therefore, the circuit can be redrawn as
      
      It can be observed that 2Ω$2\mathrm{\Omega }$ and 4Ω$4\mathrm{\Omega }$ resistors are connected in parallel in all the four loops.
      Hence, equivalent resistance (R') of each loop is given by,R′=2×42+4=86=43Ω$R^{\mathrm{\prime }}=\frac{2\times 4}{2+4}=\frac{8}{6}=\frac{4}{3}\mathrm{\Omega }$
      The circuit reduces to
      
      All the four resistors are connected in series.
      Therefore, equivalent resistance of the given circuit is 43×4=163Ω$\frac{4}{3}\times 4=\frac{16}{3}\mathrm{\Omega }$
   2. It can be observed from the given circuit that five resistors of resistance  each are connected in series.
      Therefore, equivalent resistance of the circuit = R + R + R + R + R = 5R.

---

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

Ans: The resistance of each resistor connected in the given circuit, R=1Ω$R=1\mathrm{\Omega }$
Let the equivalent resistance of the given circuit = R'
The network is infinite, Hence, equivalent resistance is given by the relation,
R′=2+R′R′+1$R^{\prime }=2+\frac{R^{\prime }}{R^{\prime }+1}$
R′2−2R′−2=0$R^{\prime 2}-2R^{\prime }-2=0$
R′=2±4+8√2$R^{\mathrm{\prime }}=\frac{2\pm \sqrt{4+8}}{2}$
=2±12√2=1±3–√$=\frac{2\pm \sqrt{12}}{2}=1\pm \sqrt{3}$
Negative value of R' cannot be accepted. Hence, equivalent resistance,
R′=(1+3–√)=1+1.73=2.73Ω$R^{\mathrm{\prime }}=(1+\sqrt{3})=1+1.73=2.73\mathrm{\Omega }$
Internal resistance of circuit is 0.5Ω$0.5\mathrm{\Omega }$
Therefore, total resistance of the given circuit =2.73+0.5=3.23Ω$=2.73+0.5=3.23\mathrm{\Omega }$
Supply Voltage, V = 12V
According to Ohm; s Law,
I=123.23=3.72A$I=\frac{12}{3.23}=3.72A$

---

22: Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 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 replace by a cell of unknown emf and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

1. What is the value ε$\epsilon$?
2. What purpose does the high resistance of 600 kΩ$k\mathrm{\Omega }$ have?
3. Is the balance point affected by this high resistance?
4. Is the balance point affected by the internal resistance of the driver cell?
5. Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
6. 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:

1. Constant emf of the given standard cell, E1 = 1.02V
   Balance point on the wire, l1 = 67.3cm
   A cell of unknown emf,, replaced the standard cell. Therefore, new balance on the wire, l = 82.3cm
   The relation connecting emf and balance point is,
   E1l1=εl$\frac{E_1}{l_1}=\frac{\epsilon }{l}$
   ε=ll1×E1$\epsilon =\frac{l}{l_1}\times E_1$
   =82.367.3×1.02=1.247V$=\frac{82.3}{67.3}\times 1.02=1.247V$
   The value of unknown emf is 1.247V.
2. The purpose of using the high resistance of 600kΩ$k\mathrm{\Omega }$ is to reduce the current passing through the galvanometer when the movable contact is far from the balance point.
3. The balance point is not affected due to the presence of high resistance.
4. The point is not affected by the internal resistance of the driver cell.
5. The method would not work if 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 as potential developed in the potentiometre wire will always be less than the emf of cell.
6. 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.
   Modification: 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.

---

23: Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0Ω$\mathrm{\Omega }$ 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$\epsilon$?

Ans: Resistance of the standard resistor, R = 10.0Ω$\mathrm{\Omega }$
Balance point for this resistance, l1 = 58.3 cm
Current in the potentiometer wire = i
Hence, potential drop across R, E1 = iR
Resistance of the unknown resistor = X
Balance point for this resistor, l2 = 68.5 cm
Hence, potential drop across X, E2 = iX
The relation connecting emf and balance point is,
E1E2=l1l2$\frac{E_1}{E_2}=\frac{l_1}{l_2}$
iRiX=l1l2$\frac{iR}{iX}=\frac{l_1}{l_2}$
X=l1l2×R$X=\frac{l_1}{l_2}\times R$
=68.558.3×10=11.749Ω$=\frac{68.5}{58.3}\times 10=11.749\mathrm{\Omega }$
Therefore, the value of the unknown resistance, X, is 11.75Ω$\mathrm{\Omega }$.

If we fail to find a balance point with the given cell of emf ,, then the potential drop across R and X must be reduced by inserting a high resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.

---

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 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: Internal resistance of the cell=r
Balance point of the cell in open circuit, l1 = 76.3cm

An external resistance (R) is connected to the circuit with R=9.5Ω$R=9.5\mathrm{\Omega }$
New balance point of the circuit, l2 = 64.8cm
Current flowing through the circuit= I
The relation connecting internal resistance of cell and emf is given by,
r=(l1−l2)l2×R$r=\frac{(l_1-l_2)}{l_2}\times R$
=76.3−64.864.8×9.5=1.68Ω$=\frac{76.3-64.8}{64.8}\times 9.5=1.68\mathrm{\Omega }$
Therefore, the internal resistance of the cell is 1.68Ω$1.68\mathrm{\Omega }$.