Solutions for Exercises of Current Electricity
Ω1. Choose the correct option.
i) Kirchhoff’s first law, i.e., ΣI = 0 at a junction, deals with the conservation of
(A) charge
(B) energy
(C) momentum
(D) mass
ii) When the balance point is obtained in the potentiometer, a current is drawn from
(A) both the cells and auxiliary battery
(B) cell only
(C) auxiliary battery only
(D) neither cell nor auxiliary battery
iii) In the following circuit diagram, an infinite series of resistances is shown. Equivalent resistance between points A and B is
(A) infinite
(B) zero
(C) 2 Ω
(D) 1.5 Ω
iv) Four resistances 10 Ω, 10 Ω, 10 Ω and 15 Ω form a Wheatstone’s network. What shunt is required across 15 Ω resistor to balance the bridge
(A) 10 Ω
(B) 15 Ω
(C) 20 Ω
(D) 30 Ω
v) A circular loop has a resistance of 40 Ω. Two points P and Q of the loop, which are one quarter of the circumference apart are connected to a 24 V battery, having an internal resistance of 0.5 Ω. What is the current flowing through the battery.
(A) 0.5 A (B) lA
(C) 2A (D) 3A
vi) To find the resistance of a gold bangle, two diametrically opposite points of the bangle are connected to the two terminals of the left gap of a metre bridge. A resistance of 4 Q is introduced in the right What is the resistance of the bangle if the null point is at 20 cm from the left end?
(A) 2 Ω
(B) 4 Ω
(C) 8 Ω
(D) 16 Ω
2. Answer in brief.
i) Define or describe a Potentiometer.
ii) Define Potential Gradient.
iii) Why should not the jockey be slided along the potentiometer wire?
iv) Are Kirchhoff’s laws applicable for both AC and DC currents?
v) In a Wheatstone’s meter-bridge experiment, the null point is obtained in middle one third portion of wire. Why is it recommended?
vi) State any two sources of errors in meter-bridge experiment. Explain how they can be minimized.
vii) What is potential gradient? How is it measured? Explain.
viii) On what factors does the potential gradient of the wire depend?
ix) Why is potentiometer preferred over a voltmeter for measuring emf?
x) State the uses of a potentiometer.
xi) What are the disadvantages of a potentiometer?
xii) Distinguish between a potentiometer and a voltmeter.
xiii) What will be the effect on the position of zero deflection if only the current flowing through the potentiometer wire is (i) increased (ii) decreased.
3. Obtain the balancing condition in case of a Wheatstone’s network.
4. Explain with neat circuit diagram, how you will determine the unknown resistance by using a meter-bridge.
5. Describe Kelvin’s method to determine the resistance of a galvanometer by using a meter bridge.
6. Describe how a potentiometer is used to compare the emfs of two cells by connecting the cells individually.
7. Describe how a potentiometer is used to compare the emfs of two cells by combination method.
8. Describe with the help of a neat circuit diagram how you will determine the internal resistance of a cell by using a potentiometer. Derive the necessary formula.
9. On what factors does the internal resistance of a cell depend?
10. A battery of emf 4 volt and internal resistance 1 Ω is connected in parallel with another battery of emf 1 V and internal resistance 1 Ω (with their like poles connected together). The combination is used to send current through an external resistance of 2 Ω. Calculate the current through the external resistance.
The connections according to the questions are shown below. Let us apply KVL to find the current through 2Ω
Let us currents through different branches as shown. Potential drops can be shown according to the current directions as shown in the figure,
Let us apply first KVL in the loop consisting of 1V, 1Ω, 4V and 4Ω by tracing clockwise. We have,
−1 + I1 − I2 + 4 = 0 ∴ I1 – I2 = -3…..(1)
Apply second KVL in the loop consisting of 1V, 1Ω and 2Ω tracing clockwise. We have,
$latex -2(I_1\;+\;I_2)\;-\;I_1\;+\;1\;=\;0\;\;\;\;\therefore\;-3I_1-2I_2=-1….(2)$
Multiply equation (1) by 3 and adding (1) and (2),
−5I2 = −10 ∴ I2 = 2A
Put I1 in equation (1) we have,
I1 − 2 = −3
∴ I1 = −1A
(-ve indicates opposite direction to our shown direction)
$latex \therefore\;I\;=\frac2{30+10}=\frac2{40}=\frac1{20}A$
$latex \therefore\;V\;=\;2\;-\;\left(\frac1{20}\times10\right)$
$latex \therefore\;V\;=\;2\;-\;0.5\;=\;1.5\;V$
11. Two cells of emf 1.5 Volt and 2 Volt having respective internal resistances of 1 Ω and 2 Ω are connected in parallel so as to send current in same direction through an external resistance of 5 Ω. Find the current through the external resistance.
The connections according to the questions are shown below. Let us apply KVL to find the current through 5Ω
Let us currents through different branches as shown. Potential drops can be shown according to the current directions as shown in figure. Let us apply first KVL in the loop consisting of 2V, 2Ω, 1.5V and 1Ω by tracing clockwise. We have,
−2 + 2I1 − I2 + 1.5 = 0 ∴ 2I1 − I2 = 0.5…. (1)
Apply second KVL in the loop consisting of 2V, 2 5Ω tracing clockwise. We have,
−5(I1 + I2) − 2I1 + 2 = 0
∴ −7I1 − 5I2 = −2 ….(2)
Multiply equation (1) by 5 and then (1) – (2),
17I1 = 4.5 ∴ I1 = 4.5/17 A
Put I1 in equation (1) we have,
$latex =\;2\;\times\;\frac{4.5}{17}\;-\;I_2\;=\;0.5$
$latex \therefore\;I_2\;=\;\frac1{34}A$
Hence, current through external resistance 5 shown in the figure, (I1 + I2)
$latex \therefore\;I_{5\Omega}\;=\;(I_1\;+\;I_2)$
$latex \therefore\;I_{5\Omega}\;=\;\frac{4.5}{17}+\frac1{34}$
$latex \therefore\;I_{5\Omega}\;=\;\frac5{17}A$
12. A voltmeter has a resistance 30 Ω. What will be its reading, when it is connected across a cell of emf 2 V having internal resistance 10 Ω?
R = 30 Ω, E = 2 V, r = 10 Ω
$latex V\;=\;E\;-\;Ir\;\;\;\;\;\;\;\;\;\therefore I=\frac E{R+r}$
We know that in potentiometer experiment for comparing emfs by sum and difference method
$latex \frac{E_1}{E_2}\;=\;\frac{I_1+I_2}{I_1-I_2}$
$latex =\;\frac{2.7+0.3}{2.7-0.3}\;=\;\frac3{2.4}$
$latex \therefore\frac{E_1}{E_2}=1.25$
13. A set of three coils having resistances 10 Ω, 12 Ω and 15 Ω are connected in This combination is connected in series with series combination of three coils of the same resistances. Calculate the total resistance and current through the circuit, if a battery of emf 4.1 Volt is used for drawing current.
R1 = 10Ω,R2 = 12Ω and R3 = 15Ω
Equivalent resistance of parallel combinations
$latex\frac1{R_P}=\frac1{R_1}+\frac1{R_2}+\frac1{R_3}$
$latex =\frac1{10}+\frac1{12}+\frac1{15}=\frac14$
∴ Rp = 4Ω
Equivalent resistance of series connections
Rs = R1 + R2 + R3 = 10 + 12 + 15
∴ Rs = 37Ω
Now these parallel combination and series combination are connected in series, hence
$latex \therefore R_{eq}=R_p+R_s=4+37=41\Omega$
$latex \therefore\;I\;=\;\frac V{R_{eq}}$
$latex \therefore\;I\;=\frac{4.1}{41}\;=\$
14. A potentiometer wire has a length of 1.5 m and resistance of 10 Ω. It is connected in series with the cell of emf 4 Volt and internal resistance 5 Ω. Calculate the potential drop per centimeter of the wire.
L = 1.5 m, Rwire = 10 Ω, E = 4 V, r = 5 Ω, K=? (in cm)
We know, that,
$latex K\;=\;\frac{V_{wire}}L,$
$latex V_{wire}\;=\;IR_{wire}\;\&\;I=\frac E{r+R_{wire}}$
$latex \therefore V_{wire}\;=\frac4{5+10}\times10=\frac{40}{15}V$
$latex \therefore\;K\;=\;\frac{V_{wire}}L=\frac{40/15}{1.5}=1.778\frac Vm$
$latex \therefore\;K\;=\;1.78\;\times\;10^{-2}\frac V{cm}$
$latex \therefore\;K\;=\;0.0178\;\frac V{cm}$
15. When two cells of emfs. εl and ε2 are connected in series so as to assist each other, their balancing length on a potentiometer is found to be 2.7 m. When the cells are connected in series so as to oppose each other, the balancing length is found to be 0.3 m. Compare the emfs of the two cells.
l1 = 2.7m , l2 = 0.3m, E1/E2=?
We know that in potentiometer experiment for comparing emfs by sum and difference method,
$latex \frac{E_1}{E_2}\;=\;\frac{I_1+I_2}{I_1-I_2}$
$latex =\;\frac{2.7+0.3}{2.7-0.3}\;=\;\frac3{2.4}$
$latex \therefore\frac{E_1}{E_2}=1.25$
16. The emf of a cell is balanced by a length of 120 cm of potentiometer wire. When the cell is shunted by a resistance of 10 Ω, the balancing length is reduced by 20 cm. Find the internal resistance of the cell.
Let E be the emf and r be the internal resistance of the emf.
l1 = 120 cm = 120 × 10-2 m, R = 10 Ω,
l2 = 100 cm = 100 × 10-2 m, r = ?
We know that for potentiometer experiment,
$latex \therefore r\;=\;R\left(\frac{I_1-I_2}{I_2}\right)$
$latex \therefore\;r\;=\;10\left(\frac{120-100}{100}\right)=\;2\Omega$
17. A potential drop per unit length along a wire is 5 x 10-3 V/m. If the emf of a cell balances against length 216 cm of this potentiometer wire, find the emf of the cell.
K = 5×10-3 V/m, l = 216 cm = 216 × 10-2 m, E=?
We know that for potentiometer, emf for given balancing is given as, E = Kl
∴ E = Kl = 5 × 10 × 216 × 10-2
∴ E = 1080 × 10-5
∴ E = 0.01080V
18. The resistance of a potentiometer wire is 8 Ω and its length is 8 m. A resistance box and a 2 V battery are connected in series with it. What should be the resistance in the box, if it is desired to have a potential drop of 1 μV/mm?
L = 8 m, Rwire = 8Ω ,E = 2 V, R = ?
$latex K\;=\;1\left(\frac{\mu V}{mm}\right)$
$latex K\;=\;\frac{10^{-6}}{10^{-3}}\;=\;10^{-3}\;=\;0.001\;\frac Vm$
We know,
$latex K\;=\frac{V_{wire}}L\;\&\;V_{wire}=IR_{wire}$
$latex I=\frac E{R+R_{wire}}=\frac2{R+8}$
$latex \therefore\;V_{wire}=\frac2{R+8}\;\times\;8\;=\;\frac{16}{8+R}$
$latex \therefore\;K\;=\;0.001\;=\frac{\displaystyle\frac{16}{8+r}}8$
$latex \therefore\;K\;=\;\frac{16}{64+8R}$
∴ 0.064 + 0.008R = 16
$latex \therefore\;R\;=\;\frac{15.936}{0.008}=1992\;\Omega$
19. Find the equivalent resistance between the terminals of A and B in the network shown in the figure below given that the resistance of each resistor is 10 ohm.
[Ans: 14 Ohm]
20. A voltmeter has a resistance of 100 Ω. What will be its reading when it is connected across a cell of emf 2 V and internal resistance 20 Ω?
R = 100 Ω, E = 2 V, r = 20 Ω
$latex f=\frac1T=\frac1{\displaystyle\frac{2\pi r}v}=\frac v{2\pi r}=\frac{qB}{2nm}$
$latex \therefore\;f\;=\;\frac{1.6\times10^{-1}\times6\times10^{-4}}{2\pi\times9\times10^{-31}}$
$latex \therefore\;f\;=\;0.1868\;\times\;10^8$
$latex \therefore\;f\;=\;18.68\;\times\;10^6\;Hz\;=\;18.68\;MHz\;$
$latex E\;=\;K.E.=\frac12mv^2$
$latex \therefore\;E\;=\frac12\times\times9\times10^{-31}\times\left(3\times10^7\right)^2$
$latex =\;40.5\;\times\;10^{-17}\;J$
$latex \therefore\;E\;=\;\frac{40.5\times10^{-17}}{1.6\times10^{-19}}eV$
$latex \therefore\;E\;=\;25.31\times10^2$
$latex \therefore\;E\;=\;2.531\;\times\;10^3\;eV$
$latex \therefore\;E\;=\;2.531\;keV\;\;$