JEE Physics Current Electricity MCQs Set 31

Multiple Choice Questions (MCQs) for JEE Physics: Current Electricity

Review structured MCQ sets for JEE Physics Current Electricity. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

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Multiple Choice Questions

Question 1: In a meter bridge experiment with the resistance box R in the left gap and the unknown X in the right gap (balance length l measured from the left end), the readings are: (1) R = 1000 \(\Omega\), l = 60 cm; (2) R = 100 \(\Omega\), l = 13 cm; (3) R = 10 \(\Omega\), l = 1.5 cm; (4) R = 1 \(\Omega\), l = 1.0 cm. Which reading is inconsistent?
(a) Reading 4
(b) Reading 3
(c) Reading 2
(d) Reading 1
Show Answer & Explanation

Answer: (a) Reading 4

Explanation:
1. \( X=R\dfrac{100-l}l \): (1) 667 \(\Omega\); (2) 669 \(\Omega\); (3) 657 \(\Omega\); (4) 99 \(\Omega\).
2. Reading 4 gives a very different X, so it is inconsistent.

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 2: A four-wire potentiometer is made of a 400 cm long wire between A and B, of resistance 0.01 \(\Omega\)/cm. It is connected in series with a 1 \(\Omega\) resistor and two 1.5 V cells in series, each of internal resistance 0.5 \(\Omega\). An ideal voltmeter is connected between A and a jockey J at 50 cm from A. The reading of the voltmeter will be
(a) 0.20 V
(b) 0.25 V
(c) 0.75 V
(d) 0.50 V
Show Answer & Explanation

Answer: (b) 0.25 V

Explanation:
1. EMF \( =3 \) V, total resistance \( =1+1+4=6\ \Omega \) (wire \( 4\ \Omega \)), so \( I=0.5 \) A.
2. The 50 cm of wire is \( 0.5\ \Omega \): \( V=0.5\times0.5=0.25 \) V.

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 3: In a meter bridge, the wire of length 1 m has a non-uniform cross-section such that the variation \( \dfrac{dR}{dl} \) of its resistance R with length l is \( \dfrac{dR}{dl}\propto\dfrac1{\sqrt l} \). Two equal resistances are connected in the gaps. The galvanometer has zero deflection when the jockey is at point P. The length AP is
(a) 0.3 m
(b) 0.2 m
(c) 0.35 m
(d) 0.25 m
Show Answer & Explanation

Answer: (d) 0.25 m

Explanation:
1. Integrating, \( R(l)\propto\sqrt l \).
2. Equal gap resistances need \( R(AP)=\dfrac12R(AB) \): \( \sqrt l=\dfrac12\sqrt1\Rightarrow l=0.25 \) m.

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 4: An ideal battery of 4 V and a resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5 \(\Omega\). The value of R to give a potential difference of 5 mV across 10 cm of potentiometer wire is
(a) \( 395\ \Omega \)
(b) \( 495\ \Omega \)
(c) \( 490\ \Omega \)
(d) \( 480\ \Omega \)
Show Answer & Explanation

Answer: (a) \( 395\ \Omega \)

Explanation:
1. 5 mV across 10 cm means 50 mV across the whole wire, so \( I=\dfrac{50\ \text{mV}}{5\ \Omega}=10 \) mA.
2. \( R+5=\dfrac4{0.01}=400\Rightarrow R=395\ \Omega \).

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 5: In a meter bridge, the null point is obtained at a distance of 40 cm from A with resistances \( R_1 \) (left) and \( R_2 \) (right). If a 10 \(\Omega\) resistor is connected in series with \( R_1 \), the null point shifts by 10 cm (to the 50 cm mark). The resistance that should be connected in parallel with \( (R_1+10)\ \Omega \) such that the null point shifts back to its initial position is
(a) \( 20\ \Omega \)
(b) \( 30\ \Omega \)
(c) \( 60\ \Omega \)
(d) \( 40\ \Omega \)
Show Answer & Explanation

Answer: (c) \( 60\ \Omega \)

Explanation:
1. \( \dfrac{R_1}{R_2}=\dfrac{40}{60} \) and \( \dfrac{R_1+10}{R_2}=1 \). So \( R_1=20\ \Omega \), \( R_2=30\ \Omega \).
2. Need \( R_1+10=30\ \Omega \) reduced to \( 20\ \Omega \): \( \dfrac1{20}=\dfrac1{30}+\dfrac1x\Rightarrow x=60\ \Omega \).

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 6: A galvanometer having a resistance of 20 \(\Omega\) and 30 divisions on both sides has a figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter up to 15 volt is
(a) \( 100\ \Omega \)
(b) \( 125\ \Omega \)
(c) \( 120\ \Omega \)
(d) \( 80\ \Omega \)
Show Answer & Explanation

Answer: (d) \( 80\ \Omega \)

Explanation:
1. Full-scale current \( =30\times0.005=0.15 \) A.
2. Total resistance \( =\dfrac{15}{0.15}=100\ \Omega \), so the series resistance \( =100-20=80\ \Omega \).

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 7: A potentiometer wire AB of length L and resistance 12r is joined to a cell D of emf \( \epsilon \) and internal resistance r. A cell C having emf \( \epsilon/2 \) and internal resistance 3r is connected (through a galvanometer) between A and the jockey J. The length AJ at which the galvanometer shows no deflection is
(a) \( \dfrac5{12}L \)
(b) \( \dfrac{13}{24}L \)
(c) \( \dfrac{11}{12}L \)
(d) \( \dfrac{11}{24}L \)
Show Answer & Explanation

Answer: (b) \( \dfrac{13}{24}L \)

Explanation:
1. Current in the wire \( I=\dfrac\epsilon{12r+r}=\dfrac\epsilon{13r} \).
2. No current in C: \( \dfrac\epsilon2=I\cdot12r\cdot\dfrac xL=\dfrac{12\epsilon}{13}\dfrac xL\Rightarrow x=\dfrac{13}{24}L \).

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 8: On interchanging the resistances in the two gaps of a meter bridge, the balance point shifts to the left by 10 cm. The resistance of their series combination is 1 k\(\Omega\). How much was the resistance on the left slot before interchanging the resistances?
(a) \( 990\ \Omega \)
(b) \( 910\ \Omega \)
(c) \( 550\ \Omega \)
(d) \( 900\ \Omega \)
Show Answer & Explanation

Answer: (c) \( 550\ \Omega \)

Explanation:
1. Initially \( \dfrac{R_L}{R_R}=\dfrac l{100-l} \); after interchange the balance length is \( 100-l \). The shift \( l-(100-l)=10\Rightarrow l=55 \) cm.
2. \( \dfrac{R_L}{R_R}=\dfrac{55}{45} \) with \( R_L+R_R=1000 \): \( R_L=550\ \Omega \).

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 9: When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 \(\Omega\), it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0-10 V is
(a) \( 1.985\times10^3\ \Omega \)
(b) \( 2.045\times10^3\ \Omega \)
(c) \( 2.535\times10^3\ \Omega \)
(d) \( 4.005\times10^3\ \Omega \)
Show Answer & Explanation

Answer: (a) \( 1.985\times10^3\ \Omega \)

Explanation:
1. Total resistance \( =\dfrac{10}{5\times10^{-3}}=2000\ \Omega \).
2. Series resistance \( =2000-15=1985\ \Omega \).

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 10: A galvanometer having a coil resistance of 100 \(\Omega\) gives a full scale deflection when a current of 1 mA is passed through it. The value of the resistance which can convert this galvanometer into an ammeter giving a full scale deflection for a current of 10 A is
(a) \( 2\ \Omega \)
(b) \( 0.01\ \Omega \)
(c) \( 0.1\ \Omega \)
(d) \( 3\ \Omega \)
Show Answer & Explanation

Answer: (b) \( 0.01\ \Omega \)

Explanation:
1. Shunt \( S=\dfrac{I_gG}{I-I_g}=\dfrac{10^{-3}\times100}{10-10^{-3}} \).
2. \( S\approx0.01\ \Omega \).

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 11: A resistance of 2 \(\Omega\) is connected across one gap of a meter bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2 \(\Omega\), is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is
(a) \( 4\ \Omega \)
(b) \( 5\ \Omega \)
(c) \( 6\ \Omega \)
(d) \( 3\ \Omega \)
Show Answer & Explanation

Answer: (d) \( 3\ \Omega \)

Explanation:
1. First: \( \dfrac{l}{100-l}=\dfrac2X \). After interchange the balance length is \( 100-l \). Shift \( =100-2l=20\Rightarrow l=40 \) cm.
2. \( \dfrac{40}{60}=\dfrac2X\Rightarrow X=3\ \Omega \).

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 12: A moving coil galvanometer of resistance 100 \(\Omega\) is used as an ammeter using a resistance 0.1 \(\Omega\). The maximum deflection current in the galvanometer is 100 \(\mu\)A. The current in the circuit, so that the ammeter shows maximum deflection, is
(a) 100.1 mA
(b) 1000.1 mA
(c) 10.01 mA
(d) 1.01 mA
Show Answer & Explanation

Answer: (a) 100.1 mA

Explanation:
1. The shunt carries \( I_s=\dfrac{I_gG}S=\dfrac{100\times10^{-6}\times100}{0.1}=0.1 \) A.
2. Total \( =0.1+10^{-4}=0.1001 \) A \( =100.1 \) mA.

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 13: A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 \(\Omega\) resistance, it can be converted into a voltmeter of range 0-30 V. If connected to a \( \dfrac{2n}{249}\ \Omega \) resistance, it becomes an ammeter of range 0-1.5 A. The value of n is
(a) 2
(b) 10
(c) 5
(d) 25
Show Answer & Explanation

Answer: (c) 5

Explanation:
1. Voltmeter: \( G+4990=\dfrac{30}{0.006}=5000\Rightarrow G=10\ \Omega \).
2. Ammeter: \( S=\dfrac{I_gG}{I-I_g}=\dfrac{0.006\times10}{1.494}=\dfrac{2\times5}{249} \), so \( n=5 \).

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 14: A microammeter has a resistance of 100 \(\Omega\) and a full scale range of 50 \(\mu\)A. It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Which of the following range and resistance combinations are correct? (a) 50 V range with 10 k\(\Omega\) resistance in series. (b) 10 V range with 200 k\(\Omega\) resistance in series. (c) 5 mA range with 1 \(\Omega\) resistance in parallel. (d) 10 mA range with 1 \(\Omega\) resistance in parallel.
(a) (a) and (b) only
(b) (c) and (d) only
(c) (a) and (d) only
(d) (b) and (c) only
Show Answer & Explanation

Answer: (d) (b) and (c) only

Explanation:
1. (a) 50 V needs \( \dfrac{50}{50\times10^{-6}}=1\ \text{M}\Omega \), not 10 k\(\Omega\): false. (b) \( \dfrac{10}{50\times10^{-6}}=200\ \text{k}\Omega \): true.
2. (c) Shunt for 5 mA: \( \dfrac{100\times50\times10^{-6}}{5\times10^{-3}}\approx1\ \Omega \): true. (d) For 10 mA it is about \( 0.5\ \Omega \): false.

Teacher's Note:
1. For instruments, write the balance or deflection condition (Wheatstone ratio, potentiometer null, shunt or series resistance) and solve for the unknown.
2. Always check whether the meter reads a series or parallel combination, since that decides the sign of the error.

Question 15: A 200 \(\Omega\) resistor has the colour code red, black, brown (and a tolerance band). If one replaces the red colour by green in the code, the new resistance will be
(a) \( 100\ \Omega \)
(b) \( 500\ \Omega \)
(c) \( 400\ \Omega \)
(d) \( 300\ \Omega \)
Show Answer & Explanation

Answer: (b) \( 500\ \Omega \)

Explanation:
1. Red-black-brown means \( 20\times10^1=200\ \Omega \).
2. Green-black-brown means \( 50\times10^1=500\ \Omega \).

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 16: A Wheatstone bridge is balanced with \( R_1/R_2=R_3/R_4 \), where \( R_2=80\ \Omega \), \( R_4=40\ \Omega \) and the carbon resistor \( R_1 \) is coloured orange, red, brown. Assuming the colour code gives accurate values, \( R_3 \) has the colour code
(a) brown, blue, black
(b) grey, black, brown
(c) brown, blue, brown
(d) red, green, brown
Show Answer & Explanation

Answer: (c) brown, blue, brown

Explanation:
1. Orange-red-brown means \( 32\times10=320\ \Omega \). Balance: \( R_3=R_1\dfrac{R_4}{R_2}=320\times\dfrac{40}{80}=160\ \Omega \).
2. \( 160=16\times10 \): brown (1), blue (6), brown (\( \times10 \)).

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 17: A 2 W carbon resistor is colour coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is
(a) 20 mA
(b) 0.4 mA
(c) 63 mA
(d) 100 mA
Show Answer & Explanation

Answer: (a) 20 mA

Explanation:
1. Green-black-red: \( 50\times10^2=5000\ \Omega \) (brown: 1% tolerance).
2. \( P=I^2R\Rightarrow I_{max}=\sqrt{\dfrac2{5000}}=0.02 \) A \( =20 \) mA.

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 18: A carbon resistance has the colour code: green, orange, yellow, golden. The value of the resistance is
(a) \( 5.3\ \text{M}\Omega\pm5\% \)
(b) \( 530\ \text{k}\Omega\pm5\% \)
(c) \( 64\ \text{k}\Omega\pm10\% \)
(d) \( 6.4\ \text{M}\Omega\pm5\% \)
Show Answer & Explanation

Answer: (b) \( 530\ \text{k}\Omega\pm5\% \)

Explanation:
1. Green 5, orange 3, yellow \( \times10^4 \): \( 53\times10^4=530\ \text{k}\Omega \).
2. Golden tolerance is \( \pm5\% \).

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 19: In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be
(a) 8 A
(b) 10 A
(c) 14 A
(d) 12 A
Show Answer & Explanation

Answer: (d) 12 A

Explanation:
1. Total power \( =15\times40+5\times100+5\times80+1000=2500 \) W.
2. \( I=\dfrac{2500}{220}=11.4 \) A, so a 12 A fuse is the minimum.

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 20: The supply voltage in a room is 120 V. The resistance of the lead wires is 6 \(\Omega\). A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to it?
(a) 10.4 V
(b) zero
(c) 2.9 V
(d) 13.3 V
Show Answer & Explanation

Answer: (a) 10.4 V

Explanation:
1. Bulb \( R=\dfrac{120^2}{60}=240\ \Omega \): its voltage \( =120\times\dfrac{240}{246}=117.1 \) V. Heater \( R=60\ \Omega \): parallel \( =48\ \Omega \), voltage \( =120\times\dfrac{48}{54}=106.7 \) V.
2. Decrease \( =117.1-106.7\approx10.4 \) V.

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

Question 21: Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances \( R_{100},R_{60} \) and \( R_{40} \) respectively, the relation between these resistances is
(a) \( R_{100}=R_{40}+R_{60} \)
(b) \( \dfrac1{R_{100}}\gt \dfrac1{R_{60}}\gt \dfrac1{R_{40}} \)
(c) \( R_{100}\gt R_{60}\gt R_{40} \)
(d) \( \dfrac1{R_{100}}=\dfrac1{R_{40}}+\dfrac1{R_{60}} \)
Show Answer & Explanation

Answer: (b) \( \dfrac1{R_{100}}\gt \dfrac1{R_{60}}\gt \dfrac1{R_{40}} \)

Explanation:
1. At the working temperature on the same mains voltage, \( R=\dfrac{V^2}P \), so \( R_{100}\lt R_{60}\lt R_{40} \).
2. The cold resistances keep the same order, so \( \dfrac1{R_{100}}\gt \dfrac1{R_{60}}\gt \dfrac1{R_{40}} \).

Teacher's Note:
1. Reduce the network step by step using series (add R) and parallel (add 1/R) rules, then apply Ohm's law V = IR or the power formulas.
2. Check the answer with a quick limiting case or by comparing voltage drops around a loop.

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