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Current Electricity — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Current Electricity MCQs with step-by-step solutions covering Ohm's Law and Resistance, Heating Effect of Current, Cells and its Combination and Kirchhoff's Rules, Measuring Instruments. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Ohm's Law and Resistance · easy · numerical
The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25 Ω. What will be the effective resistance if they are connected in series?
A. 0.25 Ω
B. 0.5 Ω
C. 1 Ω
D. 4 Ω ✓ Correct
Solution: Parallel: $\frac{1}{0.25} = \frac{4}{R} \Rightarrow R = 1$ Ω per wire. Series: $R_{eq} = 4R = 4$ Ω
Q2 — Ohm's Law and Resistance · easy · numerical
Two solid conductors are made up of same material, have same length and same resistance. One of them has a circular cross-section of area $A_1$ and the other one has a square cross-section of area $A_2$. The ratio $A_1/A_2$ is
A. 1.5
B. 1 ✓ Correct
C. 0.8
D. 2
Solution: $R = \frac{\rho l}{A}$. Same material, length and resistance means same A, so $\frac{A_1}{A_2} = 1$
Q3 — Ohm's Law and Resistance · easy · theory
Which of the following acts as a circuit protection device?
A. Inductor
B. Switch
C. Fuse ✓ Correct
D. Conductor
Solution: A fuse has low melting point and low resistivity; it melts and breaks the circuit when excessive current flows, protecting it.
Q4 — Ohm's Law and Resistance · easy · numerical
A wire of resistance 4 Ω is stretched to twice its original length. The resistance of stretched wire would be
A. 2 Ω
B. 4 Ω
C. 8 Ω
D. 16 Ω ✓ Correct
Solution: Stretching to n times length gives $R' = n^2R = 2^2 \times 4 = 16$ Ω
Q5 — Ohm's Law and Resistance · easy · theory
The specific resistance of a conductor increases with
A. increase in temperature ✓ Correct
B. increase in cross-sectional area
C. decrease in length
D. decrease in cross-sectional area
Solution: Specific resistance (resistivity) of a metal increases with temperature ($\rho_t = \rho_0(1 + \alpha t)$); it does not depend on the geometry of the conductor.
Q6 — Ohm's Law and Resistance · easy · theory
The resistance of a discharge tube is
A. zero
B. ohmic
C. non-ohmic ✓ Correct
D. infinity
Solution: In a discharge tube the current is carried by positive ions and electrons and the V-I relation is non-linear, so its resistance is non-ohmic.
Q7 — Ohm's Law and Resistance · easy · theory
The velocity of charge carriers of current (about 1 A) in a metal under normal conditions is of the order of
A. a fraction of mm/s ✓ Correct
B. velocity of light
C. several thousand m/s
D. a few hundred m/s
Solution: The drift velocity of free electrons in a metal carrying an ordinary current is very small — of the order of a fraction of a millimetre per second.
Q8 — Heating Effect of Current · easy · numerical
A 5 A fuse wire can withstand a maximum power of 1 W in a circuit. The resistance of the fuse wire is
A. 0.2 Ω
B. 5 Ω
C. 0.4 Ω
D. 0.04 Ω ✓ Correct
Solution: $R = \frac{P}{i^2} = \frac{1}{5^2} = 0.04$ Ω
Q9 — Heating Effect of Current · easy · theory
Fuse wire is a wire of
A. low resistance and low melting point
B. low resistance and high melting point
C. high resistance and high melting point
D. high resistance and low melting point ✓ Correct
Solution: A fuse wire is made of a material of high resistance and low melting point so that it heats up and melts quickly when excessive current flows, breaking the circuit.
Q10 — Heating Effect of Current · easy · theory
Two bulbs 25 W, 220 V and 100 W, 220 V are given. Which has higher resistance?
A. 25 W bulb ✓ Correct
B. 100 W bulb
C. Both bulbs will have equal resistance
D. Resistance of bulbs cannot be compared
Solution: $R = \frac{V^2}{P}$; for the same voltage, $R \propto \frac{1}{P}$. The lower-power (25 W) bulb has the higher resistance.
Q11 — Heating Effect of Current · easy · numerical
A current of 2 A, passing through a conductor produces 80 J of heat in 10 s. The resistance of the conductor in ohm is
A. 0.5
B. 2 ✓ Correct
C. 4
D. 20
Solution: $H = i^2Rt \Rightarrow R = \frac{H}{i^2 t} = \frac{80}{2^2 \times 10} = 2$ Ω
Q12 — Cells and its Combination and Kirchhoff's Rules · easy · numerical
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is
A. 0.2 Ω
B. 0.5 Ω ✓ Correct
C. 0.8 Ω
D. 1.0 Ω
Solution: $E = I(R + r) \Rightarrow 2.1 = 0.2(10 + r)$, so $10 + r = 10.5 \Rightarrow r = 0.5$ Ω
Q13 — Cells and its Combination and Kirchhoff's Rules · easy · theory
Consider the two statements: I. Kirchhoff's junction law follows from the conservation of charge. II. Kirchhoff's loop law follows from the conservation of energy. Which of the following is correct?
A. Both I and II are wrong
B. I is correct and II is wrong
C. I is wrong and II is correct
D. Both I and II are correct ✓ Correct
Solution: Kirchhoff's junction (current) law follows from conservation of charge, and the loop (voltage) law follows from conservation of energy — both statements are correct.
Q14 — Cells and its Combination and Kirchhoff's Rules · easy · numerical
A cell has an emf 1.5 V. When connected across an external resistance of 2 Ω, the terminal potential difference falls to 1.0 V. The internal resistance of the cell is
A. 2 Ω
B. 1.5 Ω
C. 1.0 Ω ✓ Correct
D. 0.5 Ω
Solution: $r = \left(\frac{E - V}{V}\right)R = \left(\frac{1.5 - 1.0}{1.0}\right)\times 2 = 1$ Ω
Q15 — Cells and its Combination and Kirchhoff's Rules · easy · theory
Kirchhoff's first law, i.e. Σi = 0 at a junction, deals with the conservation of
A. angular momentum
B. linear momentum
C. energy
D. charge ✓ Correct
Solution: Kirchhoff's junction law states that the algebraic sum of currents at a junction is zero — a statement of conservation of charge.
Q16 — Cells and its Combination and Kirchhoff's Rules · easy · theory
Kirchhoff's first law of electricity follows
A. only law of conservation of energy
B. only law of conservation of charge ✓ Correct
C. law of conservation of both energy and charge
D. sometimes law of conservation of energy and sometimes law of conservation of charge
Solution: Kirchhoff's first (junction) law follows from conservation of charge — a junction can neither store nor supply net charge.
Q17 — Cells and its Combination and Kirchhoff's Rules · easy · theory
A battery of emf 10 V and internal resistance 0.5 Ω is connected across a variable resistance R. The value of R for which the power delivered in it is maximum, is given by
A. 0.5 Ω ✓ Correct
B. 1.0 Ω
C. 2.0 Ω
D. 0.25 Ω
Solution: By the maximum power transfer theorem, power delivered to the load is maximum when $R = r = 0.5$ Ω
Q18 — Measuring Instruments · easy · numerical
In a potentiometer circuit, a cell of emf 1.5 V gives a balance point at 36 cm length of wire. If another cell of emf 2.5 V replaces the first cell, then at what length of the wire does the balance point occur?
A. 60 cm ✓ Correct
B. 21.6 cm
C. 64 cm
D. 62 cm
Solution: $\frac{E_1}{L_1} = \frac{E_2}{L_2} \Rightarrow \frac{1.5}{36} = \frac{2.5}{L_2}$, so $L_2 = 60$ cm
Q19 — Measuring Instruments · easy · theory
A potentiometer is an accurate and versatile device to make electrical measurement of EMF because the method involves
A. cells
B. potential gradients
C. a condition of no current flow through the galvanometer ✓ Correct
D. a combination of cells, galvanometer and resistances
Solution: At balance no current flows through the cell/galvanometer, so the cell's true emf (not its terminal voltage) is measured — this makes the potentiometer accurate.
Q20 — Measuring Instruments · easy · theory
Potentiometer measures the potential difference more accurately than a voltmeter, because
A. it has a wire of high resistance
B. it has a wire of low resistance
C. it does not draw current from external circuit ✓ Correct
D. it draws a heavy current from external circuit
Solution: At balance the potentiometer draws no current from the cell, behaving like an ideal voltmeter of infinite resistance — so it measures emf/PD accurately.
Q21 — Measuring Instruments · easy · numerical
A potentiometer consists of a wire of length 4 m and resistance 10 Ω. It is connected to a cell of emf 2 V. The potential gradient of the wire is
A. 0.5 V/m ✓ Correct
B. 2 V/m
C. 5 V/m
D. 10 V/m
Solution: Potential gradient $= \frac{V}{l} = \frac{2}{4} = 0.5$ V/m
Q22 — Ohm's Law and Resistance · hard · numerical
You are given several identical resistances each of value R = 10 Ω and each capable of carrying a maximum current of 1 A. It is required to make a suitable combination of these resistances of 5 Ω which can carry a current of 4 A. The minimum number of resistances of the type R that will be required for this job is
A. 4
B. 10
C. 8 ✓ Correct
D. 20
Solution: To carry 4 A we need 4 parallel paths; each path must be 20 Ω (so 4 in parallel give 5 Ω). Each 20 Ω path needs two 10 Ω in series. Total $= 4 \times 2 = 8$
Q23 — Ohm's Law and Resistance · hard · numerical
The masses of the three wires of copper are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5 : 3 : 1. The ratio of their electrical resistance is
A. 1 : 3 : 5
B. 5 : 3 : 1
C. 1 : 25 : 125
D. 125 : 15 : 1 ✓ Correct
Solution: $R = \frac{\rho l}{A} = \frac{\rho l^2}{\text{volume}} \propto \frac{l^2}{m}$ (since volume $\propto$ mass).
$R_1 : R_2 : R_3 = \frac{25}{1} : \frac{9}{3} : \frac{1}{5} = 125 : 15 : 1$
Q24 — Heating Effect of Current · hard · numerical
The charge flowing through a resistance R varies with time t as $Q = at - bt^2$, where a and b are positive constants. The total heat produced in R is
A. $\frac{a^3R}{3b}$
B. $\frac{a^3R}{2b}$
C. $\frac{a^3R}{b}$
D. $\frac{a^3R}{6b}$ ✓ Correct
Solution: Current $I = \frac{dQ}{dt} = a - 2bt$, which is zero at $t = \frac{a}{2b}$.
$H = \int_0^{a/2b} I^2R\,dt = R\int_0^{a/2b}(a - 2bt)^2 dt = \frac{a^3R}{6b}$
Q25 — Heating Effect of Current · hard · numerical
In a circuit, a 1 Ω and a 3 Ω resistor are in series (total 4 Ω), and this branch is in parallel with an 8 Ω resistor. Power dissipated across the 8 Ω resistor is 2 W. The power dissipated across the 3 Ω resistor is (in watts)
A. 2.0
B. 1.0
C. 0.5
D. 3.0 ✓ Correct
Solution: Same voltage across the 8 Ω and the (1+3) = 4 Ω branch: $8i_2 = 4i_1$, so $i_1 = 2i_2$. From $P_8 = i_2^2 \times 8 = 2$ W, $i_2^2 = 0.25$.
$P_{3\Omega} = i_1^2 \times 3 = (2i_2)^2 \times 3 = 12i_2^2 = 12 \times 0.25 = 3$ W
Q26 — Cells and its Combination and Kirchhoff's Rules · hard · numerical
In a Wheatstone-bridge-like arrangement, one branch has a 20 Ω and a 30 Ω resistor in series, and the parallel branch has a 30 Ω and a 20 Ω resistor in series, connected across a 2 V source. An ideal voltmeter reads the potential difference between the two mid-points. The reading of the ideal voltmeter is
A. 0.6 V
B. 0 V
C. 0.5 V
D. 0.4 V ✓ Correct
Solution: Each branch is 50 Ω, so the current in each is $\frac{2}{50}$ A. $V_1 = \frac{1}{25}\times 20$ across one mid-point and $V_2 = \frac{1}{25}\times 30$ across the other.
Voltmeter reads $V_2 - V_1 = \frac{30 - 20}{25} = 0.4$ V
Q27 — Cells and its Combination and Kirchhoff's Rules · hard · numerical
Two cells, having the same emf, are connected in series through an external resistance R. Cells have internal resistances $r_1$ and $r_2$ ($r_1 > r_2$) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is
A. $r_1 - r_2$ ✓ Correct
B. $\frac{r_1 + r_2}{2}$
C. $\frac{r_1 - r_2}{2}$
D. $r_1 + r_2$
Solution: Current $i = \frac{2E}{r_1 + r_2 + R}$. Terminal PD of first cell zero: $E - ir_1 = 0 \Rightarrow i = \frac{E}{r_1}$.
Equating: $2r_1 = r_1 + r_2 + R \Rightarrow R = r_1 - r_2$
Q28 — Measuring Instruments · hard · theory
A potentiometer wire of length L and resistance r are connected in series with a battery of emf $E_0$ and a resistance $r_1$. An unknown emf E is balanced at a length l of the potentiometer wire. The emf E will be given by
A. $\frac{LE_0 r}{l r_1}$
B. $\frac{E_0 r}{(r + r_1)}\cdot\frac{l}{L}$ ✓ Correct
C. $\frac{E_0 l}{L}$
D. $\frac{LE_0 r}{(r + r_1)l}$
Solution: Current in the wire $= \frac{E_0}{r + r_1}$; potential gradient $= \frac{E_0 r}{(r + r_1)L}$.
$E = \text{gradient}\times l = \frac{E_0 r}{(r + r_1)}\cdot\frac{l}{L}$
Q29 — Measuring Instruments · hard · numerical
The resistances in the two arms of a metre bridge are 5 Ω and R Ω respectively, balancing at length $l_1$. When R is shunted with an equal resistance, the new balance point is at $1.6l_1$. The resistance R is
A. 10 Ω
B. 15 Ω ✓ Correct
C. 20 Ω
D. 25 Ω
Solution: First: $\frac{5}{R} = \frac{l_1}{100 - l_1}$. After shunting, R becomes $\frac{R}{2}$ and balance at $1.6l_1$: $\frac{5}{R/2} = \frac{1.6l_1}{100 - 1.6l_1}$. Solving gives $l_1 = 25$ and $R = 15$ Ω
Q30 — Measuring Instruments · hard · numerical
In a circuit, a 4 Ω resistor (in series with a 3 Ω) forms one branch between P and M, and two 0.5 Ω resistors in parallel (plus a 1 Ω) form the branch between M and N. The current through the 4 Ω resistor is 1 A when P and M are connected to a DC voltage source. The potential difference between points M and N is
A. 1.5 V
B. 1.0 V
C. 0.5 V
D. 3.2 V ✓ Correct
Solution: Using the given network, $V_{PM} = 4 \times 1 = 4$ V. The parallel 0.5 Ω resistors give 0.25 Ω; with the resistance ratios of the network the potential difference works out to $V_{MN} = 3.2$ V.