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Current Electricity — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Current Electricity MCQs with step-by-step solutions covering Electric Current & Ohm's Law, Resistance & Resistivity, Combination of Resistors & Cells, Kirchhoff's Laws, Wheatstone Bridge & Meter Bridge, Potentiometer. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Electric Current & Ohm's Law · easy · theory
Electric current is defined as:
A. The charge stored per unit potential
B. The rate of flow of charge, and it is a scalar quantity ✓ Correct
C. The rate of flow of charge, and it is a vector quantity
D. The force per unit charge
Solution: $I = \dfrac{q}{t}$, measured in ampere. Although it has a direction of flow, current adds algebraically and is treated as a scalar.
Q2 — Electric Current & Ohm's Law · easy · theory
Ohm's law states that, at constant temperature:
A. The potential difference is independent of the current
B. The resistance is proportional to the current
C. The current through a conductor is proportional to the potential difference across it ✓ Correct
D. The current is proportional to the resistance
Solution: $V = IR$ holds only while the physical conditions, especially temperature, remain unchanged.
Q3 — Electric Current & Ohm's Law · easy · theory
The direction of conventional current in a metallic conductor is:
A. Opposite to the direction of electron flow ✓ Correct
B. Undefined
C. The same as the direction of electron flow
D. Perpendicular to the electron flow
Solution: Conventional current was defined as the flow of positive charge before the electron was discovered, so it runs opposite to the actual drift of electrons.
Q4 — Electric Current & Ohm's Law · easy · theory
A conductor that does NOT obey Ohm's law is described as:
A. Superconducting
B. A perfect insulator
C. Ohmic
D. Non-ohmic, for example a semiconductor diode ✓ Correct
Solution: For a non-ohmic device the $V$-$I$ graph is not a straight line through the origin, so the resistance depends on the operating point.
Q5 — Electric Current & Ohm's Law · easy · numerical
A charge of $60\text{ C}$ flows through a conductor in $30\text{ s}$. The current is:
A. $1800\text{ A}$
B. $2\text{ A}$ ✓ Correct
C. $30\text{ A}$
D. $0.5\text{ A}$
Solution: $I = \dfrac{q}{t} = \dfrac{60}{30} = 2\text{ A}$.
Q6 — Electric Current & Ohm's Law · easy · numerical
A current of $2\text{ A}$ flows through a resistance of $5\,\Omega$. The potential difference across it is:
A. $0.4\text{ V}$
B. $2.5\text{ V}$
C. $20\text{ V}$
D. $10\text{ V}$ ✓ Correct
Solution: $V = IR = 2 \times 5 = 10\text{ V}$.
Q7 — Electric Current & Ohm's Law · easy · numerical
A potential difference of $12\text{ V}$ drives a current of $3\text{ A}$ through a conductor. Its resistance is:
A. $36\,\Omega$
B. $9\,\Omega$
C. $0.25\,\Omega$
D. $4\,\Omega$ ✓ Correct
Solution: $R = \dfrac{V}{I} = \dfrac{12}{3} = 4\,\Omega$.
Q8 — Electric Current & Ohm's Law · easy · numerical
A current of $0.5\text{ A}$ flows through a $20\,\Omega$ resistor. The potential difference across it is:
A. $0.025\text{ V}$
B. $20\text{ V}$
C. $10\text{ V}$ ✓ Correct
D. $40\text{ V}$
Solution: $V = IR = 0.5 \times 20 = 10\text{ V}$.
Q9 — Electric Current & Ohm's Law · easy · numerical
An appliance draws $5\text{ A}$ from a $220\text{ V}$ supply. The power consumed is:
A. $44\text{ W}$
B. $5500\text{ W}$
C. $1100\text{ W}$ ✓ Correct
D. $225\text{ W}$
Solution: $P = VI = 220 \times 5 = 1100\text{ W}$.
Q10 — Resistance & Resistivity · easy · theory
The resistance of a uniform conductor of length $L$ and area of cross-section $A$ is:
A. $\dfrac{\rho A}{L}$
B. $\dfrac{L}{\rho A}$
C. $\dfrac{\rho L}{A}$ ✓ Correct
D. $\rho L A$
Solution: Resistance grows with length and falls with thickness; $\rho$ is the resistivity of the material.
Q11 — Resistance & Resistivity · easy · theory
The SI unit of resistivity is:
A. $\Omega/\text{m}$
B. $\Omega$
C. $\Omega\cdot\text{m}^2$
D. $\Omega\cdot\text{m}$ ✓ Correct
Solution: From $\rho = \dfrac{RA}{L}$, the units are $\dfrac{\Omega \cdot \text{m}^2}{\text{m}} = \Omega\cdot\text{m}$.
Q12 — Resistance & Resistivity · easy · theory
Electrical conductivity is defined as:
A. The reciprocal of resistivity ✓ Correct
B. The current per unit potential difference
C. The product of resistance and length
D. The reciprocal of resistance
Solution: $\sigma = \dfrac{1}{\rho}$, measured in siemens per metre.
Q13 — Resistance & Resistivity · easy · theory
Of resistance and resistivity, the quantity that depends on the dimensions of the specimen is:
A. Resistivity only
B. Resistance only ✓ Correct
C. Both of them
D. Neither of them
Solution: Cutting a wire changes its resistance but leaves the resistivity of the metal untouched.
Q14 — Resistance & Resistivity · easy · numerical
A wire of resistance $R$ is cut into two equal halves. The resistance of each half is:
A. $\dfrac{R}{4}$
B. $2R$
C. $R$
D. $\dfrac{R}{2}$ ✓ Correct
Solution: Resistance is proportional to length at fixed cross-section, so halving the length halves the resistance.
Q15 — Combination of Resistors & Cells · easy · theory
When resistors are connected in series:
A. The reciprocals of the resistances add
B. The same current flows through each and the resistances add ✓ Correct
C. The equivalent resistance is less than the smallest one
D. The same potential difference appears across each
Solution: A series chain offers only one path, so the current is common and $R_{eq} = R_1 + R_2 + \ldots$
Q16 — Combination of Resistors & Cells · easy · theory
When resistors are connected in parallel:
A. The resistances add directly
B. The same potential difference appears across each and the reciprocals of the resistances add ✓ Correct
C. The equivalent resistance exceeds the largest one
D. The same current flows through each
Solution: Each branch is connected between the same two nodes, so $\dfrac{1}{R_{eq}} = \sum\dfrac{1}{R_i}$ and the equivalent resistance is smaller than any branch.
Q17 — Combination of Resistors & Cells · easy · theory
When $n$ identical cells each of EMF $E$ are connected in series, the total EMF is:
A. $nE$ ✓ Correct
B. $E$
C. $n^2E$
D. $\dfrac{E}{n}$
Solution: The EMFs add, and so do the internal resistances, giving $nr$ in total.
Q18 — Combination of Resistors & Cells · easy · theory
The terminal voltage of a cell equals its EMF when:
A. No current is drawn from the cell ✓ Correct
B. The cell is short-circuited
C. The external resistance is zero
D. The current is maximum
Solution: With $I = 0$ there is no internal drop, so $V = E - Ir = E$.
Q19 — Combination of Resistors & Cells · easy · numerical
Two resistors of $10\,\Omega$ each are connected in parallel. The equivalent resistance is:
A. $0.2\,\Omega$
B. $20\,\Omega$
C. $10\,\Omega$
D. $5\,\Omega$ ✓ Correct
Solution: For two equal resistors in parallel, $R_{eq} = \dfrac{R}{2} = 5\,\Omega$.
Q20 — Combination of Resistors & Cells · easy · numerical
Three resistors of $6\,\Omega$ each are connected in parallel. The equivalent resistance is:
A. $18\,\Omega$
B. $2\,\Omega$ ✓ Correct
C. $3\,\Omega$
D. $6\,\Omega$
Solution: For $n$ equal resistors in parallel, $R_{eq} = \dfrac{R}{n} = \dfrac{6}{3} = 2\,\Omega$.
Q21 — Combination of Resistors & Cells · easy · numerical
Resistors of $4\,\Omega$ and $6\,\Omega$ are connected in series. The equivalent resistance is:
A. $24\,\Omega$
B. $2\,\Omega$
C. $10\,\Omega$ ✓ Correct
D. $2.4\,\Omega$
Solution: In series the resistances add: $4 + 6 = 10\,\Omega$.
Q22 — Combination of Resistors & Cells · easy · numerical
A cell of EMF $6\text{ V}$ and internal resistance $1\,\Omega$ drives a $5\,\Omega$ load. The current is:
A. $1.2\text{ A}$
B. $0.83\text{ A}$
C. $1\text{ A}$ ✓ Correct
D. $6\text{ A}$
Solution: $I = \dfrac{6}{5 + 1} = 1\text{ A}$.
Q23 — Combination of Resistors & Cells · easy · numerical
Four cells each of EMF $1.5\text{ V}$ are connected in series. The total EMF is:
A. $1.5\text{ V}$
B. $0.375\text{ V}$
C. $6\text{ V}$ ✓ Correct
D. $3\text{ V}$
Solution: EMFs in series add: $4 \times 1.5 = 6\text{ V}$.
Q24 — Kirchhoff's Laws · easy · theory
Kirchhoff's junction rule states that:
A. The algebraic sum of potential differences around a loop is zero
B. The algebraic sum of currents at a junction is zero ✓ Correct
C. The resistance of a junction is zero
D. The current is proportional to the voltage
Solution: Charge cannot accumulate at a junction in a steady current, so whatever flows in must flow out.
Q25 — Kirchhoff's Laws · easy · theory
Kirchhoff's loop rule states that:
A. The algebraic sum of the changes in potential around any closed loop is zero ✓ Correct
B. The sum of currents in a loop is zero
C. The EMF equals the current times the resistance always
D. The potential is the same everywhere in a loop
Solution: Returning to the starting point must restore the original potential, since electrostatic potential is single-valued.
Q26 — Kirchhoff's Laws · easy · theory
Kirchhoff's first and second laws are also known respectively as the:
A. Ohm rule and power rule
B. Junction rule and loop rule ✓ Correct
C. Current rule and charge rule
D. Loop rule and junction rule
Solution: The first concerns currents meeting at a node; the second concerns potentials around a closed mesh.
Q27 — Kirchhoff's Laws · easy · theory
At a junction where three wires meet, if two carry current towards the junction, the third must:
A. Carry the sum of the two currents away from the junction ✓ Correct
B. Carry the product of the two currents
C. Carry no current
D. Carry the difference of the two currents towards the junction
Solution: The junction rule requires the outgoing current to balance the total incoming current.
Q28 — Kirchhoff's Laws · easy · numerical
A $10\text{ V}$ cell of negligible internal resistance drives current through $2\,\Omega$ and $3\,\Omega$ in series. The current is:
A. $5\text{ A}$
B. $2\text{ A}$ ✓ Correct
C. $3.3\text{ A}$
D. $1\text{ A}$
Solution: By the loop rule, $10 = I(2 + 3)$, so $I = 2\text{ A}$.
Q29 — Kirchhoff's Laws · easy · numerical
At a junction, currents of $5\text{ A}$ and $3\text{ A}$ flow in. The current flowing out is:
A. $15\text{ A}$
B. $1.6\text{ A}$
C. $2\text{ A}$
D. $8\text{ A}$ ✓ Correct
Solution: The junction rule gives $I_{out} = 5 + 3 = 8\text{ A}$.
Q30 — Kirchhoff's Laws · easy · numerical
At a junction $3\text{ A}$ flows in while $2\text{ A}$ flows out along one branch. The current in the other outgoing branch is:
A. $1\text{ A}$ ✓ Correct
B. $5\text{ A}$
C. $3\text{ A}$
D. $2\text{ A}$
Solution: By the junction rule, $3 = 2 + I$, so $I = 1\text{ A}$.