Kirchhoff's Laws — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Kirchhoff's Laws MCQs with step-by-step solutions (21 questions). Part of Current Electricity. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — 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.
Q2 — 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.
Q3 — Kirchhoff's Laws · medium · theory
Kirchhoff's junction rule is a consequence of the conservation of:
A. Charge ✓ Correct
B. Energy
C. Mass
D. Momentum
Solution: The rule simply says that charge is neither created nor destroyed at a junction.
Q4 — Kirchhoff's Laws · medium · theory
Kirchhoff's loop rule is a consequence of the conservation of:
A. Angular momentum
B. Charge
C. Energy ✓ Correct
D. Momentum
Solution: A unit charge carried once around a loop must return with the same energy, so the net work done on it is zero.
Q5 — Kirchhoff's Laws · medium · theory
In applying Kirchhoff's loop rule, the potential change across a resistor traversed in the direction of the current is taken as:
A. Positive, a rise of $IR$
B. Zero
C. Equal to the EMF
D. Negative, a drop of $IR$ ✓ Correct
Solution: Moving with the current goes from higher to lower potential, so the change is $-IR$.
Q6 — Kirchhoff's Laws · medium · theory
Kirchhoff's laws can be applied to:
A. Any network, including those containing non-ohmic elements ✓ Correct
B. Only purely resistive networks
C. Only networks with a single cell
D. Only balanced bridge circuits
Solution: Both rules follow from conservation laws rather than from Ohm's law, so they hold for any circuit element.
Q7 — 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.
Q8 — 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.
Q9 — Kirchhoff's Laws · medium · numerical
At a junction, currents of $2\text{ A}$ and $3\text{ A}$ flow in and $1\text{ A}$ flows out along one branch. The current in the remaining branch is:
A. $4\text{ A}$ flowing out ✓ Correct
B. $2\text{ A}$ flowing out
C. $6\text{ A}$ flowing out
D. $4\text{ A}$ flowing in
Solution: Total in $= 5\text{ A}$ and one branch takes $1\text{ A}$ out, so the remaining branch must carry $5 - 1 = 4\text{ A}$ out.
Q10 — 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 $I = 2\text{ A}$.
Q11 — 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 $I_{out} = 5 + 3 = 8\text{ A}$.
Q12 — Kirchhoff's Laws · medium · numerical
Two cells of $12\text{ V}$ and $6\text{ V}$ are connected in opposition through a $3\,\Omega$ resistor of negligible internal resistance. The current is:
A. $0.5\text{ A}$
B. $3\text{ A}$
C. $2\text{ A}$ ✓ Correct
D. $6\text{ A}$
Solution: The net EMF is $12 - 6 = 6\text{ V}$, so $I = \dfrac{6}{3} = 2\text{ A}$.
Q13 — Kirchhoff's Laws · medium · numerical
Two cells of $6\text{ V}$ and $4\text{ V}$ are connected so as to aid each other through a $5\,\Omega$ resistor. The current is:
A. $0.4\text{ A}$
B. $10\text{ A}$
C. $2\text{ A}$ ✓ Correct
D. $1.2\text{ A}$
Solution: The EMFs add to $10\text{ V}$, so $I = \dfrac{10}{5} = 2\text{ A}$.
Q14 — Kirchhoff's Laws · medium · numerical
A $12\text{ V}$ supply drives current through $4\,\Omega$ and $2\,\Omega$ in series. The potential difference across the $4\,\Omega$ resistor is:
A. $6\text{ V}$
B. $12\text{ V}$
C. $4\text{ V}$
D. $8\text{ V}$ ✓ Correct
Solution: $I = \dfrac{12}{6} = 2\text{ A}$, so $V_{4\Omega} = 2 \times 4 = 8\text{ V}$.
Q15 — 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 $I = 1\text{ A}$.
Q16 — Kirchhoff's Laws · medium · numerical
Cells of EMF $10\text{ V}$ and $4\text{ V}$ oppose each other in a loop containing a $2\,\Omega$ resistor of negligible internal resistance. The current is:
A. $2\text{ A}$
B. $3\text{ A}$ ✓ Correct
C. $7\text{ A}$
D. $5\text{ A}$
Solution: Net EMF $= 10 - 4 = 6\text{ V}$, so $I = \dfrac{6}{2} = 3\text{ A}$.
Q17 — Kirchhoff's Laws · medium · numerical
A current of $2\text{ A}$ flows through a $3\,\Omega$ resistor in a network. The power dissipated in it is:
A. $6\text{ W}$
B. $18\text{ W}$
C. $12\text{ W}$ ✓ Correct
D. $1.5\text{ W}$
Solution: $P = I^2R = 4 \times 3 = 12\text{ W}$.
Q18 — Kirchhoff's Laws · easy · numerical
A current of $1.5\text{ A}$ flows through a $5\,\Omega$ resistor. The potential drop across it is:
A. $0.3\text{ V}$
B. $3.3\text{ V}$
C. $7.5\text{ V}$ ✓ Correct
D. $11.25\text{ V}$
Solution: $V = IR = 1.5 \times 5 = 7.5\text{ V}$.
Q19 — Kirchhoff's Laws · easy · numerical
A $20\text{ V}$ source drives current through two $5\,\Omega$ resistors in series. The current is:
A. $8\text{ A}$
B. $2\text{ A}$ ✓ Correct
C. $1\text{ A}$
D. $4\text{ A}$
Solution: $I = \dfrac{20}{5 + 5} = 2\text{ A}$.
Q20 — Kirchhoff's Laws · easy · numerical
At a junction $4\text{ A}$ flows in, while $1.5\text{ A}$ leaves by one branch. The current leaving by the second branch is:
A. $2.5\text{ A}$ ✓ Correct
B. $4\text{ A}$
C. $5.5\text{ A}$
D. $1.5\text{ A}$
Solution: The junction rule gives $4 = 1.5 + I$, so $I = 2.5\text{ A}$.
Q21 — Kirchhoff's Laws · medium · numerical
A cell of EMF $9\text{ V}$ and internal resistance $1\,\Omega$ drives current through $2\,\Omega$ and $6\,\Omega$ in series. The current is:
A. $1.5\text{ A}$
B. $1\text{ A}$ ✓ Correct
C. $0.75\text{ A}$
D. $3\text{ A}$
Solution: $I = \dfrac{9}{2 + 6 + 1} = \dfrac{9}{9} = 1\text{ A}$.