Kirchhoff's Laws — NEET Physics MCQs with Solutions
Free NEET Physics Kirchhoff's Laws MCQs with step-by-step solutions (39 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
A cell of e.m.f. $E$ is connected with an external resistance $R$, then p.d. across cell is $V$. The internal resistance of cell will be
A. $\dfrac{(E - V)R}{V}$ ✓ Correct
B. $\dfrac{(E - V)R}{E}$
C. $\dfrac{(V - E)R}{E}$
D. $\dfrac{(V - E)R}{V}$
Solution: The current is $i = V/R$, and the lost volts are $E - V = ir$. So $r = (E - V)R/V$.
Q2 — Kirchhoff's Laws · easy · theory
Two cells, e.m.f. of each is $E$ and internal resistance $r$ are connected in parallel between the resistance $R$. The maximum energy given to the resistor will be, only when
A. $R = r/2$ ✓ Correct
B. $R = r$
C. $R = 2r$
D. $R = 0$
Solution: Two such cells in parallel behave as a single cell of e.m.f. $E$ and internal resistance $r/2$. Maximum power is delivered when the external resistance equals this internal resistance, i.e. $R = r/2$.
Q3 — Kirchhoff's Laws · easy · theory
The terminal potential difference of a cell when short-circuited is ($E$ = E.M.F. of the cell)
A. $E/3$
B. $E/2$
C. $E$
D. Zero ✓ Correct
Solution: On short circuit the external resistance is zero, so $V = iR = 0$ — the whole e.m.f. is dropped across the internal resistance.
Q4 — Kirchhoff's Laws · easy · numerical
A 50 V battery is connected across a 10 ohm resistor. The current is 4.5 amperes. The internal resistance of the battery is
A. 1.1 ohm ✓ Correct
B. 0.5 ohm
C. Zero
D. 5.0 ohm
Solution: From $E = i(R + r)$: $50 = 4.5(10 + r)$, so $10 + r = 11.11$ and $r \approx 1.1\ \Omega$.
Q5 — Kirchhoff's Laws · easy · numerical
A new flashlight cell of e.m.f. 1.5 volts gives a current of 15 amps, when connected directly to an ammeter of resistance $0.04\ \Omega$. The internal resistance of cell is
A. $0.10\ \Omega$
B. $10\ \Omega$
C. $0.04\ \Omega$
D. $0.06\ \Omega$ ✓ Correct
Solution: From $E = i(R + r)$: $1.5 = 15(0.04 + r)$, so $0.04 + r = 0.1$ and $r = 0.06\ \Omega$.
Q6 — Kirchhoff's Laws · easy · theory
When cells are connected in parallel, then
A. The current decreases
B. The e.m.f. increases
C. The current increases ✓ Correct
D. The e.m.f. decreases
Solution: Connecting cells in parallel keeps the e.m.f. the same but lowers the total internal resistance, so the circuit current increases.
Q7 — Kirchhoff's Laws · easy · theory
To get the maximum current from a parallel combination of $n$ identical cells each of internal resistance $r$ in an external resistance $R$, when
A. $R = r$
B. None of these
C. $R \ll r$ ✓ Correct
D. $R \gg r$
Solution: A parallel grouping is worthwhile when the internal resistance dominates, i.e. when the external resistance is much smaller than the internal resistance of a cell ($R \ll r$).
Q8 — Kirchhoff's Laws · easy · numerical
Two identical cells send the same current in $2\ \Omega$ resistance, whether connected in series or in parallel. The internal resistance of the cell should be
A. $2.5\ \Omega$
B. $\dfrac{1}{2}\ \Omega$
C. $2\ \Omega$ ✓ Correct
D. $1\ \Omega$
Solution: In series $i = \dfrac{2E}{2r + R}$ and in parallel $i = \dfrac{E}{r/2 + R}$. Equating gives $R = r$, so $r = 2\ \Omega$.
Q9 — Kirchhoff's Laws · easy · numerical
A torch battery consisting of two cells of 1.45 volts and an internal resistance $0.15\ \Omega$, each cell sending currents through the filament of the lamps having resistance 1.5 ohms. The value of current will be
A. 16.11 amp
B. 0.1611 amp
C. 2.6 amp
D. 1.611 amp ✓ Correct
Solution: The cells are in series: $i = \dfrac{2 \times 1.45}{1.5 + 2 \times 0.15} = \dfrac{2.9}{1.8} \approx 1.611\ A$.
Q10 — Kirchhoff's Laws · easy · numerical
The electromotive force of a primary cell is 2 volts. When it is short-circuited it gives a current of 4 amperes. Its internal resistance in ohms is
A. 0.5 ✓ Correct
B. 5.0
C. 2.0
D. 8.0
Solution: On short circuit $i = E/r$, so $r = E/i = 2/4 = 0.5\ \Omega$.
Q11 — Kirchhoff's Laws · easy · numerical
A current of two amperes is flowing through a cell of e.m.f. 5 volts and internal resistance 0.5 ohm from negative to positive electrode. If the potential of negative electrode is 10 V, the potential of positive electrode will be
A. 15 V
B. 5 V
C. 16 V
D. 14 V ✓ Correct
Solution: The terminal potential difference is $V = E - ir = 5 - 2 \times 0.5 = 4\ V$, so the positive electrode is at $10 + 4 = 14\ V$.
Q12 — Kirchhoff's Laws · easy · theory
Emf is most closely related to
A. Mechanical force
B. Magnetic field
C. Potential difference ✓ Correct
D. Electric field
Solution: E.m.f. is the work done per unit charge by the source, so like potential difference it is measured in volts.
Q13 — Kirchhoff's Laws · easy · numerical
For driving a current of 2 A for 6 minutes in a circuit, 1000 J of work is to be done. The e.m.f. of the source in the circuit is
A. 1.68 V
B. 1.38 V ✓ Correct
C. 3.10 V
D. 2.04 V
Solution: Charge $q = it = 2 \times 360 = 720\ C$, so $E = W/q = 1000/720 \approx 1.38\ V$.
Q14 — Kirchhoff's Laws · easy · theory
Four identical cells each having an electromotive force (e.m.f.) of 12 V, are connected in parallel. The resultant electromotive force (e.m.f.) of the combination is
A. 12 V ✓ Correct
B. 48 V
C. 3 V
D. 4 V
Solution: Identical cells in parallel give the same e.m.f. as a single cell; only the internal resistance drops. So the resultant e.m.f. is 12 V.
Q15 — Kirchhoff's Laws · easy · theory
Electromotive force is the force, which is able to maintain a constant
A. Resistance
B. Potential difference ✓ Correct
C. Current
D. Power
Solution: The source of e.m.f. does work on the charges to keep a steady potential difference across its terminals.
Q16 — Kirchhoff's Laws · easy · numerical
A cell of emf 6 V and resistance 0.5 ohm is short circuited. The current in the cell is
A. 24 amp
B. 12 amp ✓ Correct
C. 3 amp
D. 6 amp
Solution: On short circuit $i = E/r = 6/0.5 = 12\ A$.
Q17 — Kirchhoff's Laws · easy · numerical
A storage cell is charged by 5 amp D.C. for 18 hours. Its strength after charging will be
A. 90 AH ✓ Correct
B. 18 AH
C. 15 AH
D. 5 AH
Solution: Capacity $= $ current $\times$ time $= 5 \times 18 = 90$ ampere-hours.
Q18 — Kirchhoff's Laws · easy · numerical
When a resistor of $11\ \Omega$ is connected in series with an electric cell, the current flowing in it is 0.5 A. Instead, when a resistor of $5\ \Omega$ is connected to the same electric cell in series, the current increases by 0.4 A. The internal resistance of the cell is
A. $2\ \Omega$
B. $2.5\ \Omega$ ✓ Correct
C. $1.5\ \Omega$
D. $3.5\ \Omega$
Solution: The e.m.f. is the same in both cases: $0.5(11 + r) = 0.9(5 + r)$, giving $5.5 + 0.5r = 4.5 + 0.9r$, so $r = 2.5\ \Omega$.
Q19 — Kirchhoff's Laws · easy · theory
The internal resistance of a cell is the resistance of
A. Electrodes of the cell
B. Material used in the cell
C. Vessel of the cell
D. Electrolyte used in the cell ✓ Correct
Solution: Inside the cell the current is carried by ions through the electrolyte, and it is the electrolyte that opposes this flow.
Q20 — Kirchhoff's Laws · easy · numerical
How much work is required to carry a $6\ \mu C$ charge from the negative terminal to the positive terminal of a 9 V battery
A. $54 \times 10^{-6}\ J$ ✓ Correct
B. $54 \times 10^{-12}\ J$
C. $54 \times 10^{-9}\ J$
D. $54 \times 10^{-3}\ J$
Solution: $W = qV = 6 \times 10^{-6} \times 9 = 54 \times 10^{-6}\ J$.
Q21 — Kirchhoff's Laws · easy · theory
Kirchoff's I law and II law of current, proves the
A. Conservation of current and energy
B. Conservation of charge and energy ✓ Correct
C. Conservation of mass and charge
D. None of these
Solution: The junction rule follows from conservation of charge, and the loop rule from conservation of energy.
Q22 — Kirchhoff's Laws · easy · theory
A capacitor is connected to a cell of emf $E$ having some internal resistance $r$. The potential difference across the
A. Cell is $E$ ✓ Correct
B. Capacitor is $< E$
C. Capacitor is $> E$
D. Cell is $< E$
Solution: In the steady state no current flows through a capacitor branch, so there is no drop across $r$ and the terminal p.d. of the cell (and hence across the capacitor) equals $E$.
Q23 — Kirchhoff's Laws · easy · theory
To draw maximum current from a combination of cells, how should the cells be grouped
A. Series
B. Mixed
C. Parallel
D. Depends upon the relative values of external and internal resistance ✓ Correct
Solution: Series grouping suits $R \gg r$, parallel suits $R \ll r$ and mixed grouping suits comparable values — so the best arrangement depends on how $R$ compares with $r$.
Q24 — Kirchhoff's Laws · easy · numerical
The $n$ rows each containing $m$ cells in series are joined in parallel. Maximum current is taken from this combination across an external resistance of $3\ \Omega$ resistance. If the total number of cells used are 24 and internal resistance of each cell is $0.5\ \Omega$ then
A. $m = 6, n = 4$
B. $m = 2, n = 12$
C. $m = 8, n = 3$
D. $m = 12, n = 2$ ✓ Correct
Solution: Maximum current requires $R = mr/n$, so $3 = 0.5m/n$, i.e. $m = 6n$. With $mn = 24$ we get $6n^{2} = 24$, so $n = 2$ and $m = 12$.
Q25 — Kirchhoff's Laws · easy · numerical
The magnitude and direction of the current in the circuit shown will be
A. $\dfrac{7}{3}\ A$ from $a$ to $b$ through $e$
B. $1\ A$ from $b$ to $a$ through $e$
C. $1\ A$ from $a$ to $b$ through $e$ ✓ Correct
D. $\dfrac{7}{3}\ A$ from $b$ to $a$ through $e$
Solution: The two cells oppose, so the net e.m.f. is $10 - 4 = 6\ V$ and the total resistance is $1 + 2 + 3 = 6\ \Omega$. Hence $i = 1\ A$, flowing from $a$ to $b$ through $e$ since the 10 V cell dominates.
Q26 — Kirchhoff's Laws · easy · numerical
The figure below shows currents in a part of electric circuit. The current $i$ is
A. 1.7 amp ✓ Correct
B. 1.3 amp
C. 1 amp
D. 3.7 amp
Solution: By Kirchhoff's junction rule the current in equals the current out: $2 + 2 = 1 + 1.3 + i$, so $i = 1.7\ A$.
Q27 — Kirchhoff's Laws · easy · theory
In the circuit shown, A and V are ideal ammeter and voltmeter respectively. Reading of the voltmeter will be
A. $0.5\ V$
B. $1\ V$
C. Zero ✓ Correct
D. $2\ V$
Solution: The ideal ammeter has zero resistance, so it short-circuits the branch the voltmeter is across. With no p.d. across it, the voltmeter reads zero.
Q28 — Kirchhoff's Laws · easy · numerical
The figure shows a network of currents. The magnitude of currents is shown here. The current $i$ will be
A. $3\ A$
B. $13\ A$
C. $23\ A$ ✓ Correct
D. $-3\ A$
Solution: Applying Kirchhoff's current law to the network — the total current entering must equal the total leaving — gives $i = 23\ A$.
Q29 — Kirchhoff's Laws · easy · theory
A battery of e.m.f. $E$ and internal resistance $r$ is connected to a variable resistor $R$ as shown here. Which one of the following is true
A. Current in the circuit is maximum when $R \gg r$
B. Potential difference across the terminals of the battery is maximum when $R = r$
C. Current in the circuit is maximum when $R = r$
D. Power delivered to the resistor is maximum when $R = r$ ✓ Correct
Solution: Maximum power transfer occurs when the external resistance equals the internal resistance, i.e. $R = r$. (The current is largest when $R$ is smallest, and the terminal p.d. is largest when $R$ is largest.)
Q30 — Kirchhoff's Laws · easy · numerical
Consider the circuit given here with the following parameters: E.M.F. of the cell = 12 V, internal resistance of the cell $= 2\ \Omega$, resistance $R = 4\ \Omega$. Which one of the following statements is true
A. Power output in is $= 8\ W$
B. Rate of energy loss in the source is $= 8\ W$ ✓ Correct
C. Rate of energy conversion in the source is $16\ W$
D. Potential drop across $R$ is $= 16\ V$
Solution: The current is $i = 12/(4 + 2) = 2\ A$, so the energy lost inside the source is $i^{2}r = 2^{2} \times 2 = 8\ W$.