Cells and its Combination and Kirchhoff's Rules — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Cells and its Combination and Kirchhoff's Rules MCQs with step-by-step solutions (27 questions). Part of Current Electricity. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
In a single loop, a 2 V cell (internal resistance 1 Ω), a 4 V cell (internal resistance 1 Ω) and a 4 Ω resistor are connected in series (the two cells aiding). The current I in the circuit will be
A. 0.75 A
B. 1 A ✓ Correct
C. 1.5 A
D. 0.5 A
Solution: Applying KVL: $2 + 4 = I(1 + 1 + 4)$, so $6 = 6I \Rightarrow I = 1$ A
Q2 — Cells and its Combination and Kirchhoff's Rules · medium · theory
For a circuit with three branches (currents $i_1$, $i_2$, $i_3$; resistors $R_1$, $R_2$; emfs $E_1$, $E_2$, $E_3$), the Kirchhoff's loop rule for the loop BCDEB is given by the equation
A. $-i_2R_2 + E_2 - E_3 + i_3R_1 = 0$
B. $i_2R_2 + E_2 - E_3 - i_3R_1 = 0$ ✓ Correct
C. $i_2R_2 + E_2 + E_3 + i_3R_1 = 0$
D. $-i_2R_2 + E_2 + E_3 + i_3R_1 = 0$
Solution: Traversing the loop BCDEB and summing potential changes to zero gives $i_2R_2 + E_2 - E_3 - i_3R_1 = 0$
Q3 — 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
Q4 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then, the current drawn from battery becomes 10I. The value of 'n' is
A. 20
B. 11
C. 10 ✓ Correct
D. 9
Solution: Series: $I = \frac{E}{nR + R} = \frac{E}{R(n+1)}$. Parallel: $10I = \frac{E}{\frac{R}{n} + R} = \frac{nE}{R(n+1)}$.
Dividing: $10 = n \Rightarrow n = 10$
Q5 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
In a circuit, points A and B are separated by a 2 Ω resistor (carrying 2 A), a 3 V cell and a 1 Ω resistor in series. The potential difference $(V_A - V_B)$ is
A. −3 V
B. +3 V
C. +6 V
D. +9 V ✓ Correct
Solution: Adding the potential changes from A to B: $V_A + 3 = V_B + 2 \times 2 + 2 \times 1$, so $V_A - V_B = 4 + 2 + 3 = 9$ V
Q6 — 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
Q7 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
In a circuit, cells A and B (negligible internal resistance) are connected through a galvanometer and resistors, with $R_1 = 500$ Ω in the A-branch and $R = 100$ Ω. For $V_A = 12$ V, the galvanometer shows no deflection. The value of $V_B$ is
A. 4 V
B. 2 V ✓ Correct
C. 12 V
D. 6 V
Solution: No deflection means the current $I = \frac{12}{500 + 100} = 2 \times 10^{-2}$ A flows through the $R_1$-R chain, and $V_B$ equals the drop across R: $V_B = 100 \times 2 \times 10^{-2} = 2$ V
Q8 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
A current of 2 A flows through a 2 Ω resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 Ω resistor. The internal resistance of the battery is
A. $\frac{1}{3}$ Ω ✓ Correct
B. $\frac{1}{4}$ Ω
C. 1 Ω
D. 0.5 Ω
Solution: $E = 2(2 + r)$ and $E = 0.5(9 + r)$. Equating: $2(2 + r) = 0.5(9 + r) \Rightarrow r = \frac{1}{3}$ Ω
Q9 — 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.
Q10 — Cells and its Combination and Kirchhoff's Rules · medium · theory
In a mesh with an emf $\varepsilon_1$, resistor $R$ carrying current $(i_1 + i_2)$, and a resistor $r_1$ carrying current $i_1$, which of the following is a correct loop equation?
A. $\varepsilon_1 - (i_1 + i_2)R - i_1r_1 = 0$ ✓ Correct
B. $\varepsilon_2 - i_2r_2 - \varepsilon_1 - i_1r_1 = 0$
C. $-\varepsilon_2 - (i_1 + i_2)R + i_2r_2 = 0$
D. $\varepsilon_1 - (i_1 + i_2)R + i_1r_1 = 0$
Solution: Summing the potential changes around the mesh to zero: $\varepsilon_1 - (i_1 + i_2)R - i_1r_1 = 0$
Q11 — Cells and its Combination and Kirchhoff's Rules · medium · theory
A student measures the terminal potential difference (V) of a cell (of emf ε and internal resistance r) as a function of the current (I) flowing through it. The slope and intercept of the graph between V and I, respectively, equal
A. ε and −r
B. −r and ε ✓ Correct
C. r and −ε
D. −ε and r
Solution: $V = \varepsilon - Ir$. Compared with $y = mx + c$: slope $= -r$, intercept $= \varepsilon$
Q12 — 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: $2r_1 = r_1 + r_2 + R \Rightarrow R = r_1 - r_2$
Q13 — Cells and its Combination and Kirchhoff's Rules · medium · theory
In a circuit, a 4 Ω and a 4 Ω resistor are in series in the upper branch, and a 1 Ω and a 3 Ω resistor are in series in the lower branch, both across a source V, with points A and B at the two mid-points. If a conducting wire is connected between points A and B, the current in this wire will
A. flow from A to B
B. flow in the direction which will be decided by the value of V
C. be zero
D. flow from B to A ✓ Correct
Solution: The potential at A (mid-point of the 4-4 branch) is $\frac{V}{2}$, while B (mid-point of the 1-3 branch) is at $\frac{3V}{4}$. Since B is at higher potential, current flows from B to A.
Q14 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
Two batteries, one of emf 18 V and internal resistance 2 Ω and the other of emf 12 V and internal resistance 1 Ω, are connected so that they oppose each other, with a voltmeter across them. The voltmeter V will record a reading of
A. 15 V
B. 30 V
C. 14 V ✓ Correct
D. 18 V
Solution: Net emf $= 18 - 12 = 6$ V, total resistance $= 3$ Ω, so $I = 2$ A. Terminal voltage $= 18 - 2 \times 2 = 14$ V (equivalently $12 + 2 \times 1 = 14$ V).
Q15 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
A 6 V battery is connected to the terminals of a 3 m long wire of uniform thickness and resistance of 100 Ω. The difference of potential between two points on the wire separated by a distance of 50 cm will be
A. 2 V
B. 3 V
C. 1 V ✓ Correct
D. 1.5 V
Solution: Current $= \frac{6}{100} = 0.06$ A. Resistance of 50 cm $= 100 \times \frac{0.5}{3} = \frac{50}{3}$ Ω.
$V = 0.06 \times \frac{50}{3} = 1$ V
Q16 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
A battery is charged at a potential of 15 V for 8 h when the current flowing is 10 A. The battery on discharge supplies a current of 5 A for 15 h. The mean terminal voltage during discharge is 14 V. The watt-hour efficiency of the battery is
A. 82.5%
B. 80%
C. 90%
D. 87.5% ✓ Correct
Solution: Input $= 15 \times 10 \times 8 = 1200$ Wh; output $= 14 \times 5 \times 15 = 1050$ Wh.
Efficiency $= \frac{1050}{1200} = 87.5\%$
Q17 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
For a cell, the terminal potential difference is 2.2 V when the circuit is open and reduces to 1.8 V when the cell is connected to a resistance R = 5 Ω. The internal resistance (r) of the cell is
A. $\frac{10}{9}$ Ω ✓ Correct
B. $\frac{9}{10}$ Ω
C. $\frac{11}{9}$ Ω
D. $\frac{5}{9}$ Ω
Solution: $r = \left(\frac{E}{V} - 1\right)R = \left(\frac{2.2}{1.8} - 1\right)\times 5 = \frac{2}{9}\times 5 = \frac{10}{9}$ Ω
Q18 — 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
Q19 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
In a circuit, $R_A = 3$ Ω is in parallel with the series combination of $R_B = 6$ Ω and $R_C = 6$ Ω, connected across a 4.8 V source. The current i drawn from the source is
A. 1.6 A
B. 2 A ✓ Correct
C. 0.32 A
D. 3.2 A
Solution: $R_B + R_C = 12$ Ω; this in parallel with $R_A = 3$ Ω gives $\frac{12 \times 3}{15} = 2.4$ Ω.
$i = \frac{4.8}{2.4} = 2$ A
Q20 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
In a circuit, two 3 Ω resistors (arms AC and CB) are in series, and this combination is in parallel with a 3 Ω resistor (arm AB), across a 2 V source. The current i drawn from the source is
A. 1 A ✓ Correct
B. $\frac{2}{3}$ A
C. $\frac{2}{9}$ A
D. $\frac{1}{8}$ A
Solution: AC + CB $= 6$ Ω, in parallel with AB (3 Ω): $\frac{6 \times 3}{9} = 2$ Ω.
$i = \frac{2}{2} = 1$ A
Q21 — 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.
Q22 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
In a circuit, a 4 Ω resistor is in parallel with an 8 Ω resistor, and this combination is in series with a 2 Ω resistor (between B and C). If the current in the 4 Ω resistance is 1.2 A, the potential difference between B and C is
A. 3.6 V ✓ Correct
B. 6.3 V
C. 1.8 V
D. 2.4 V
Solution: PD across the 4 Ω (and hence the 8 Ω) = $4 \times 1.2 = 4.8$ V. Current in 8 Ω $= \frac{4.8}{8} = 0.6$ A. Total $= 1.2 + 0.6 = 1.8$ A.
$V_{BC} = 1.8 \times 2 = 3.6$ V
Q23 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
In a circuit, a 3 Ω resistor is in parallel with a 6 Ω resistor, and this combination is in series with a 4 Ω resistor. If the current through the 3 Ω resistor is 0.8 A, then the potential drop through the 4 Ω resistor is
A. 9.6 V
B. 2.6 V
C. 4.8 V ✓ Correct
D. 1.2 V
Solution: PD across 3 Ω (and 6 Ω) $= 3 \times 0.8 = 2.4$ V. Current in 6 Ω $= 0.4$ A. Total current $= 0.8 + 0.4 = 1.2$ A.
Drop across 4 Ω $= 1.2 \times 4 = 4.8$ V
Q24 — 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.
Q25 — 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
Q26 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
Two identical batteries each of emf 2 V and internal resistance 1 Ω are available to produce heat in an external resistance by passing a current through it. The maximum power that can be developed across R using these batteries is
A. 3.2 W
B. 2 W ✓ Correct
C. 1.28 W
D. $\frac{8}{9}$ W
Solution: Connect the two cells in series: $E_{eq} = 4$ V, $r_{eq} = 2$ Ω. Maximum power when $R = 2$ Ω: $P = \left(\frac{4}{2 + 2}\right)^2 \times 2 = 2$ W
Q27 — Cells and its Combination and Kirchhoff's Rules · medium · numerical
Two batteries of emf 4 V and 8 V with internal resistances 1 Ω and 2 Ω are connected in a circuit with a resistance of 9 Ω (the batteries opposing). The current and potential difference between the points P and Q are
A. $\frac{1}{3}$ A and 3 V ✓ Correct
B. $\frac{1}{6}$ A and 4 V
C. $\frac{1}{9}$ A and 9 V
D. $\frac{1}{12}$ A and 12 V
Solution: Applying KVL: $-2i + 8 - 4 - i - 9i = 0 \Rightarrow i = \frac{1}{3}$ A.
PD across PQ $= \frac{1}{3}\times 9 = 3$ V