Combination of Resistors & Cells — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Combination of Resistors & Cells 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 — 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 $R_{eq} = R_1 + R_2 + \ldots$
Q2 — 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.
Q3 — Combination of Resistors & Cells · medium · theory
The electromotive force of a cell is:
A. The potential difference across its terminals when no current is drawn ✓ Correct
B. Always equal to the terminal voltage
C. The potential difference when the cell is short-circuited
D. The current it can supply
Solution: On open circuit no current flows, so there is no drop across the internal resistance and $V = E$.
Q4 — Combination of Resistors & Cells · medium · theory
The terminal potential difference of a cell of EMF $E$ and internal resistance $r$ delivering current $I$ is:
A. $E + Ir$
B. $Ir - E$
C. $\dfrac{E}{Ir}$
D. $E - Ir$ ✓ Correct
Solution: Part of the EMF is spent driving the current through the cell itself, so the terminal voltage falls below $E$ as the current grows.
Q5 — 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.
Q6 — Combination of Resistors & Cells · medium · theory
When $n$ identical cells each of EMF $E$ and internal resistance $r$ are connected in parallel, the equivalent EMF and internal resistance are:
A. $E$ and $\dfrac{r}{n}$ ✓ Correct
B. $nE$ and $\dfrac{r}{n}$
C. $\dfrac{E}{n}$ and $nr$
D. $nE$ and $nr$
Solution: Parallel identical cells behave as a single cell of the same EMF but with the internal resistances in parallel.
Q7 — Combination of Resistors & Cells · medium · theory
Maximum power is transferred from a cell of internal resistance $r$ to an external load $R$ when:
A. $R = \dfrac{r}{2}$
B. $R \to \infty$
C. $R = 2r$
D. $R = r$ ✓ Correct
Solution: This is the maximum power transfer theorem; at this matching the efficiency is only $50\%$.
Q8 — 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 $V = E - Ir = E$.
Q9 — 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, $R_{eq} = \dfrac{R}{2} = 5\,\Omega$.
Q10 — 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, $R_{eq} = \dfrac{R}{n} = \dfrac{6}{3} = 2\,\Omega$.
Q11 — 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: $4 + 6 = 10\,\Omega$.
Q12 — Combination of Resistors & Cells · medium · numerical
Resistors of $4\,\Omega$ and $6\,\Omega$ are connected in parallel. The equivalent resistance is:
A. $5\,\Omega$
B. $0.42\,\Omega$
C. $10\,\Omega$
D. $2.4\,\Omega$ ✓ Correct
Solution: $R_{eq} = \dfrac{4 \times 6}{4 + 6} = \dfrac{24}{10} = 2.4\,\Omega$.
Q13 — Combination of Resistors & Cells · hard · numerical
A uniform wire of resistance $20\,\Omega$ is bent into a closed circle. The resistance between two diametrically opposite points is:
A. $5\,\Omega$ ✓ Correct
B. $2.5\,\Omega$
C. $10\,\Omega$
D. $20\,\Omega$
Solution: The circle forms two semicircular arcs of $10\,\Omega$ each in parallel, giving $\dfrac{10 \times 10}{20} = 5\,\Omega$.
Q14 — Combination of Resistors & Cells · medium · numerical
A cell of EMF $2\text{ V}$ and internal resistance $0.5\,\Omega$ is connected to an external resistance of $1.5\,\Omega$. The current is:
A. $1.33\text{ A}$
B. $0.5\text{ A}$
C. $1\text{ A}$ ✓ Correct
D. $4\text{ A}$
Solution: $I = \dfrac{E}{R + r} = \dfrac{2}{1.5 + 0.5} = 1\text{ A}$.
Q15 — 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: $I = \dfrac{6}{5 + 1} = 1\text{ A}$.
Q16 — 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: $4 \times 1.5 = 6\text{ V}$.
Q17 — Combination of Resistors & Cells · medium · numerical
Two cells each of EMF $2\text{ V}$ and internal resistance $0.5\,\Omega$ are connected in series with an external resistance of $3\,\Omega$. The current is:
A. $0.5\text{ A}$
B. $1\text{ A}$ ✓ Correct
C. $1.33\text{ A}$
D. $2\text{ A}$
Solution: $I = \dfrac{2 + 2}{3 + 0.5 + 0.5} = \dfrac{4}{4} = 1\text{ A}$.
Q18 — Combination of Resistors & Cells · hard · numerical
A wire of resistance $12\,\Omega$ is cut into three equal parts, which are then joined in parallel. The equivalent resistance is:
A. $4\,\Omega$
B. $36\,\Omega$
C. $\dfrac{4}{3}\,\Omega$ ✓ Correct
D. $12\,\Omega$
Solution: Each part has $4\,\Omega$, and three in parallel give $\dfrac{4}{3}\,\Omega$.
Q19 — Combination of Resistors & Cells · medium · numerical
A cell of EMF $10\text{ V}$ shows a terminal voltage of $8\text{ V}$ while supplying $2\text{ A}$. Its internal resistance is:
A. $0.5\,\Omega$
B. $4\,\Omega$
C. $5\,\Omega$
D. $1\,\Omega$ ✓ Correct
Solution: $r = \dfrac{E - V}{I} = \dfrac{10 - 8}{2} = 1\,\Omega$.
Q20 — Combination of Resistors & Cells · medium · numerical
Resistors of $2\,\Omega$, $3\,\Omega$ and $6\,\Omega$ are connected in parallel. The equivalent resistance is:
A. $2\,\Omega$
B. $11\,\Omega$
C. $0.5\,\Omega$
D. $1\,\Omega$ ✓ Correct
Solution: $\dfrac{1}{R} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = 1$, so $R = 1\,\Omega$.
Q21 — Combination of Resistors & Cells · medium · numerical
A cell of EMF $12\text{ V}$ and internal resistance $2\,\Omega$ delivers maximum power to an external resistance of:
A. $2\,\Omega$ ✓ Correct
B. $1\,\Omega$
C. $4\,\Omega$
D. $12\,\Omega$
Solution: Maximum power transfer requires the load to match the internal resistance, $R = r = 2\,\Omega$.