There is a current of 1.0 A in the circuit shown below. What is the resistance of P ? ![A single-loop circuit with a 10 V battery, a 3 ohm resistor, a 2.5 ohm resistor and a resistor P carrying a current of 1.0 A — Tamil Nadu Class 12 Physics question](/figures/current-electricity/single-loop-10v-3ohm-2p5ohm-p-current-1a-fig-pyq-8c3b49.svg) (a) 3.5 Ω (b) 1.5 Ω (c) 4.5 Ω (d) 2.5 Ω

There is a current of 1.0 A in the circuit shown below. What is the resistance of P ? (a) 3.5 Ω (b) 1.5 Ω (c)

Question 87 of 102

Q. There is a current of 1.0 A in the circuit shown below. What is the resistance of P ?

A single-loop circuit with a 10 V battery, a 3 ohm resistor, a 2.5 ohm resistor and a resistor P carrying a current of 1.0 A — Tamil Nadu Class 12 Physics question — Figure

(a) 3.5 Ω

(b) 1.5 Ω

(d) 2.5 Ω

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Concept understanding — Resistors in Series and Parallel

Series combination. When resistors are connected end to end so the same current I is forced through every one of them in turn (since charge cannot accumulate anywhere along the chain), the voltage across each one differs according to Ohm's law ( $V_1=IR_1$ , etc.), and the supply voltage V is the SUM of these individual drops: $V=V_1+V_2+V_3=I(R_1+R_2+R_3)$ . This defines the equivalent series resistance,

R_S = R_1+R_2+R_3+\ldots

which is always GREATER than even the largest individual resistor in the chain.

Parallel combination. When resistors are instead all connected directly across the same two points, each one experiences the identical voltage V, but the total current I splits into separate branch currents ( $I_1=V/R_1$ , etc.) that must add back up to the total, $I=I_1+I_2+I_3$ . This gives the equivalent parallel resistance through the reciprocal-sum rule,

\dfrac{1}{R_P} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\ldots

which is always LESS than even the smallest individual resistor -- adding another parallel branch can only open up additional room for current to flow, never restrict it.

Why household wiring is parallel. Every appliance in a house is deliberately connected in parallel across the mains supply, precisely because a parallel connection guarantees each device gets the full supply voltage and operates completely independently of the others -- switching any one appliance off leaves the voltage across, and the current through, every other appliance completely unaffected. A series connection, by contrast, would mean switching off any single appliance breaks the circuit for every appliance downstream of it. …

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Applying $R_{total}=V/I$ to the single series loop and subtracting the two known resistors gives $P = 4.5\ \Omega$ .

Working

The circuit is a single series loop: a 10 V battery, a $3\ \Omega$ resistor, a $2.5\ \Omega$ resistor, and the unknown resistor $P$ , carrying a current $I = 1.0$ A.

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 55/6/11 markMCQ
$R_s$
›Reveal solutionSolution
For $n$ identical resistors $R$ , series equivalent $R_s = nR$ and parallel equivalent $R_p = R/n$ . Their difference is $R_s - R_p = \frac{n^2 - 1}{n}R$ , which matches option (C).

The problem is about combining identical resistors — a classic exercise in spotting how series and parallel formulas scale. When you put $n$ resistors of the same value $R$ in series, the total resistance just adds up: $R_s = nR$ . When you put them in parallel, the reciprocal adds up, giving $R_p = R/n$ . The question then asks for the difference $R_s - R_p$ , which is a simple subtraction — but the trick is in the algebra and in recognising which of the given options matches the result.

Let’s walk through it step by step.

Series combination: For $n$ resistors each of resistance $R$ connected end-to-end, the equivalent resistance is the sum:

$R_s = R + R + \cdots + R \quad (n \text{ times}) = nR.$

Parallel combination: For $n$ identical resistors connected across the same two points, the equivalent resistance is given by:

$\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \cdots + \frac{1}{R} \quad (n \text{ times}) = \frac{n}{R}.$
Taking the reciprocal:
$R_p = \frac{R}{n}.$

Find the difference: Subtract the parallel equivalent from the series equivalent:

$R_s - R_p = nR - \frac{R}{n}.$

Simplify the expression: Factor out $R$ and combine the terms over a common denominator: $R_s - R_p = R\left(n - \frac{1}{n}\right) = R\left(\frac{n^2 - 1}{n}\right) = \frac{n^2 - 1}{n}R.$ …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. There is a current of 1.0 A in the circuit shown below. What is the resistance of P ?$
(a) 3.5 Ω
(b) 1.5 Ω
(c) 4.5 Ω
(d) 2.5 Ω
›Reveal solutionSolution

Applying $R_{total}=V/I$ to the single series loop and subtracting the two known resistors gives $P = 4.5\ \Omega$ .

Working

The circuit is a single series loop: a 10 V battery, a $3\ \Omega$ resistor, a $2.5\ \Omega$ resistor, and the unknown resistor $P$ , carrying a current $I = 1.0$ A.

…

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CBSE 2020Set ANNUAL1 markMCQ
$Q. The current in the circuit is : [figure: a 15 V battery connected across three 15 Ω resistors, all three resistors in parallel with each other and with the battery]$
(a) 4 A
(b) 1 A
(c) 2 A
(d) 3 A
›Reveal solutionSolution

Three equal $15\ \Omega$ resistors in parallel combine to $5\ \Omega$ ; Ohm's law with the 15 V battery then gives a total current of 3 A.

Working

For three equal resistors $R=15\ \Omega$ in parallel:

$\dfrac1{R_{eq}}=\dfrac1{15}+\dfrac1{15}+\dfrac1{15}=\dfrac3{15}\ \Rightarrow\ R_{eq}=5\ \Omega$

…

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Q.There is a current of 1.0 A in the circuit shown below. What is the resistance of P ?

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