JEE MainPhysics

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Current Electricity: JEE Main Physics Question with Solution

Q: Why use P = I^2 R instead of P = V^2 / R?

The diagram indicates current entering and leaving each network. Because is the common given quantity across all circuits, directly relates to .

Different combination of

3

resistors of equal resistance

\mathrm{R}

are shown in the figures. The increasing order for power dissipation is:

Four resistor configurations (A), (B), (C), (D) with current I entering each network of three identical resistors of resistance R.

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What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A

\mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{B}}<\mathrm{P}_{\mathrm{C}}<\mathrm{P}_{\mathrm{D}}

B

\mathrm{P}_{\mathrm{C}}<\mathrm{P}_{\mathrm{D}}<\mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{B}}

C

\mathrm{P}_{\mathrm{B}}<\mathrm{P}_{\mathrm{C}}<\mathrm{P}_{\mathrm{D}}<\mathrm{P}_{\mathrm{A}}

D

\mathrm{P}_{\mathrm{C}}<\mathrm{P}_{\mathrm{B}}<\mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{D}}

Step-by-step solutionView

Correct answer

For a constant applied voltage $V$ , power dissipation $P = \frac{V^2}{R_{\text{eq}}}$ is inversely proportional to equivalent resistance, giving the order $P_C < P_B < P_A < P_D$ .

Option analysis

Why each option works or fails

A ·

\mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{B}}<\mathrm{P}_{\mathrm{C}}<\mathrm{P}_{\mathrm{D}}

Assumes that power dissipation is directly proportional to equivalent resistance ( $P \propto R_{\text{eq}}$ ) rather than inversely proportional ( $P \propto 1/R_{\text{eq}}$ ) under a constant voltage source. When circuits are connected across the same potential difference $V$ , use $P = \frac{V^2}{R_{\text{eq}}}$ , which means higher equivalent resistance results in lower total power dissipation.

B ·

\mathrm{P}_{\mathrm{C}}<\mathrm{P}_{\mathrm{D}}<\mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{B}}

Incorrectly evaluates the equivalent resistance of the series-parallel combinations, misranking circuits A and B relative to D. Calculate $R_{\text{eq}}$ systematically: $R_A = \frac{2}{3}R$ , $R_B = \frac{3}{2}R$ , $R_C = 3R$ , and $R_D = \frac{R}{3}$ .

C ·

\mathrm{P}_{\mathrm{B}}<\mathrm{P}_{\mathrm{C}}<\mathrm{P}_{\mathrm{D}}<\mathrm{P}_{\mathrm{A}}

Mixes up the formula for power with constant current ( $P = I^2 R$ ) instead of constant voltage ( $P = V^2 / R$ ), leading to an inverted ranking. Since the potential difference across each whole circuit is fixed to the same voltage source $V$ , the total power is strictly $P = \frac{V^2}{R_{\text{eq}}}$ .

D ·

\mathrm{P}_{\mathrm{C}}<\mathrm{P}_{\mathrm{B}}<\mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{D}}

None. This is the correct option. Correctly determines the equivalent resistances ( $R_C = 3R > R_B = 1.5R > R_A \approx 0.67R > R_D \approx 0.33R$ ) and inverts the order for power: $P_C < P_B < P_A < P_D$ .

Reviewed route

Solution

StepWorking

01given

Three identical resistors each of resistance $R$ are connected in 4 configurations with the same input current $I$ entering each combination: (A) Two in parallel with one in series: $R_A = R + \frac{R}{2} = \frac{3R}{2}$ (B) Two in series in parallel with one: $R_B = \frac{2R \cdot R}{2R + R} = \frac{2R}{3}$ (C) All three in parallel: $R_C = \frac{R}{3}$ (D) All three in series: $R_D = 3R$

02find

The increasing order of power dissipation among configurations A, B, C, and D.

03strategise

Since current $I$ enters each network, the total power dissipated in each circuit is $P = I^2 R_{eq}$ . Thus, $P \propto R_{eq}$ . The increasing order of power directly follows the increasing order of equivalent resistance.

04execute

Comparing the equivalent resistances: $R_C = \frac{1}{3}R \approx 0.33R$ $R_B = \frac{2}{3}R \approx 0.67R$ $R_A = \frac{3}{2}R = 1.5R$ $R_D = 3R$ Therefore, $R_C < R_B < R_A < R_D$ , which implies $P_C < P_B < P_A < P_D$ .

✓verify

All-parallel always gives the minimum resistance, and all-series gives the maximum. The mixed circuits naturally sit in between with $2R/3 < 3R/2$ . Hence $P_C < P_B < P_A < P_D$ is logically sound.

Hints that build this answer step by step

For multiple circuits connected to the same voltage source $V$ , how is the total dissipated power $P$ related to the equivalent resistance $R_{\text{eq}}$ ?
$P = \frac{V^2}{R_{\text{eq}}}$ , so $P$ is inversely proportional to $R_{\text{eq}}$
What are the equivalent resistances $R_A, R_B, R_C, R_D$ for the four configurations?
$R_A = \frac{2}{3}R,\; R_B = \frac{3}{2}R,\; R_C = 3R,\; R_D = \frac{1}{3}R$
Given $R_C = 3R > R_B = 1.5R > R_A \approx 0.67R > R_D \approx 0.33R$ , what is the increasing order of power dissipation?
$P_C < P_B < P_A < P_D$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why use P = I^2 R instead of P = V^2 / R?

The diagram indicates current $I$ entering and leaving each network. Because $I$ is the common given quantity across all circuits, $P = I^2 R_{eq}$ directly relates $P$ to $R_{eq}$ .

Answer

For a constant applied voltage $V$ , power dissipation $P = \frac{V^2}{R_{\text{eq}}}$ is inversely proportional to equivalent resistance, giving the order $P_C < P_B < P_A < P_D$ .

Why each option works or fails

A: $\mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{B}}<\mathrm{P}_{\mathrm{C}}<\mathrm{P}_{\mathrm{D}}$ - Assumes that power dissipation is directly proportional to equivalent resistance ( $P \propto R_{\text{eq}}$ ) rather than inversely proportional ( $P \propto 1/R_{\text{eq}}$ ) under a constant voltage source. When circuits are connected across the same potential difference $V$ , use $P = \frac{V^2}{R_{\text{eq}}}$ , which means higher equivalent resistance results in lower total power dissipation.
B: $\mathrm{P}_{\mathrm{C}}<\mathrm{P}_{\mathrm{D}}<\mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{B}}$ - Incorrectly evaluates the equivalent resistance of the series-parallel combinations, misranking circuits A and B relative to D. Calculate $R_{\text{eq}}$ systematically: $R_A = \frac{2}{3}R$ , $R_B = \frac{3}{2}R$ , $R_C = 3R$ , and $R_D = \frac{R}{3}$ .
C: $\mathrm{P}_{\mathrm{B}}<\mathrm{P}_{\mathrm{C}}<\mathrm{P}_{\mathrm{D}}<\mathrm{P}_{\mathrm{A}}$ - Mixes up the formula for power with constant current ( $P = I^2 R$ ) instead of constant voltage ( $P = V^2 / R$ ), leading to an inverted ranking. Since the potential difference across each whole circuit is fixed to the same voltage source $V$ , the total power is strictly $P = \frac{V^2}{R_{\text{eq}}}$ .
D · correct: $\mathrm{P}_{\mathrm{C}}<\mathrm{P}_{\mathrm{B}}<\mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{D}}$ - None. This is the correct option. Correctly determines the equivalent resistances ( $R_C = 3R > R_B = 1.5R > R_A \approx 0.67R > R_D \approx 0.33R$ ) and inverts the order for power: $P_C < P_B < P_A < P_D$ .

Step-by-step solution

given: Three identical resistors each of resistance $R$ are connected in 4 configurations with the same input current $I$ entering each combination: (A) Two in parallel with one in series: $R_A = R + \frac{R}{2} = \frac{3R}{2}$ (B) Two in series in parallel with one: $R_B = \frac{2R \cdot R}{2R + R} = \frac{2R}{3}$ (C) All three in parallel: $R_C = \frac{R}{3}$ (D) All three in series: $R_D = 3R$
find: The increasing order of power dissipation among configurations A, B, C, and D.
strategise: Since current $I$ enters each network, the total power dissipated in each circuit is $P = I^2 R_{eq}$ . Thus, $P \propto R_{eq}$ . The increasing order of power directly follows the increasing order of equivalent resistance.
execute: Comparing the equivalent resistances: $R_C = \frac{1}{3}R \approx 0.33R$ $R_B = \frac{2}{3}R \approx 0.67R$ $R_A = \frac{3}{2}R = 1.5R$ $R_D = 3R$ Therefore, $R_C < R_B < R_A < R_D$ , which implies $P_C < P_B < P_A < P_D$ .
verify: All-parallel always gives the minimum resistance, and all-series gives the maximum. The mixed circuits naturally sit in between with $2R/3 < 3R/2$ . Hence $P_C < P_B < P_A < P_D$ is logically sound.

Current Electricity: JEE Main Physics Question with Solution

What feels right?

For a constant applied voltage $V$ , power dissipation $P = \frac{V^2}{R_{\text{eq}}}$ is inversely proportional to equivalent resistance, giving the order $P_C < P_B < P_A < P_D$ .

Why each option works or fails

Solution

01given

02find

03strategise

04execute

✓verify

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