JEE Main 2026 Apr 6 Shift 2, Physics Q24: Kirchhoff’s Laws

JEE Main2026Apr 6, Shift 2Physics

Two cells of emfs 1 V and 2 V and internal resistances $2 Ω$ and $1 Ω$ , respectively connected in parallel, give a current of 1 A through an external resistance. If the polarity of one cell is reversed, then the value of current through the external resistance will be $\frac{α}{5}$ A. The value of $α$ is _____.

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Solution

Step 1 — original configuration (both emfs aiding). Replace the two parallel cells by a single Thevenin source:

$E_{e q} = \frac{E _{1} / r _{1} + E _{2} / r _{2}}{1/ r _{1} + 1/ r _{2}} = \frac{1/2 + 2/1}{1/2 + 1/1} = \frac{5/2}{3/2} = \frac{5}{3}$ V.

$r_{e q} = \frac{r _{1} r _{2}}{r _{1} + r _{2}} = \frac{2 \cdot 1}{2 + 1} = \frac{2}{3} Ω$ .

Given external current $= 1$ A: $1 = \frac{5/3}{2/3 + R}$ $\Rightarrow$ $R = 1 Ω$ .

Step 2 — polarity of one cell reversed. Using the 2 V cell as positive and the 1 V cell flipped:

$E_{e q}^{'} = \frac{2/1 - 1/2}{1/1 + 1/2} = \frac{3/2}{3/2} = 1$ V (internal resistance unchanged at $r_{e q} = 2/3 Ω$ ).

New current: $i = \frac{1}{2/3 + 1} = \frac{1}{5/3} = \frac{3}{5}$ A.

Comparing with $\frac{α}{5}$ , we get $α = 3$ .

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JEE Main 2026 Apr 6 Shift 2, Physics Q24: Kirchhoff’s Laws

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