JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Current Electricity

Two cells of emfs 1 V and 2 V and internal resistance

2Ω

and

1Ω

, respectively connected in parallel, gave current of 1 A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be

\frac{α}{5} A

. The value of

α

is ______.

Answer: 3

Solution

Two cells in parallel reduce to one equivalent source. With both cells aiding,

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}{3} V

and

r_{e q} = \frac{1}{1/2 + 1} = \frac{2}{3} Ω

. The stated current fixes the external resistance:

1 = \frac{5/3}{2/3 + R}

, so

R = 1 Ω

. When one cell is reversed the two emfs oppose, giving

E_{e q}^{'} = \frac{E _{2} / r _{2} - E _{1} / r _{1}}{1/ r _{1} + 1/ r _{2}} = \frac{2 - 1/2}{3/2} = 1 V

, while

r_{e q}

stays

\frac{2}{3} Ω

. The current becomes

i = \frac{1}{2/3 + 1} = \frac{3}{5} A

. Comparing with

\frac{α}{5} A

gives

α = 3

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Integer answer

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JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Current Electricity

Solution

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