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Physics · Current Electricity

JEE Main 2026 — 6 April, Evening Shift — Question 19

Two cells of emfs 1V and 2V and internal resistance 2Ω and 1Ω, respectively connected in parallel, gave current of 1A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be α5\frac{\alpha}{5} A. The value of α\alpha is ______.

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

Let external resistance R. For parallel same polarity: Eeq=(1/2+2/1)/(1/2+1)=(0.5+2)/(0.5+1)=2.5/1.5=5/3E_{eq} = (1/2+2/1)/(1/2+1) = (0.5+2)/(0.5+1)=2.5/1.5=5/3 V, req=1/(0.5+1)=2/3Ωr_{eq}=1/(0.5+1)=2/3\Omega.

Current I=1=(5/3)/(R+2/3)⇒R=1ΩI=1 = (5/3)/(R+2/3) \Rightarrow R=1\Omega.

Reverse polarity: Eeq=(1/2−2/1)/(1.5)=(−1.5)/1.5=−1E_{eq} = (1/2 - 2/1)/(1.5) = (-1.5)/1.5 = -1 V (magnitude 1V). Then I′=1/(1+2/3)=1/(5/3)=3/5I' = 1/(1+2/3)=1/(5/3)=3/5 A. So α=3\alpha=3.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Evening Shift
Subject
Physics
Chapter
Current Electricity
Topic
Combination of Resistors and cells, Wheatstone Bridge

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