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JEE Main 20265 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
5 April
Shift
Evening Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2026 · 5 April · Evening Shift · Physics · Current Electricity

When an external resistance of 5\ is connected across terminals of...

When an external resistance of 5 Ω5\ \Omega is connected across terminals of a cell, a current of 0.25 A0.25\text{ A} flows through it. When the 5 Ω5\ \Omega resistor is replaced by a 2 Ω2\ \Omega resistor, a current of 0.5 A0.5\text{ A} flows through it. The internal resistance of the cell is _______ Ω\Omega.

Correct answer

0

Step-by-step solution

Let the EMF of the cell be EE and its internal resistance be rr.

The current II in the circuit is given by I=ER+rI = \dfrac{E}{R + r}.

For the first case, R1=5 ΩR_1 = 5\ \Omega and I1=0.25 AI_1 = 0.25\text{ A}:

0.25=E5+r⇒E=0.25(5+r)0.25 = \dfrac{E}{5 + r} \Rightarrow E = 0.25(5 + r)

For the second case, R2=2 ΩR_2 = 2\ \Omega and I2=0.5 AI_2 = 0.5\text{ A}:

0.5=E2+r⇒E=0.5(2+r)0.5 = \dfrac{E}{2 + r} \Rightarrow E = 0.5(2 + r)

Equating the expressions for EE:

0.25(5+r)=0.5(2+r)0.25(5 + r) = 0.5(2 + r)

5+r=2(2+r)5 + r = 2(2 + r)

5+r=4+2r5 + r = 4 + 2r

r=1 Ωr = 1\ \Omega

Answer: 11
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