Current Electricity

JEE Main level, numerical answer, medium

169 of 380

Question 169 of 380

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.

Show answer and solution

Answer: 1

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

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Circuit builder simulationDrag the sliders and watch the numbers in this question change.Revise Current Electricity firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Current Electricity (Electric Cell or Battery), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.