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Current Electricity

JEE Main level, single correct, medium

189 of 380

Question 189 of 380

A cell ${E}_{1}$ of emf $6V$ and internal resistance $2\Omega$ is connected with another cell ${E}_{2}$ of emf $4V$ and internal resistance $8\Omega$ (as shown in the figure). The potential difference across points $X$ and $Y$ is:

Show answer and solution

Answer: (B) $5.6V$

I=\dfrac{6-4}{10}=\dfrac{1}{5}A

{V}_{x}+4+8\times \dfrac{1}{5}-{V}_{y}=0

{V}_{x}-{V}_{y}=-5.6\Rightarrow \left|{V}_{x}-{V}_{y}\right|=5.6V

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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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.

A cell E1{E}_{1}E1​ of emf 6V6V6V and internal resistance 2Ω2\Omega2Ω is connected with another cell E2{E}_{2}E2​ of emf 4V4V4V and internal resistance 8Ω8\Omega8Ω (as shown in the figure). The potential difference across points XXX and YYY is:

JEE Main question paper PDF

More Physics practice

A cell ${E}_{1}$ of emf $6V$ and internal resistance $2\Omega$ is connected with another cell ${E}_{2}$ of emf $4V$ and internal resistance $8\Omega$ (as shown in the figure). The potential difference across points $X$ and $Y$ is: