Verified Solution • 2025–26 Session moderate

QJEEM21Phf4e5: JEE Main 2021 Physics - Class 12 JEEM21Phf4e5 Solution: A cell E 1 of emf 6V and internal resistance 2 Ω is connected…

A cell E1 of emf 6V and internal resistance 2Ω is connected with another cell E2 of emf 4V and internal resistance 8Ω (as shown in the figure). The potential difference across points X and Y is :

JEE Main 2021 (Online) 24th February Morning Shift Physics - Current Electricity Question 206 English

Multiple Choice Options

  • A 10.0 V
  • B 2.0 V
  • C 5.6 V
  • D 3.6 V

Step-by-Step Verified Solution

The circuit can be shown as below

JEE Main 2021 (Online) 24th February Morning Shift Physics - Current Electricity Question 206 English Explanation
The current through the circuit,

I=ε1−ε2r1+r2=6−410=15A

∴ Potential difference across points X and Y is

VXY=E2+Ir2=4+15×8=5.6V

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring