Physics · Current Electricity

JEE Main 2024 — 30 January, Shift 1 — Question 47

Two cells are connected in opposition as shown. Cell $\mathrm{E}_{1}$ is of 8 Vemf and $2 \Omega$ internal resistance; the cell $E_{2}$ is of 2 V emf and $4 \Omega$ internal resistance. The terminal potential difference of cell $\mathrm{E}_{2}$ is:

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

$I=\frac{8-2}{2+4}=\frac{6}{6}=1 A$

Applying Kirchhoff from C to B

$V_{C}-2-4 \times 1=V_{B}$

$V_{C}-V_{B}=6 V$

$=6 \mathrm{~V}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Physics
Chapter: Current Electricity
Topic: Circuit Analysis, Kirchhoff's Law and Nodal Analysis