Physics · Capacitors and R-C Circuits

JEE Main 2024 — 27 January, Shift 1 — Question 56

The charge accumulated on the capacitor connected in the following circuit $\mu \mathrm{C}$ (Given $\mathrm{C}=150 \mu \mathrm{~F}$ )

Answer: 400

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{V}_{\mathrm{A}}+\frac{10}{3}(1)-6(1)=\mathrm{V}_{\mathrm{B}}$

$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=6-\frac{10}{3}=\frac{8}{3}$ volt $\mathrm{Q}=\mathrm{C}\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}\right)$

$=150 \times \frac{8}{3}=400 \mu \mathrm{C}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 1
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Charging and Discharging of R-C Circuits