In the given figure, the charge stored in 6 μ F capacitor, when points A and B are joined by a connecting wire is \ \ \ \ \ \ \ μ C .

Physics · Capacitors and R-C Circuits

JEE Main 2024 — 29 January, Shift 2 — Question 54

In the given figure, the charge stored in $6 \mu \mathrm{~F}$ capacitor, when points A and B are joined by a connecting wire is $\_\_\_\_\_\_\_$ $\mu \mathrm{C}$ .

Answer: 36

Numerical answer — enter this value.

Step-by-step solution

At steady state, capacitor behaves as an open circuit and current flows in circuit as shown in the diagram.

${{\rm{R}}_{{\rm{eq}}}} = 9{\rm{\Omega }}$${\rm{i}} = \frac{{9{\rm{\;V}}}}{{9{\rm{\Omega }}}} = 1{\rm{\;A}}$${\rm{\Delta }}{{\rm{V}}_{6{\rm{\Omega }}}} = 1 \times 6 = 6{\rm{\;V}}$${{\rm{V}}_{\rm{A}}} = 3{\rm{\;V}}$

So, potential difference across $6\mu {\rm{\;F}}$ is 6 V . Hence ${\rm{Q}} = {\rm{C\Delta V}}$

$\begin{array}{*{20}{r}}{}&{\; = 6 \times 6 \times {{10}^{ - 6}}{\rm{C}}}\\{}&{\; = 36\mu {\rm{C}}}\end{array}$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Charging and Discharging of R-C Circuits