Alternating Current Question 7
Question 7 - 2024 (29 Jan Shift 2)
In the given figure, the charge stored in $6 \mu \mathrm{F}$ capacitor, when points $\mathrm{A}$ and $\mathrm{B}$ are joined by a connecting wire is $\mu \mathrm{C}$.
Show Answer
Answer: (36)
Solution:
At steady state, capacitor behaves as an open circuit and current flows in circuit as shown in the diagram.
$\mathrm{R}_{\text {eq }}=9 \Omega$
$\mathrm{i}=\frac{9 \mathrm{~V}}{9 \Omega}=1 \mathrm{~A}$
$\Delta \mathrm{V}_{6 \Omega}=1 \times 6=6 \mathrm{~V}$
$\mathrm{V}_{\mathrm{A}}=3 \mathrm{~V}$
So, potential difference across $6 \mu \mathrm{F}$ is $6 \mathrm{~V}$.
Hence
$$ \begin{aligned} Q & =C \Delta V \ & =6 \times 6 \times 10^{-6} \mathrm{C} \ & =36 \mu \mathrm{C} \end{aligned} $$
