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