JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 22)
In the given figure, the charge stored in $$6 \mu F$$ capacitor, when points $$A$$ and $$B$$ are joined by a connecting wire is __________ $$\mu C$$.
_29th_January_Evening_Shift_en_22_1.png)
Answer
36
Explanation
At steady state, capacitor behaves as an open circuit and current flows in circuit as shown in the diagram.
_29th_January_Evening_Shift_en_22_2.png)
$$\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=9 \Omega \\ & \mathrm{i}=\frac{9 \mathrm{~V}}{9 \Omega}=1 \mathrm{~A} \\ & \Delta \mathrm{V}_{6 \Omega}=1 \times 6=6 \mathrm{~V} \\ & \mathrm{~V}_{\mathrm{A}}=3 \mathrm{~V} \end{aligned}$$
So, potential difference across 6$$\mu$$F is 6 V.
$$\begin{aligned} \text { Hence } \mathrm{Q} & =\mathrm{C} \Delta \mathrm{V} \\ & =6 \times 6 \times 10^{-6} \mathrm{C} \\ & =36 \mu \mathrm{C} \end{aligned}$$
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