JEE MAIN - Physics (2023 - 13th April Evening Shift - No. 9)
In the network shown below, the charge accumulated in the capacitor in steady state will be:
_13th_April_Evening_Shift_en_9_1.png)
10.3 $$\mu$$C
7.2 $$\mu$$C
4.8 $$\mu$$C
12 $$\mu$$C
Explanation
In steady state, capacitor behaves as open circuit.
$$ \begin{aligned} \Rightarrow \Delta V_C & =i \times 6 \Omega \\\\ & =\frac{3}{10} \times 6 \mathrm{~V} \\\\ & =1.8 \mathrm{~V} \\\\ \Rightarrow Q= & C V=7.2 \mu \mathrm{C} \end{aligned} $$
$$ \begin{aligned} \Rightarrow \Delta V_C & =i \times 6 \Omega \\\\ & =\frac{3}{10} \times 6 \mathrm{~V} \\\\ & =1.8 \mathrm{~V} \\\\ \Rightarrow Q= & C V=7.2 \mu \mathrm{C} \end{aligned} $$
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