Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A capacitor of capacitance $5 \mu F$ is connected as shown in the figure. The internal resistance of the cell is $0.5\Omega$ . The amount of charge on the capacitor plate is
Text Solution
Verified by ExpertsThe correct answer is:
C
In steady state condition. No current flows through line (1). Hence total current $i = \frac{2.5}{(1+1+0.5)} = 1 \mathrm{A}$
Potential difference a cross line (2) = potential difference a cross capacitor
$= 1 \times 2 = 2 \sqrt{\theta t}$
So, charge on capacitor = 5 × × 2 = 10 μ μ C
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