Q.
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
A) $0\mu C$
B) $5\mu C$
C) $10\mu C$
D) $25\mu C$
Explanation
(c) In steady state condition. No current flows through line
(1). Hence total current $i = \frac{2.5}{(1 + 1 + 0.5)} = 1\ A$
Potential difference a cross line (2) = potential difference across capacitor
$= 1 \times 2 = 2\ Volt$
So, charge on capacitor $= 5 \times 2 = 10\ \mu C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.