Q. A capacitor of capacitance 2$\mu$F is connected as shown in figure. The internal resistance of the cell is $0.5\Omega$. The amount of charge on the capacitor plates is
A) Zero
B) $2\mu\ C$
C) $4\mu C$
D) $6\mu C$
Explanation
(c) According to loop rule,
$2.5 - 0.5I - 2I = 0$
$\Rightarrow I = 1A$
$V_{A} - V_{B} = \frac{q_{o}}{C} = 2I = 2V$
$q_{o} = C \times 2 = 2 \times 10^{- 6} \times 2 = 4\mu C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.