Q.
In the given circuit of figure with steady current, the potential drop across the capacitor must be
A) $V$
B) $\frac{V}{2}$
C) $\frac{V}{3}$
D) $\frac{2V}{3}$
Explanation
(c) In steady state no current flows through the capacitor segment. The steady current in remaining loop $I = \frac{2V - V}{2R + R} = \frac{V}{3R}\ \ $ (anti-clockwise). Now applying Kirchhoff’s second law to loop containing $\ 2V,2R,C\ $and $V,$ we have $V_{c} = 2V - 1.2V - \frac{V}{3R}.2R - V = \frac{V}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.