Q. In the figure below, what is the potential difference between the points $A$ and $B$ and between $B$ and $C$ respectively in steady state
A) $V_{AB} = V_{BC} = 100\ V$
B) $V_{AB} = 75\ V,V_{BC} = 25\ V$
C) $V_{AB} = 25\ V,V_{BC} = 75\ V$
D) $V_{AB} = V_{BC} = 50\ V$
Explanation
(c) $C_{eq} = \frac{(3 + 3) \times (1 + 1)}{(3 + 3) + (1 + 1)} + 1 = \left( \frac{6 \times 2}{6 + 2} \right) + 1 = \frac{5}{2}\mu F$
$\therefore Q = C \times V = \frac{5}{2} \times 100 = 250\mu C$
Charge in $6\mu F$ branch $= VC = \left( \frac{6 \times 2}{6 + 2} \right)100 = 150\ \mu C$
$V_{AB} = \frac{150}{6} = 25V$ and $V_{BC} = 100 - V_{AB} = 75V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.