Q.
Four capacitors are connected as shown in figure. The equivalent capacitance between $A$ and $B$ is
A) 4 $\mu$F
B) 0.25 $\mu$F
C) 0.75 $\mu$F
D) 1.33 $\mu$F
Explanation
(d) $\frac{1}{C_{s}} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 3$
$C_{s} = \frac{1}{3}$
Capacitance between A and B
$C_{p} = \frac{1}{3} + 1$
$\frac{4}{3}\mu F = 1.33\mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.