Q.
Four capacitors are connected in a circuit as shown in the following figure. Calculate the effective capacitance between the points $A$ and $B$
A) $\frac{4}{3}\mu F$
B) $\frac{24}{5}\mu F$
C) $9\mu F$
D) $5\mu F$
Explanation
(a) $C_{2}$ and $C_{3}$ are connected in series, parallel with $C_{1},$ then their equivalent capacitance $C^{''} = \frac{2 \times 2}{2 + 2} + 1 = 2\mu F$
[image:1_image412.emf]
$C_{4}$ and $C^{''}$ are in series
$\therefore Effective\ capacity\ C_{eff} = \frac{4}{3}\mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.