Q.
Four identical capacitors are connected as shown in diagram. When a battery of $6\ V$ is connected between $A$ and $B,$ the charge stored is found to be $1.5\ \mu C.$ The value of $C_{1}$ is
A) $2.5\ \mu F$
B) $15\ \mu F$
C) $1.5\ \mu F$
D) $0.1\ \mu F$
Explanation
(d) The capacitance across $A$ and $B$
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$= \frac{C_{1}}{2} + C_{1} + C_{1} = \frac{5}{2}C_{1}$
As $Q = CV,$
$1.5\mu C = \frac{5}{2}C_{1} \times 6$
$\Rightarrow C_{1} = \frac{1.5}{15} \times 10^{- 6} = 0.1 \times 10^{- 6}F = 0.1\mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.