Question

In the given network capacitance $C_{2}=10\, \mu F,\, C_{1}=5\, \mu F$ and $C_{3}=4\, \mu F$. The resultant capacitance between $P$ and $Q$ will be :

• A. $ 4.7\,\mu F $
• B. $ 1.2\,\mu F $
• C. $ 3.2\,\mu F $
• D. $ 2.2\,\mu F $

Answer 1

Answer: (C)

Seeing in the given circuit $C_{1}$ and $C_{2}$ are connected in parallel. Hence, their equivalent capacitance
$C_{e q} =C_{1}+C_{2}=5+10$
$=15\, \mu F$
As $C_{eq}$ and $C_{3}$ are connected in series, Hence, resultant capacitance between P and Q is given by
$\frac{1}{C_{P Q}}=\frac{1}{C_{e q}}+\frac{1}{C_{3}}$
$=\frac{1}{15}+\frac{1}{4}=\frac{19}{60}$
$C_{P Q}=\frac{60}{19}=3.2\, \mu F$