Question

$4\mu F,6\mu F$ and $12\mu F$ condensers are in series across 90V. The p.d. across$12\mu F$ condenser is :

• A. 30V
• B. 90V
• C. 15V
• D. 45V

Answer 1

Answer: (C)

15V

We know that

$q=CV$
$\frac{1}{C_{ef}}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$
$C_{eff}=2\mu F$
$\therefore q=C_{ef}{V}_{ef}$
$q=\left(2\right)\left(90\right)$ $\times {10}^{-6}$
$q=180$$\mu C$
$\therefore V=\frac{q}{c}$
$V_3=\frac{q}{c_3}$
$V_3=\frac{45}{3}$
$\therefore V_3=15$