Question

Four capacitors $4\mu F,6\mu F,5\mu F,3\mu F$ are connected in series to a source of $150V$. The potential difference in volt, across the $5\mu F$ capacitor will be

• A. $30$ volt
• B. $40$ volt
• C. $50$ volt
• D. $60$ volt

Answer 1

Answer: (A)

Equivalent capacitance $\frac{1}{C_{\text{eq }}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\frac{1}{C_4}$
$\frac{1}{C_{\text{eq }}}=\frac{1}{4}+\frac{1}{6}+\frac{1}{5}+\frac{1}{3}$
$\frac{1}{C_{\text{eq }}}=\frac{15+10+12+20}{60}=\frac{57}{60}$
$C_{eq}=1.05\mu F$
$q=C_{eq}V$
$\therefore q=1.05\times 10^{-6}\times 150=157.5\times 10^{-6}$
$q=1.5\times 10^{-4}C$
$\therefore V=\frac{q}{C}=\frac{1.5\times 10^{-4}}{5\times 10^{-6}}$
$=0.3\times 10^2$
$V=30$ volt