Question

A series combination of two capacitances of value $0.1muF$ and $1\mu F$ is connected with a source of voltage $500volts$. The potential difference in volts across the capacitor of value $0.1muF$ will be :

• A. $50$
• B. $500$
• C. $45.5$
• D. $454.5$

Answer 1

The correct option is D $454.5$

Given,

Capacitance, $C_1=0.1\mu Fand{C}_2=1\mu F$

In series charge is equal

$Q=C_1{V}_1=C_2{V}_2$

$V_2=\frac{C_1{V}_1}{C_2}$

In series total potential difference is sum of all paternal difference

$V=V_1+V_2$

$V=V_1+\frac{C_1{V}_1}{C_2}=V_1\left(\frac{C_2+C_1}{C_2}\right)$

$V_1=\frac{C_{2}V}{C_2+C_1}=\frac{1\times 500}{1+0.1}=454.54V$

Hence, Potential difference across $0.1\mu Fis454.5V$