Question

Two capacitors of caacity $C_1$ and $C_2$ are connected in series and potential difference V is applied across it. Then the potential difference across$C_1$ will be

• A. $\frac{C_{2}V}{C_1+C_2}$
• B. $\frac{C_{1}V}{C_1+C_2}$
• C. $(1+\frac{C_2}{C_1})V$
• D. $(1-\frac{C_2}{C_1})V$

Answer 1

The correct option is A $\frac{C_{2}V}{C_1+C_2}$

Given, Two capacitors of capacity $C_1$ and $C_2$ are connected in series then,

$C=\frac{C_1{C}_2}{C_1+C_2}$ (resultant $C$)

Potential difference $=V$

$V=V\times \frac{C_1}{C_1{C}_2}\left(\frac{\frac{C_1{C}_2}{C_1+C_2}}{C_1}\right)=V\times \frac{C_2}{C_1+C_2}$