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. $V\frac{C_2}{C_1}$
• B. $V\frac{C_2}{C_1+C_2}$
• C. $V\frac{C_1+C_2}{C_1}$
• D. $V\frac{C_1}{C_1+C_2}$

Answer 1

Answer: (B)

$Given:$ two capacitors of capacity $C_1$ and $C_2$ are connected in series and potential difference V is applied across it.

$Solution:$ We know that,

$Q=C_{\text{net}}V$

$Q=\left(\frac{C_1{C}_2}{c_1+C_2}\right)V$

$\therefore$ $,$Voltage across $C_1=\frac{Q}{C_1}$

$=\frac{\frac{\left(c_1{c}_2\right)v}{(c_1+c_2)}}{c_1}$

$=\frac{c_2}{c_1+c_2}v$

$So,the$ $correct$ $option:C$