Question

A $2\mu F$ capacitor is charged to a potential difference of $30V$ The battery is removed and it is connected to a series combination of two uncharged capacitors as shown in figure. The final potential difference across $C_1$ will be -

• A. $10V$
• B. $20V$
• C. $30V$
• D. $5V$

Answer 1

Answer: (A)

$q_{\text{final}}$ on the equivalent of $3\mu F\&6\mu F$
$=\left(\frac{C_1{V}_1}{C_1+C_2}\right)C_2$
$=\frac{60C}{4F}\times 2\mu F$
$=30\mu C$

Since this charge will be same on both $3\mu F$ & $6\mu F$.
Hence potential difference across $C_1$ will be
$V_1=\frac{q}{C_1}$
$\Rightarrow V_1=\frac{30}{3}$
$=10V$