Question

A $2 \,\mu F$ capacitor is charged to a potential difference of $30\,V$ 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. $10\,V$
• B. $20\,V$
• C. $30\,V$
• D. $5\,V$

Answer 1

Answer: (A)

$q_{\text {final}}$ on the equivalent of $3 \mu F \& 6 \mu F$
$=\left(\frac{C_{1} V_{1}}{C_{1}+C_{2}}\right) C_{2}$
$=\frac{60\, C }{4\, F } \times 2 \mu F$
$ =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
$V _{1}=\frac{ q }{ C _{1}}$
$\Rightarrow V _{1}=\frac{30}{3}$
$=10\, V$