Q. A $10\, \mu F$ capacitor is fully charged to a potential difference of $50\, V$. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes $20\, V$. The capacitance of the second capacitor is:

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A

$10\, \mu F$

0%

B

$15\, \mu F$

83%

C

$20\, \mu F$

17%

D

$30\, \mu F$

0%

Solution:

Initially

$•$ Charge on capacitor $10 \mu F$
$Q=C V=(10 \mu F)(50 V )$
$Q =500 \mu C$

$•$ Final Charge on $10 \mu F$ capacitor
$Q = CV =(10 \mu F )(20 V )$
$Q =200 \mu C$
$•$ From charge conservation,
Charge on unknown capacitor
$C =500 \mu C -200 \mu C =300 \mu C$
$\Rightarrow $ Capacitance $( C )=\frac{ Q }{ V }=\frac{300 \mu C }{20 V }=15 \mu F$

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