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......\(\mu F\)

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Verified Solution

Answer & Solution

Correct Answer

(B) \(15\)

Detailed explanation

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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