JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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......\(\mu F\)
- A \(10\)
- B \(15\)
- C \(20\)
- D \(30\)
Answer & Solution
Correct Answer
(B) \(15\)
Step-by-step Solution
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 \(C =500 \mu C -200 \mu C =300 \mu C\)…
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