JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A \(2\, \mu F\) capacitor \(C _{1}\) is first charged to a potential difference of \(10\, V\) using a battery.Then the battery is removed and the capacitor is connected to an uncharged capacitor \(C _{2}\) of \(8\, \mu F\). The charge in \(C _{2}\) on equilibrium condition is \(\ldots\,\mu C\). (Round off to the Nearest Integer)
- A \(9\)
- B \(25\)
- C \(20\)
- D \(16\)
Answer & Solution
Correct Answer
(D) \(16\)
Step-by-step Solution
Detailed explanation
\(20=\left( C _{1}+ C _{2}\right) V \Rightarrow V =2\, volt\) \(Q _{2}= C _{2} V =16\, \mu C\) \(=16\)
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