A \(5\,\mu F\) capacitor is fully charged by a \(12\,V\) battery and then disconnected. If it is connected now in parallel to an uncharged capacitor, the voltage across it is \(3\,V\). Then the capacity of the uncharged capacitor is
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- \(5\,\mu F\)
- \(15\,\mu F\)
- \(50\,\mu F\)
- \(10\,\mu F\)
- \(25\,\mu F\)
The Correct Option is B
Solution and Explanation
Step 1: Initial charge on capacitor. \[ Q = CV = 5 \times 12 = 60\,\mu C \]
Step 2: After connection, voltage becomes 3 V. Total capacitance = \(5 + C\)
Step 3: Apply charge conservation. \[ Q = (5 + C)\times 3 \]
Step 4: Substitute. \[ 60 = 3(5 + C) \Rightarrow 60 = 15 + 3C \]
Step 5: Solve. \[ 3C = 45 \Rightarrow C = 15\,\mu F \]
Step 6: Final Answer. \[ \boxed{15\,\mu F} \]
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