- 2017
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Question 1: A capacitor of capacitance 10µF is charged to a potential difference of 50 V. Now, the charging battery is disconnected and an uncharged capacitor of capacitance 2µF is connected to it. Calculate the common potential difference across the two capacitors.
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Solution: The charge on the first capacitor is Q = CV = 10µF × 50 V = 500µC. When the two capacitors are connected, the total charge remains the same. So, the charge on the second capacitor is also 500µC. The common potential difference is V = Q/C = 500µC/(10µF + 2µF) = 25 V.
- 2016
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Question 1: A capacitor of capacitance C is charged to a potential V. The energy stored in the capacitor is U. The capacitor is now connected to another uncharged capacitor of capacitance 2C in parallel. The energy stored in the system of two capacitors is now
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Solution: The total energy stored in the system of two capacitors
