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Please answer 13 question with detailed explanation........ .....thankiu in advance

Please answer 13 question with detailed explanation........
.....thankiu in advance

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Grade:12th pass

1 Answers

Eshan
askIITians Faculty 2095 Points
2 years ago
Dear student,

Since charge on2\mu Fcapacitor is24\mu C, the voltage across it is12V. Since it is in parallel to capapcitor C, C also has 12V across it.

Since the battery provides 18V, the voltage on6\mu Fcapacitor is18V-12V=6V.
Hence the charge stored on it is36\mu C. This current must have divided into the two capacitors in series. Hence charge stored on C is36\mu C-24\mu C=12\mu C. But potential difference across it is12V. Hence the capacitance C is\dfrac{12\mu C}{12V}=1\mu F

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