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Question 15

Two identical capacitors $$A$$ and $$B$$, charged to the same potential $$5V$$ are connected in two different circuits as shown below at time $$t = 0$$. If the charge on capacitors $$A$$ and $$B$$ at time $$t = CR$$ is $$Q_A$$ and $$Q_B$$ respectively, then (Here $$e$$ is the base of natural logarithm)

Circuit A (Diode is reverse biased):  $$I = 0 \implies Q_A = Q_0 = CV$$

Circuit B (Diode is forward biased):  $$Q_B = Q_0 e^{-\frac{t}{CR}} = CV e^{-\frac{CR}{CR}} = \frac{VC}{e}$$

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