JEE PYQ: Capacitance Question 24
Question 24 - 2020 (09 Jan Shift 2)
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: capacitor discharges through R, Circuit B: capacitor discharges through R with diode in forward bias)
(1) $Q_A = \frac{VC}{e}$, $Q_B = \frac{CV}{2}$
(2) $Q_A = VC$, $Q_B = CV$
(3) $Q_A = VC$, $Q_B = \frac{VC}{e}$
(4) $Q_A = \frac{CV}{2}$, $Q_B = \frac{VC}{e}$
Show Answer
Answer: (3)
Solution
In case I, diode is reverse biased, so no current flows. $Q_A = CV$. In case II, current will flow as diode is forward biased. Charge decays exponentially: $V’ = Ve^{-t/CR}$. At $t = CR$: $V’ = Ve^{-1} = \frac{V}{e}$. $Q_B = CV’ = \frac{CV}{e}$.
BETA
