Capacitor Discharge Voltage Formula

The voltage $V(t)$ across a capacitor during discharge is modeled by the exponential decay formula:

$V(t) = E e^{-t/\tau}$

where $E$ is the initial voltage and $\tau$ represents the time constant of the circuit.

Calculating Voltage Percentages

To match the lists, we calculate the remaining voltage percentage at specific time intervals (multiples of $\tau$):

• For $t = 1\tau$: The voltage is $V(1\tau) = E e^{-1\tau/\tau} = E e^{-1} \approx 0.3679 E$. This is approximately 36.8% of the initial voltage $E$.
• For $t = 2\tau$: The voltage is $V(2\tau) = E e^{-2\tau/\tau} = E e^{-2} \approx 0.1353 E$. This is approximately 13.5% of $E$.
• For $t = 3\tau$: The voltage is $V(3\tau) = E e^{-3\tau/\tau} = E e^{-3} \approx 0.0498 E$. This is approximately 5% of $E$.
• For $t = 4\tau$: The voltage is $V(4\tau) = E e^{-4\tau/\tau} = E e^{-4} \approx 0.0183 E$. This is approximately 1.87% of $E$.
• For $t = 5\tau$: The voltage is $V(5\tau) = E e^{-5\tau/\tau} = E e^{-5} \approx 0.0067 E$. This is approximately 0.67% of $E$.

Matching List-I with List-II

Based on the calculations, we match the time constant multiples (List-I) with the corresponding voltage percentages (List-II):

• P ($1\tau$) corresponds to 36.8% of E, which is item 2 in List-II.
• Q ($2\tau$) corresponds to 13.5% of E, which is item 3 in List-II.
• R ($3\tau$) corresponds to 5% of E, which is item 4 in List-II.
• S ($4\tau$) corresponds to 1.87% of E, which is item 5 in List-II.
• T ($5\tau$) corresponds to 0.67% of E, which is item 1 in List-II.

The correct sequence is P-2, Q-3, R-4, S-5, T-1.

This matches the selection code provided in Option 4 (D).