Question

A charged capacitor is allowed to discharge though a resistor by closing the key at the instant $\tau = 0$ (see figure). At the instant $\tau = (\ln 4) \mu s$, the reading of the ammeter falls to half the initial value. The resistance of the ammeter is equal to:

• A. 1 M$\Omega$
• B. 1 $\Omega$
• C. 2 $\Omega$
• D. 2 M$\Omega$

Answer 1

Answer: (C)

2 $\Omega$

Answer 2

The discharge of a capacitor through a resistor follows an exponential decay. The current $I(t)$ in an RC circuit discharging at time $t$ is given by: $I(t) = I_0 e^{-t/(R_{total}C)}$, where $I_0$ is the initial current, $R_{total}$ is the total resistance, and $C$ is the capacitance.

The problem states that at time $t = (\ln 4) \mu s$, the current falls to half its initial value, i.e., $I(t) = I_0/2$. So, $\frac{I_0}{2} = I_0 e^{-t/(R_{total}C)}$. This simplifies to $\frac{1}{2} = e^{-t/(R_{total}C)}$.

Taking the natural logarithm of both sides: $\ln\left(\frac{1}{2}\right) = -\frac{t}{R_{total}C}$ $-\ln 2 = -\frac{t}{R_{total}C}$ $\ln 2 = \frac{t}{R_{total}C}$

This implies that the time taken for the current to halve, known as the half-life of the discharge, is $t_{1/2} = (\ln 2) R_{total}C$.

We are given: Capacitance $C = 0.5 \mu F = 0.5 \times 10^{-6} F$. Time $t = (\ln 4) \mu s$. Since $\ln 4 = \ln(2^2) = 2 \ln 2$, we have $t = (2 \ln 2) \mu s = (2 \ln 2) \times 10^{-6} s$. The resistor's resistance is $R_{resistor} = 2 \Omega$. Let the resistance of the ammeter be $R_A$. The total resistance in the circuit is $R_{total} = R_{resistor} + R_A = 2 + R_A$.

Substituting these values into the half-life equation $\ln 2 = \frac{t}{R_{total}C}$: $\ln 2 = \frac{(2 \ln 2) \times 10^{-6} s}{(2 + R_A) \times (0.5 \times 10^{-6} F)}$

Cancel $\ln 2$ from both sides: $1 = \frac{2 \times 10^{-6}}{(2 + R_A) \times 0.5 \times 10^{-6}}$

Cancel $10^{-6}$: $1 = \frac{2}{(2 + R_A) \times 0.5}$ $1 = \frac{2}{1 + 0.5 R_A}$

Multiply both sides by $(1 + 0.5 R_A)$: $1 + 0.5 R_A = 2$

Subtract 1 from both sides: $0.5 R_A = 1$

Multiply by 2: $R_A = 2 \Omega$

The resistance of the ammeter is $2 \Omega$.