CURRENT BYU 43

The problem involves a “black box” containing a resistor $R_{\text{in}}$ and a capacitor $C_{\text{in}}$. When an external charged capacitor $C = 1.0 \, \mu\text{F}$ with charge $q_0 = 50 \, \mu\text{C}$ is connected, the charge decays to a steady non-zero value. This implies that the internal circuit is a capacitor $C_{\text{in}}$ in series with a resistor $R_{\text{in}}$. The external capacitor shares its charge with the internal capacitor.

1. Finding $C_{\text{in}}$

From charge conservation, the initial charge $q_0$ is eventually redistributed between the external capacitor $C$ and the internal capacitor $C_{\text{in}}$. In the steady state ($t \to \infty$), the current is zero, so the voltage across both capacitors must be equal ($V = V_{\text{in}}$).

$$ \frac{q_{\infty}}{C} = \frac{q_{\text{in,}\infty}}{C_{\text{in}}} $$

The total charge is conserved:

$$ q_{\infty} + q_{\text{in,}\infty} = q_0 $$

From the graph, the asymptotic charge on the external capacitor is $q_{\infty} = 10 \, \mu\text{C}$. Given $q_0 = 50 \, \mu\text{C}$:

$$ q_{\text{in,}\infty} = 50 – 10 = 40 \, \mu\text{C} $$

Substituting into the voltage equality:

$$ \frac{10 \, \mu\text{C}}{1.0 \, \mu\text{F}} = \frac{40 \, \mu\text{C}}{C_{\text{in}}} $$ $$ C_{\text{in}} = 4.0 \, \mu\text{F} $$

2. Finding $R_{\text{in}}$

The circuit behaves as an RC discharge circuit where the equivalent capacitance is the series combination of $C$ and $C_{\text{in}}$ (since they are in a single loop).

$$ C_{\text{eq}} = \frac{C C_{\text{in}}}{C + C_{\text{in}}} = \frac{1.0 \times 4.0}{1.0 + 4.0} = 0.8 \, \mu\text{F} $$

The time constant of the decay is $\tau = R_{\text{in}} C_{\text{eq}}$. The charge equation is:

$$ q(t) – q_{\infty} = (q_0 – q_{\infty}) e^{-t/\tau} $$

From the graph, we can estimate $\tau$. At $t = \tau$, the transient part drops to $1/e \approx 0.37$ of its initial value. Alternatively, we can use the initial slope (current).

Initial current $i_0$ at $t=0$ is determined by the full voltage of the external capacitor applied across the resistor (since internal capacitor is initially uncharged):

$$ i_0 = \frac{V_0}{R_{\text{in}}} = \frac{q_0 / C}{R_{\text{in}}} $$

Also, $i_0 = -\left(\frac{dq}{dt}\right)_{t=0}$. From the decay equation derivative:

$$ i_0 = \frac{q_0 – q_{\infty}}{\tau} $$

Equating the expressions for current:

$$ \frac{q_0}{C R_{\text{in}}} = \frac{q_0 – q_{\infty}}{R_{\text{in}} C_{\text{eq}}} $$

This confirms our $C_{\text{eq}}$ derivation. Let’s solve for $R_{\text{in}}$ using the time constant visible in the graph. The curve drops from $50$ to roughly $25$ (which is $10 + 40/e \approx 24.7$) at $t \approx 1.6 \, \text{ms}$.

$$ \tau \approx 1.6 \, \text{ms} $$ $$ R_{\text{in}} = \frac{\tau}{C_{\text{eq}}} = \frac{1.6 \times 10^{-3} \, \text{s}}{0.8 \times 10^{-6} \, \text{F}} = 2.0 \times 10^3 \, \Omega $$