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Step-by-Step Solution
Step 1: Identify the Known Parameters
• Battery voltage, $V_0 = 20 \text{ V}$
• Measured capacitor voltage at time $t$, $V = 2 \text{ V}$
• Time taken for the capacitor to reach 2 V, $t = 1\,\mu\text{s} = 1 \times 10^{-6} \text{ s}$
• Resistance in series, $R = 10\,\Omega$
• Given value: $\ln\!\bigl(\tfrac{10}{9}\bigr) = 0.105$
Step 2: Write the Charging Formula for an RC Circuit
When a capacitor of capacitance $C$ is charged through a resistor $R$ connected to a battery of voltage $V_0$, the voltage across the capacitor at time $t$ is given by:
$$
V = V_0 \Bigl(1 - e^{-t/(RC)}\Bigr).
$$
Step 3: Substitute the Known Values
Plug in $V = 2 \text{ V}$ and $V_0 = 20 \text{ V}$ into the formula:
$$
2 = 20 \Bigl(1 - e^{-t/(RC)}\Bigr).
$$
Simplify this equation:
$$
\frac{2}{20} = 1 - e^{-t/(RC)} \quad \Longrightarrow \quad 0.1 = 1 - e^{-t/(RC)}.
$$
Hence,
$$
e^{-t/(RC)} = 1 - 0.1 = 0.9.
$$
Step 4: Solve for $-\frac{t}{RC}$
Taking natural logarithm on both sides:
$$
-\frac{t}{RC} = \ln(0.9) = \ln\!\bigl(\tfrac{9}{10}\bigr) = -\ln\!\bigl(\tfrac{10}{9}\bigr).
$$
So,
$$
\frac{t}{RC} = \ln\!\Bigl(\tfrac{10}{9}\Bigr).
$$
Step 5: Substitute the Numeric Values to Find $C$
We know $t = 1 \times 10^{-6} \text{ s}$, $R = 10\,\Omega$, and $\ln\!\bigl(\tfrac{10}{9}\bigr) = 0.105$. Thus,
$$
\frac{1 \times 10^{-6}}{10 \cdot C} = 0.105.
$$
Rearranging to solve for $C$:
$$
C = \frac{1 \times 10^{-6}}{10 \times 0.105} \,.
$$
Calculate this value:
$$
C = \frac{1 \times 10^{-6}}{1.05} = 0.95 \times 10^{-6} \text{ F} = 0.95 \,\mu\text{F}.
$$
Final Answer
The capacitance of the capacitor is $0.95\,\mu\text{F}$.
