To solve this problem, we need to analyze the series RC circuit provided in the diagram. The circuit consists of a resistor $R = 10 \, \Omega$ and a capacitor $C = 0.1 \, \text{F}$, with a constant voltage $V = 20 \, \text{V}$ applied at $t = 0$.

Key Information:

1. Initial conditions: The capacitor is uncharged at $t = 0$.
2. Capacitor charging: As the circuit is powered, current flows, and the capacitor begins to charge according to the standard charging equation for RC circuits.

The current through the capacitor $i_C(t)$ in an RC charging circuit is given by the equation: $i_C(t) = \frac{V}{R} e^{-t/RC}$ Where:

• $V$ is the applied voltage (20 V),
• $R$ is the resistance (10 Ω),
• $C$ is the capacitance (0.1 F),
• $t$ is time in seconds.

The time constant for an RC circuit is given by: $\tau = RC = 10 \, \Omega \times 0.1 \, \text{F} = 1 \, \text{second}$

Thus, the equation for the current becomes: $i_C(t) = \frac{20}{10} e^{-t/1} = 2 e^{-t} \, \text{A}$

Therefore, the correct expression for the current through the capacitor is: $\boxed{i_C(t) = 2 e^{-t} \, \text{A}}$

The correct answer from the given options is: $\boxed{d. \, i_C = 2 e^{-t} \, \text{A}}$

Would you like further explanation on RC circuits or have any specific questions?

Here are five related questions:

1. How does the time constant $RC$ affect the charging time of the capacitor?
2. What happens to the current after a long time (as $t \to \infty$) in an RC circuit?
3. Can you describe the voltage across the capacitor during charging?
4. How would increasing the resistance affect the current in the circuit?
5. What would the current expression be if the initial voltage were different?

Tip: The time constant $\tau = RC$ is a key factor in determining how fast a capacitor charges or discharges in an RC circuit.