We are given an electrical circuit with two voltage sources (10V and 5V), a capacitor with capacitance $\frac{1}{2} \, \text{F}$, and two resistors (5Ω and 10Ω). The problem asks us to:

1. Solve for the capacitor voltage ($v_c$) and current ($i_c$) at times $t=0^-$ (before the switch is closed), $t=0^+$ (just after the switch is closed), and $t \to \infty$ (long after the switch is closed).
2. Determine whether the capacitor is charging or discharging for $t > 0$.

Step-by-Step Solution

Step 1: Analyze the Circuit for $t=0^-$

At $t=0^-$, the switch is open. The capacitor has been disconnected from the 10V source for a long time, so it has been in the right-hand part of the circuit with the 5V source and 10Ω resistor.

• Since the capacitor will have reached steady-state, the voltage across it will be equal to the 5V from the right-hand source.
• No current flows through the capacitor because it's fully charged in steady-state conditions (infinite resistance for DC at steady state).

Therefore, for $t=0^-$: $v_c(0^-) = 5 \, \text{V}, \quad i_c(0^-) = 0 \, \text{A}.$

Step 2: Analyze the Circuit for $t=0^+$

At $t=0^+$, the switch closes, connecting the capacitor to the 10V source and the 5Ω resistor.

• The capacitor initially has a voltage of 5V across it.
• There will be a sudden change in current, but the voltage across the capacitor cannot change instantaneously (it stays at 5V immediately after $t=0$).

At $t=0^+$, the current through the capacitor can be found using Kirchhoff’s Voltage Law (KVL): $10V - 5\Omega \cdot i_c(0^+) - v_c(0^+) = 0$ Substitute $v_c(0^+) = 5V$: $10V - 5\Omega \cdot i_c(0^+) - 5V = 0$ $5 = 5\Omega \cdot i_c(0^+)$ $i_c(0^+) = 1 \, \text{A}.$

Therefore, at $t=0^+$: $v_c(0^+) = 5 \, \text{V}, \quad i_c(0^+) = 1 \, \text{A}.$

Step 3: Analyze the Circuit for $t \to \infty$

As $t \to \infty$, the capacitor will again reach steady-state, meaning no current flows through it, and it acts like an open circuit.

• In this case, the capacitor will charge until its voltage matches the voltage of the 10V source on the left side.
• No current will flow in steady-state because the capacitor acts like an open circuit for DC.

Therefore, for $t \to \infty$: $v_c(\infty) = 10 \, \text{V}, \quad i_c(\infty) = 0 \, \text{A}.$

Step 4: Is the Capacitor Charging or Discharging for $t>0$?

Since the capacitor's voltage increases from 5V to 10V after the switch is closed, it is charging for $t > 0$.

Summary of Results

• $v_c(0^-) = 5V$, $i_c(0^-) = 0A$
• $v_c(0^+) = 5V$, $i_c(0^+) = 1A$
• $v_c(\infty) = 10V$, $i_c(\infty) = 0A$
• The capacitor is charging for $t > 0$.

Would you like more details on any part of this process or further clarifications?

Here are five related questions to expand on this:

1. How would the time constant of this RC circuit be calculated?
2. What is the general equation for the capacitor voltage $v_c(t)$ for $t > 0$?
3. How does the current in the circuit behave over time after the switch is closed?
4. What happens if the capacitance were doubled?
5. How would the circuit behave if the resistor values were interchanged?

Tip: When solving transient circuits, always remember that capacitor voltage cannot change instantaneously, but the current can.