The graph above shows $v_0(t) = 8 e^{-\frac{t}{60}}$ for $t \geq 0$. As expected, the voltage $v_0(t)$ decreases exponentially from 8 volts towards 0 volts as time progresses.

Step 3: Find $i_0(t)$ at $t = 3s$

The current $i_0(t)$ through the $30 \, k\Omega$ resistor can be found using: $i_0(t) = \frac{v_0(t)}{R}$ At $t = 3s$: $v_0(3) = 8 e^{-\frac{3}{60}} = 8 e^{-0.05}$ Let’s calculate $v_0(3)$ and then find $i_0(3)$.At $t = 3 \, s$:

• The voltage $v_0(3)$ is approximately $7.61 \, V$.
• The current $i_0(3)$ is approximately $0.0002537 \, A$ or $253.7 \, \mu A$.

Final Answers:

1. $v_0(t) = 8 e^{-\frac{t}{60}}$ for $t \geq 0$
2. $i_0(t)$ at $t = 3 \, s$ is approximately $253.7 \, \mu A$.

Would you like further details or explanations?

5 Additional Questions for Expansion:

1. How does changing the resistance $R$ affect the time constant $\tau$ and the behavior of $v_0(t)$?
2. What happens to $v_0(t)$ if the initial capacitor voltage is different?
3. How would the circuit behavior change if a different capacitor value $C$ were used?
4. What is the steady-state behavior of $v_0(t)$ as $t \to \infty$?
5. Can you find the total energy dissipated in the resistor $30 \, k\Omega$ over time?

Tip

For RC circuits, the time constant $\tau = RC$ determines how quickly the capacitor charges or discharges. A larger $\tau$ means a slower change in voltage, while a smaller $\tau$ means a faster change.