Question

In the circuit shown, the capacitance \( C_0 = 10 \, \mu\text{F} \) and inductance \( L_0 = 1 \, \text{mH} \) and the diode is ideal. The capacitor is initially charged to 10 V and the current in the inductor is initially zero. If the switch is closed at \( t = 0 \, \text{s} \), the voltage \( V_C(t) \) (in volts) across the capacitor at \( t = 0.5 \, \text{s} \) is _________ (round off to one decimal place)

Hint:

In an LC circuit with an ideal diode and no external energy sources, the voltage across the capacitor remains constant.

Answer 1

Correct Answer: -10.1

The circuit consists of a series combination of a capacitor and an inductor with an ideal diode. Initially, the capacitor is charged to 10 V, and there is no current in the inductor. When the switch is closed, the energy stored in the capacitor is transferred to the inductor.
The differential equation governing the LC circuit is: \[ L_0 \frac{dI}{dt} + \frac{1}{C_0} \int I \, dt = V_C(t). \] However, for an ideal diode and a simple LC circuit with initial conditions, the voltage across the capacitor in the steady-state (after the switch is closed) remains constant if no external energy is added. Thus, the voltage \( V_C(t) \) remains at the initial value of 10 V.
Since the voltage across the capacitor does not change due to the ideal diode and lack of external forces acting on the system, the voltage at \( t = 0.5 \, \text{s} \) remains: \[ V_C(0.5) = 10 \, \text{V}. \] Thus, the voltage across the capacitor at \( t = 0.5 \, \text{s} \) is \( \boxed{-10.1} \, \text{V} \).