Question

As shown in the circuit, the initial voltage across the capacitor is 10 V, with the switch being open. The switch is then closed at t 0. The total energy dissipated = in the ideal Zener diode (V_z=5V) after the switch is____.

Hint:

When capacitors discharge into Zener diodes, calculate the energy dissipated as the difference in stored energy before and after clamping. Always check the voltage clamping level for accurate results.

Answer 1

Correct Answer: 0.25

Step 1: Compute the initial energy stored in the capacitor.
The energy stored in a capacitor is given by: \[ E = \frac{1}{2} C V^2 \] Substitute \( C = 10 \,\mu F = 10 \times 10^{-6} \,F \) and \( V = 10 \,V \): \[ E_{\text{initial}} = \frac{1}{2} \times 10 \times 10^{-6} \times (10)^2 = 0.0005 \,J \]
Step 2: Compute the final energy stored in the capacitor.
After the switch is closed, the voltage across the capacitor is clamped to \( V_Z = 5 \,V \). The energy stored becomes: \[ E_{\text{final}} = \frac{1}{2} \times 10 \times 10^{-6} \times (5)^2 = 0.000250 \,J \]
Step 3: Calculate the energy dissipated in the Zener diode.
The energy dissipated is the difference between the initial and final energies: \[ E_{\text{dissipated}} = E_{\text{initial}} - E_{\text{final}} = 0.0005 - 0.000250 = 0.000250 \,J \] Convert this to millijoules: \[ E_{\text{dissipated}} = 0.250 \,mJ \]
Final Answer:
\[ \boxed{0.250} \]