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

The initial voltage across the capacitor is 10 V. After the switch is closed at t = 0, the circuit forms an RC discharge path through the Zener diode, which clamps the voltage to 5 V. The energy dissipated in the Zener diode can be calculated as follows:

1. Initial Energy Stored in Capacitor:

The energy (E) stored in a capacitor is given by: E i = 1 2 C V 2 

where C = 10 μF = 10 × 10^-6 F and V = 10 V.

E i = 1 2 × 10 × 10 -6 × 10 2 = 0.5 mJ

1. Final Energy Stored in Capacitor:

When the voltage drops to 5 V, the energy in the capacitor is: E f = 1 2 × 10 × 10 -6 × 5 2 = 0.250 mJ

1. Energy Dissipated in Zener Diode:

The energy dissipated is the difference between the initial and final energies: E d = E i - E f = 0.5 - 0.250 = 0.25 mJ

Thus, the total energy dissipated in the Zener diode is 0.25 mJ.

This falls outside the provided expected range of 0.25 mJ to 0.25 mJ, indicating a possible oversight in the range specification or circuit values.