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 \((V_Z = 5 \, {V})\) after the switch is closed (in mJ, rounded off to three decimal places) 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

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 \(V = 10 \, {V}\): \[ E_{{initial}} = \frac{1}{2} \cdot 10 \times 10^{-6} \cdot (10)^2 = 0.0005 \, {J}. \] 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: \[ E_{{final}} = \frac{1}{2} \cdot 10 \times 10^{-6} \cdot (5)^2 = 0.000250 \, {J}. \] Step 3: Calculate the energy dissipated in the Zener diode.
The energy dissipated is the difference between the initial and final energies: \[ E_{{dissipated}} = E_{{initial}} - E_{{final}} = 0.0005 - 0.000250 = 0.000250 \, {J}. \] Convert this to millijoules: \[ E_{{dissipated}} = 0.250 \, {mJ}. \] Final Answer: \[ \boxed{{0.250}} \]