In the circuit given, the reverse breakdown voltage of the Zener diode is 4.8 V. The current through the Zener and the power dissipation in Zener is:

Step-by-step solution

Given: Input voltage, V = 12 V Zener voltage, V_z = 4.8 V Series resistance, R_s = 250 Load resistance, R_L = 750 First, we check if the Zener diode is in the breakdown region. The voltage across the load without the Zener diode is: V' = R_L R_s + R_L V = 750 250 + 750 12 = 9 V Since V' > V_z , the Zener diode is in breakdown and the voltage across the load is clamped at V_z = 4.8 V . The current through the series resistor R_s is: I = V - V_z R_s = 12 - 4.8 250 = 7.2 250 = 0.0288 A = 28.8 mA The current through th