A Zener voltage regulator circuit is constructed using a DC supply of 30 V, a series resistor of 100 , and a Zener diode with a breakdown voltage of 10 V. If a load resistor of 250 is connected in parallel with the Zener diode, the power dissipated by the Zener diode in the steady state is ______ mW.

Step-by-step solution

Given supply voltage V_S = 30 V, Zener voltage V_Z = 10 V, series resistance R_S = 100 , and load resistance R_L = 250 . The voltage drop across the series resistor is: V_ RS = V_S - V_Z = 30 - 10 = 20 V. The total current supplied by the source is: I_S = V_ RS R_S = 20 100 = 0.2 A. The current flowing through the load resistor is: I_L = V_Z R_L = 10 250 = 0.04 A. Applying Kirchhoff's Current Law, the current through the Zener diode is: I_Z = I_S - I_L = 0.2 - 0.04 = 0.16 A. The power dissipated by the Zener diode