A Zener diode is connected to a battery and a load as shown. The currents I, IZ and IL are respectively?

As long as the Vin is greater than the zener Voltage 10 V, the zener is in breakdown region and hence the voltage across the load remains constant. The series limiting resistance 4 kΩ limits the input current. Current in load IL = 10/2 × 10³ = 5 mA Voltage drop across the 4 kΩ is 60 − 10 = 50 V Hence I = 50/4 × 10³ = 12.5 mA Hence, IZ = I − IL = 12.5 − 5 = 7.5 mA