A Zener diode of voltage VZ(=6 V) is used to maintain a constant voltage across a load resistance RL (=1000Ω) by using a series resistance Rs(=100Ω). If the e.m.f. of source is E(=9 V), what is the power (in watt) being dissipated in Zener diode ?

Here, E=9 V;Vz=6V;RL=1000Ω and Rs=100Ω Potential drop across series resistor V=E−VZ=9−6=3 V Current through series resistance RS is I=VR=3100=0.03 A Current through load resistance RL is IL=VZRL=61000=0.006 A Current through Zener diode is IZ=I−IL=0.03−0.006=0.024amp Power dissipated in Zener diode is PZ=VZIZ=6×0.024=0.144watt