A zener diode of voltage V _ Z (=6 ~V ) is used to maintain a constant voltage across a load resistance R _ L (=1000 ) by using a series resistance R_ s (=100 ). If the e.m.f. of source is E (=9 ~V ), what is the power being dissipated in Zener diode?
A. 0.144 watt
Here, E=9 ~V ; V _ z =6 ; R _ L =1000 and R _ s =100 , Potential drop across series resistor V = E - V _ Z =9-6=3 ~V Current through series resistance R_ S is I = V R = 3 100 =0.03 ~A Current through load resistance R_ L is I _ L = V _ Z R _ L = 6 1000 =0.006 ~A Current through Zener diode is I _ Z = I - I _ L =0.03-0.006=0.024 amp . Power dissipated in Zener diode is P _ Z = V _ Z I _ Z =6 0.024=0.144 Watt
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