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

Question

A zener diode of voltage VZ(=6 V ) is used to maintain a constant voltage across a load resistance RL(=1000 Ω ) by using a series resistance Rs(=100 Ω). If the e.m.f. of source is E(=9 V ), what is the power being dissipated in Zener diode? (A) 0.144 watt (B) 0.324 watt (C) 0.244 watt (D) 0.544 watt

Tardigrade Admin · Accepted Answer

Here, E =9 V ; V z =6 ; R L =1000 W and R s =100 W Potential drop across series resistor V = E - V Z =9-6=3 V Current through series resistance RS is I =( V / R )=(3/100)=0.03 A Current through load resistance RL is I L =( V Z / R L )=(6/1000)=0.006 A Current through Zener diode is IZ=I-IL=0.03-0.006=0.024 amp. Power dissipated in Zener diode is PZ=VZ IZ=6 × 0.024=0.144 Watt

Q. A zener diode of voltage $V_{Z} (= 6 V)$ is used to maintain a constant voltage across a load resistance $R_{L} (= 1000 Ω$ ) by using a series resistance $R_{s} (= 100 Ω)$ . If the e.m.f. of source is $E (= 9 V)$ , what is the power being dissipated in Zener diode?

$0.144$ watt

$0.324$ watt

$0.244$ watt

$0.544$ watt

Q. A zener diode of voltage VZ​(=6V) 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(=9V), what is the power being dissipated in Zener diode?

0.144 watt

0.324 watt

0.244 watt

0.544 watt

Q. A zener diode of voltage $V_{Z} (= 6 V)$ is used to maintain a constant voltage across a load resistance $R_{L} (= 1000 Ω$ ) by using a series resistance $R_{s} (= 100 Ω)$ . If the e.m.f. of source is $E (= 9 V)$ , what is the power being dissipated in Zener diode?

$0.144$ watt

$0.324$ watt

$0.244$ watt

$0.544$ watt