A zener diode of voltage Vz (=6V) is used to maintain a constant voltage across a load resistance RL (=1000ohm) by using a series resistance Rs (=100ohm) .If the emf of source is E (=9V),calculate the value of current through series resistance,Zener diode and load resistance. What is the power being dissipated in Zener diode?
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see the diagram.
E = 9 V = main battery emf supply
Vz = zener diode regulated voltage = 6 V across load
load = RL = 1, 000 Ω
Shunt resistance = Rs = 100 Ω
Let resistance of the zener = r Ω
Since voltage across RL = 6V, current I1 = 6V/1000Ω = 6 mA
Current I = (E - Vz)/Rs
= (9 - 6) / 100 = 30 mA
So current through the Zener is 30 - 6 = 24 m A = Iz
Equivalent Resistance of zener = 6 V/24 mA = 250 Ω = r
Power dissipation in zener diode = Vz Iz
P = 6 V * 24 mA = 0.144 W
E = 9 V = main battery emf supply
Vz = zener diode regulated voltage = 6 V across load
load = RL = 1, 000 Ω
Shunt resistance = Rs = 100 Ω
Let resistance of the zener = r Ω
Since voltage across RL = 6V, current I1 = 6V/1000Ω = 6 mA
Current I = (E - Vz)/Rs
= (9 - 6) / 100 = 30 mA
So current through the Zener is 30 - 6 = 24 m A = Iz
Equivalent Resistance of zener = 6 V/24 mA = 250 Ω = r
Power dissipation in zener diode = Vz Iz
P = 6 V * 24 mA = 0.144 W
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hii mate you anewer is 0.44 watt
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