Numerical - Power delivered to the load circuit for firing angles

Q.  Estimate the power delivered to the load circuit for firing angles of 450  & 900  respectively in a controlled form of half-rectifier circuit with peak supply voltage of about 300 V across the load resistor of 2 kΩ.
- Published on 20 Oct 15

a. 0.502 W & 0.244 W respectively
b. 1.240 W & 0.062 W respectively
c. 2.120 W & 1.670 W respectively
d. 3.240 W & 1.097 W respectively

ANSWER: 3.240 W & 1.097 W respectively
 
> Given data:
Vm = 300 V , RL = 2 kΩ

For θ = 450 , the power delivered can be given by,
P = Vdc x Idc

But, for a half-wave controlled rectifier,

Vdc = Vm / 2Л (1 + cos θ )

= Vm / 2Л (1+ cos 45)

= Vm / 2Л ( 1+ 0.707)

Vdc = 0.27 Vm

= 0.27 x 300

Vdc = 81 V

Idc = Vdc / RL = 81 / 2000 = 40 mA

Therefore, P = Vdc x Idc = 81 x 40 = 3240 mW = 3.240 W

Similarly, the power delivered at an angle θ = 900 can be estimated as,

Vdc = Vm / 2Л (1 + cos θ )

= Vm / 2Л (1 + cos 900)

= Vm / 2Л (1 + 0)

= Vm / 2Л

= 0.159 Vm

= 0.159 x 300

Vdc = 47.7 V

Idc = Vdc / RL = 47.7 / 2000 = 23 mA

Hence, P = Vdc x Idc = 47.7 x 23 = 1097.1mW = 1.097 W

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