# Numerical - Power delivered to the load circuit for firing angles

## 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.

Options:

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

Correct Answer: d. 3.240 W & 1.097 W respectively

Explanation:

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