# 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 , R_{L} = 2 kΩ

For θ = 450 , the power delivered can be given by,

P = V_{dc} x I_{dc}

But, for a half-wave controlled rectifier,

V_{dc} = V_{m} / 2Л ( 1 + cos θ )

= V_{m} / 2Л ( 1+ cos 45)

= V_{m} / 2Л ( 1+ 0.707)

V_{dc} = 0.27 V_{m}

= 0.27 x 300

V_{dc} = 81 V

I_{dc} = V_{dc} / R_{L} = 81 / 2000 = 40 mA

Therefore, P = V_{dc} x I_{dc} = 81 x 40 = 3240 mW = 3.240 W

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

V_{dc} = V_{m} / 2Л ( 1 + cos θ )

= V_{m} / 2Л ( 1 + cos 900)

= V_{m} / 2Л ( 1 + 0 )

= V_{m} / 2Л

= 0.159 V_{m}

= 0.159 x 300

V_{dc} = 47.7 V

I_{dc} = V_{dc} / R_{L } = 47.7 / 2000 = 23 mA

Hence, P = V_{dc} x I_{dc} = 47.7 x 23 = 1097.1mW = 1.097 W