A single phase 220V, 1kW electric room radiator is connected across 220V supply through a triac. For a delay angle of 90°, the value of power dissipated by the radiator is:
32. A single phase 220V, 1kW electric room radiator is connected across 220V supply through a triac. For a delay angle of 90°, the value of power dissipated by the radiator is:
1. 49.985 W
2. 2500 W
3. 398.9 W
4. 500 W
Answer: 4
Power Dissipated by a Radiator Controlled by a TRIAC
Given
- Supply Voltage, V = 220 V
- Rated Power of Radiator, Pfull = 1000 W
- Delay (Firing) Angle, α = 90°
- Load Type = Purely Resistive
Formula
For a single-phase AC voltage controller using a TRIAC feeding a resistive load:
P = Pfull × [1 − (α/π) + sin(2α)/(2π)]
Calculation
Convert the firing angle into radians:
α = 90° = π/2
Substitute into the power equation:
P = 1000 × [1 − (π/2)/π + sin(π)/(2π)]
Since:
- (π/2)/π = 1/2
- sin(π) = 0
Therefore:
P = 1000 × (1 − 1/2)
P = 1000 × 0.5
P = 500 W
Answer
Power dissipated by the radiator = 500 W
Quick Shortcut
For a resistive load with a firing angle of 90°:
sin(2α) = sin(180°) = 0
Hence:
P = Pfull × (1 − 1/2) = Pfull/2
Therefore, a 1 kW radiator dissipates:
500 W