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, P_full = 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 = P_full × [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 = P_full × (1 − 1/2) = P_full/2

Therefore, a 1 kW radiator dissipates:

500 W

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