A 220 V, 50 Hz AC source is connected to a 25 V, 5 W lamp and an additional resistance R in series (as shown in figure) to run the lamp at its peak brightness, then the value of R (in ohm) will be _____________.

Solution

Physics sir · 4 steps

Peak brightness means the lamp gets its rated current, so first find that current from the rating.

1. Rated current. A lamp glows at full brightness when it carries its rated current. From P = VI, that is I = 5/25 = 0.2 A. The 220 V is an rms value, so this 0.2 A is an rms current too.
2. Lamp resistance. Rₗₐₘₚ = V²/P = 25²/5 = 125 Ω.
3. Total resistance. The same 0.2 A flows through the whole series circuit, driven by 220 V. The circuit has only resistors, so there is no reactance and the two resistances add directly.

   R + 125 = 220/0.2 = 1100 Ω

4. So. R = 1100 - 125 = 975 Ω. As a check, R drops 220 - 25 = 195 V, and 195/0.2 = 975 Ω.

The common slip: 1100. Taking 220/0.2 as the answer gives 1100, the total resistance. You still have to subtract the lamp's own 125 Ω.

The idea to keep. To run a device at its rating, fix the current from I = P/V. Then the series resistor must drop the extra voltage at that current. With only resistors, the voltages add like plain numbers.

Answer from the NTA final answer key. Solution by PrepWiser.