An electric bulb is rated as 220 V and 80 W. When it is operated on 110 V, the power rating would be:

The correct answer is option "4"

Concept:

• Electric Power: The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.,

\(P = \frac{W}{t} = VI = {I^2}R = \frac{{{V^2}}}{R}\)

Where V = Potential difference, R = Resistance and I = current.

Calculation:

Given - Potential difference (V) = 220 V, power of the bulb (P) = 80 W and actual voltage (V') = 110 V

• The resistance of the bulb can be calculated as,

​\(\Rightarrow R=\frac{V^2}{P}=\frac{(220)^2}{80}=605 \,\Omega\)

• The power consumed by the bulb.

\(\Rightarrow P=\frac{V^2}{R}=\frac{(110)^2}{605}=20 \,W\)