Question

An electric bulb rated 220V, 100W is connected to a source of 220 (0.8)V then actual power will be

• A. 100(0.8) W
• B. $100(0.8{)}^{2}W$
• C. > 100(0.8) W but < 100 W
• D. > $100(0.8{)}^{2}Wbut<100(0.8)W$

Answer 1

Answer: (D)

Using P = $\frac{V^2}{R}$, we get R = $\frac{V^2}{P}$ $\therefore$ resistance of bulb $\frac{220^2}{100}\Omega$
Power of bulb across changed voltage, P ' = $\frac{[220(0.8){]}^2}{220^2/100}$ = 100(0.8)${}^2$ W
But in fact resistance of bulb does not remain constant due to change in temperature of the filament. Therefore, at reduced voltage the temperature of filament will be less i.e. resistance of filament will be less and giving power more than 100(0.8)${}^2$ W.