Resistance of an electric bulb changes with temperature. If an electric bulb rated 220 V ,100 w is connected to a source voltage 220*0.8 volt then actual power will be ?

$$P_1=\frac{{(220)}^2}{R_1}$$ and $$P_2=\frac{{(220\times 0.8)}^2}{R_2}$$ $$\frac{P_2}{P_1}=\frac{{(220\times 0.8)}^2}{{(220)}^2}\times \frac{R_1}{R_2}$$ Þ$$\frac{P_2}{P_1}={(0.8)}^2\times \frac{R_1}{R_2}$$ Here R2 1\] Þ $$P_2>{(0.8)}^2{P}_1$$ Þ $$P_2>{(0.8)}^2\times 100W$$ Also $$\frac{P_2}{P_1}=\frac{(220\times 0.8)i_2}{220i_1},$$ Since \[{{i}_{2}}

Rated Power(P)=100WRated voltage(V)=220VTherefore, Current flowing through it will be 100/220A (Because P=VI)Now, source voltage =0.8*220VTherefore,actual power will be (0.8*220)(100/220)=80W