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 \[{{P}_{2}}>{{(0.8)}^{2}}\times 100\,W\] Also \[\frac{{{P}_{2}}}{{{P}_{1}}}=\frac{(220\times 0.8){{i}_{2}}}{220\,{{i}_{1}}},\] 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