The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated $$220$$ volt and $$100$$ watt is connected to a (220$$ x $$.8) volt source, then the actual power would be

Question

The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated $$220$$ volt and $$100$$ watt is connected to a (220$$  x $$.8) volt source, then the actual power would be

Infinity Learn · Accepted Answer

$$P1=$$ $$2202R1$$ and $$P1=$$ $$220$$ × $$0.82R2P2P1$$ $$=$$ $$220$$ × $$0.822202$$ × $$R1R2$$⇒ $$P2P1$$ $$=$$ $$0.82$$ × $$R1R2Here$$ $$R2$$ < $$R1(because$$ voltage decreases from $$220$$ V → $$220$$ x $$0.8$$ V. It means heat produced → $$decreases)So$$ $$R1R2$$ > $$1$$ ⇒ $$P2$$ $$0.82$$ $$P1$$ ⇒ $$P2$$ > $$0.82$$ × $$100$$ WAlso $$P2P1$$ $$=$$ $$220$$ × $$0.8i2220$$ $$i1$$, Since $$i2$$ < $$i1$$ we expectSo $$P2P1$$ < $$0.8$$ ⇒ $$P2$$ < $$100$$ × $$0.8Hence$$, the actual power would be between $$100$$ x (0.8)2$$ W and $$(100$$ x $$0.8) W

The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated 220 volt and 100 watt is connected to a (220 x .8) volt source, then the actual power would be

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