Question

the filament?
An electric bulb is rated at 220V – 200W. What is the resistance of filament​

Answer 1

The power dissipated by an electric bulb is given as P = VI.

In this case, Power P = 100 W and Voltage V = 220 V.

So, current I is I = P/V = 100/220 = 0.45 A.

When the supplied voltage is 110 V, the power consumed will be P=VI = 110 * 0.45 = 50 W.

Hence, the power consumed when operated at 110 V is 50 W.

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