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A light bulb is at 100 W for a 220 V supply . Find (a) the resistance of the bulb , (b) the peak voltage of the source , and (c ) the rms current through the bulb. |
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Answer» <P> Solution :a. We are GIVEN P = 100 W and V = 220V. The resistance of the bulb is`R = (V^2)/(P) = ((220)^2)/(100) = 484 Omega` B. The peak voltage of the source is `v_m =sqrt2 V = 311 V` C . Since P = IV `I = P/V= 100/220 = 0.450 A` |
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