Two bulbs of 100, W and 200, W working at 220 | vedclass

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Question

(a) ${P_S} = \frac{{{P_1}{P_2}}}{{{P_1} + {P_2}}} = \frac{{100 	imes 200}}{{100 + 200}} = \frac{{200}}{3} \approx 65\,watt$

Vedclass · Accepted Answer

$65$

Vedclass · Answer

(a) ${P_S} = \frac{{{P_1}{P_2}}}{{{P_1} + {P_2}}} = \frac{{100 	imes 200}}{{100 + 200}} = \frac{{200}}{3} \approx 65\,watt$

Two bulbs of $100\, W$ and $200\, W$ working at $220$ $volt$ are joined in series with $220$ $volt$ supply. Total power consumed will be approximately ........... $watt$