Two bulbs of 500, W and 200, W are manufactur | vedclass

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(A) The resistance $R$ of a bulb is given by $R = V^2/P$. For the $500\, W$ bulb,$R_1 = (220)^2 / 500$. For the $200\, W$ bulb,$R_2 = (220)^2 / 200$.

Vedclass · Accepted Answer

$5/2, 2/5$

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(A) The resistance $R$ of a bulb is given by $R = V^2/P$. For the $500\, W$ bulb,$R_1 = (220)^2 / 500$. For the $200\, W$ bulb,$R_2 = (220)^2 / 200$. Thus,$R_1/R_2 = 200/500 = 2/5$,which implies $R_2/R_1 = 5/2$. In parallel connection,the potential difference $V$ across both bulbs is the same. The heat produced $H = V^2t/R$. The ratio of heat produced is $H_1/H_2 = (V^2/R_1) / (V^2/R_2) = R_2/R_1 = 5/2$. In series connection,the current $I$ through both bulbs is the same. The heat produced $H = I^2Rt$. The ratio of heat produced is $H_1/H_2 = (I^2R_1) / (I^2R_2) = R_1/R_2 = 2/5$.

Two bulbs of $500\, W$ and $200\, W$ are manufactured to operate on a $220\, V$ line. What is the ratio of heat produced in the $500\, W$ bulb to the $200\, W$ bulb when they are connected in parallel,and when they are connected in series?

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