Two electric bulbs with ratings 25, W - 220, | vedclass

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(B) First,calculate the resistance of each bulb using $R = V^2 / P$.For the first bulb $(P_1 = 25\, W, V = 220\, V)$: $R_1 = (220)^2 / 25 = 48400 / 25

Vedclass · Accepted Answer

$25\, W$

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(B) First,calculate the resistance of each bulb using $R = V^2 / P$.For the first bulb $(P_1 = 25\, W, V = 220\, V)$: $R_1 = (220)^2 / 25 = 48400 / 25 = 1936\, \Omega$.For the second bulb $(P_2 = 100\, W, V = 220\, V)$: $R_2 = (220)^2 / 100 = 48400 / 100 = 484\, \Omega$.When connected in series to a $440\, V$ supply,the total resistance is $R_{eq} = R_1 + R_2 = 1936 + 484 = 2420\, \Omega$.The current in the series circuit is $I = V / R_{eq} = 440 / 2420 = 44 / 242 = 2 / 11\, A$.The voltage drop across the first bulb is $V_1 = I 	imes R_1 = (2 / 11) 	imes 1936 = 2 	imes 176 = 352\, V$.The voltage drop across the second bulb is $V_2 = I 	imes R_2 = (2 / 11) 	imes 484 = 2 	imes 44 = 88\, V$.Since the voltage drop across the $25\, W$ bulb $(352\, V)$ exceeds its rated voltage $(220\, V)$,it will fuse.

Two electric bulbs with ratings $25\, W - 220\, V$ and $100\, W - 220\, V$ are connected in series to a $440\, V$ supply. Which bulb will fuse?

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