Three 60, W light bulbs are mistakenly wired | vedclass

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Question

(A) The rated power $P_r = 60\, W$ at rated voltage $V_r = 120\, V$. The resistance of each bulb is $R = \frac{V_r^2}{P_r} = \frac{120^2}{60} = 240\,

Vedclass · Accepted Answer

$6.7$

Vedclass · Answer

(A) The rated power $P_r = 60\, W$ at rated voltage $V_r = 120\, V$. The resistance of each bulb is $R = \frac{V_r^2}{P_r} = \frac{120^2}{60} = 240\, \Omega$.When three identical bulbs are connected in series to a $120\, V$ supply,the total resistance is $R_{eq} = 3R = 3 	imes 240 = 720\, \Omega$.The current flowing through the series circuit is $I = \frac{V}{R_{eq}} = \frac{120}{720} = \frac{1}{6}\, A$.The power dissipated by each bulb is $P = I^2 R = (\frac{1}{6})^2 	imes 240 = \frac{1}{36} 	imes 240 = \frac{240}{36} = 6.67\, W \approx 6.7\, W$.

Three $60\, W$ light bulbs are mistakenly wired in series and connected to a $120\, V$ power supply. Assume the light bulbs are rated for single connection to $120\, V$. With the mistaken connection,the power dissipated by each bulb is: .................. $W$

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