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: .................. $W$
Medium
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Here, three $60 \mathrm{W}$ bulbs are connected in series,
Resistance of each bulb is
$R_{1}=\frac{(120)^{2}}{60}=\frac{14400}{60}=240 \Omega$
In series connection
$R=R_{1}+R_{2}+R_{3}=720 \Omega....\left(\text {sincce } R_{1}=R_{2}=R_{3}\right)$
Hence, current will be
$I=\frac{120}{720}=0.166 A$
Power dissipated by each bulb
$P=I^{2} R=(0.166)^{2} \times 240=6.7 W$
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