(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\,

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(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 $R_{eq} = 3R = 3 \times 240 = 720\, \Omega$.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 $P = I^2 R = (\frac{1}{6})^2 \times 240 = \frac{1}{36} \times 240 = \frac{240}{36} = 6.67\, W \approx 6.7\, W$.

Class 12 Physics Current Electricity Mix Examples-Current Electricity

Medium

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: .................. $W$

A
$6.7$
B
$13.3$
C
$20$
D
$40$

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Mix Examples-Current Electricity

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$A$. Mobility	$P$. $\frac{m}{ne^2 \rho}$
$B$. Electrical conductivity	$Q$. $neAv_{d}$
$C$. Relaxation period	$R$. $\frac{v_{d}}{E}$
$D$. Current	$S$. $\frac{J}{E}$

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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If two wires having resistance $R$ and $2R$ are joined in series and then in parallel,what is the ratio of heat generated in these two situations,assuming the same voltage is applied?

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