Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time combination is connected to a 220 volt a.c. supply line. The power drawn by the combination in each case respectively will be:

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{R}_1=\frac{\mathrm{V}^2}{\mathrm{P}_1}=\frac{(220)^2}{100}=484 \Omega \ & \mathrm{R}_2=\frac{\mathrm{V}^2}{\mathrm{P}_2}=\frac{(220)^2}{100}=484 \Omega \ & \mathrm{I}=\frac{220}{484 \times 2}=\frac{5}{22} \ & \mathrm{P}=\mathrm{I}^2\left(\mathrm{R}_1+\mathrm{R}_2\right)=\frac{25}{22 \times 22} \times(484 \times 2) \ & =50 \mathrm{~W}\end{aligned}$ $\begin{aligned} & P=\frac{V^2}{R_1}+\frac{V^2}{R_2} \ & =\frac{2 \times(220)^2}{484}=200 \mathrm{~W}\end{aligned}$

Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time combination is…

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