Two electric bulbs, rated at (25, W, 220, V) | vedclass

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Question

Resistance, $R=\frac{V^{2}}{P}$

$\Rightarrow \quad R_{1}=\frac{(220)^{2}}{25}=1936 \,\Omega$

$R_{2}=\frac{(220)^{2}}{100}=484\, \Omega$

$\mat

Vedclass · Accepted Answer

$P_1 = 16\, W, P_2 = 4\, W$

Vedclass · Answer

Resistance, $R=\frac{V^{2}}{P}$

$\Rightarrow \quad R_{1}=\frac{(220)^{2}}{25}=1936 \,\Omega$

$R_{2}=\frac{(220)^{2}}{100}=484\, \Omega$

$\mathrm{i}=\frac{220}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{1}{11}\, \mathrm{A}$

Power dissipated through $\mathrm{R}_{1}=\mathrm{P}_{1}=\mathrm{i}^{2} \mathrm{R}_{1}=16\, \mathrm{W}$

Power dissipated through $\mathrm{R}_{2}=\mathrm{P}_{2}=\mathrm{i}^{2} \mathrm{R}_{2}=4 \,\mathrm{W}$

Two electric bulbs, rated at $(25\, W, 220\, V)$ and $(100\, W, 220\, V)$, are connected in series acroos a $220\, V$ voltage source. If the $25\, W$ and $100\, W$ bulbs draw powers $P_1$ and $P_2$ respectively, then

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