Two electric bulbs marked $40\,W,$ $220\,V$ and $60\,W,\,\,220\,V$ when connected in series across same voltage supply of $220\,V,$ the effective power is $P_1$ and when connected in parallel, the effective power is $P_2.$ Then $\frac {P_1}{P_2}$ is
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Resistance of bulb $ = \frac{{{{{\rm{(Rated\,\, voltage)}}}^2}}}{{{\rm{ (Rated\,\, power) }}}}$
when the bulbs are connected in series
${{\rm{R}}_{\rm{s}}} = {{\rm{R}}_{{{\rm{B}}_1}}} + {{\rm{R}}_{{{\rm{B}}_2}}} = \frac{{{{(220)}^2}}}{{40}} + \frac{{(220)}}{{60}}$
${=(220)^{2}\left[\frac{1}{40}+\frac{1}{60}\right]=(220)^{2}\left(\frac{100}{2400}\right)=\frac{(220)^{2}}{24}}$
$\therefore \quad \mathrm{P}_{1}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}_{2}}=(220)^{2} \times \frac{24}{(220)^{2}}=24 \mathrm{\,W}$
$P_{2}=60+40=100 \mathrm{\,W}$
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