Two electric bulbs marked 40,W, 220,V and 60, | vedclass

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Question

Resistance of bulb $ = \frac{{{{{\rm{(Rated\,\,  voltage)}}}^2}}}{{{\rm{ (Rated\,\, power) }}}}$

when the bulbs are connected in series

${{

Vedclass · Accepted Answer

$0.24$

Vedclass · Answer

Resistance of bulb $ = \frac{{{{{m{(Rated\,\,  voltage)}}}^2}}}{{{m{ (Rated\,\, power) }}}}$

when the bulbs are connected in series

${{m{R}}_{m{s}}} = {{m{R}}_{{{m{B}}_1}}} + {{m{R}}_{{{m{B}}_2}}} = \frac{{{{(220)}^2}}}{{40}} + \frac{{(220)}}{{60}}$

${=(220)^{2}\left[\frac{1}{40}+\frac{1}{60}ight]=(220)^{2}\left(\frac{100}{2400}ight)=\frac{(220)^{2}}{24}}$

$	herefore \quad \mathrm{P}_{1}=\frac{\mathrm{V}_{1}^{2}}{\mathrm{R}_{2}}=(220)^{2} 	imes \frac{24}{(220)^{2}}=24 \mathrm{\,W}$

$P_{2}=60+40=100 \mathrm{\,W}$

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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