An electrical bulb rated 220,V , 100,W, is co | vedclass

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Question

$R _{1}=\frac{ V ^{2}}{ P }=\frac{220^{2}}{100}=484$

$R _{2}=\frac{ V ^{2}}{ P }=\frac{220^{2}}{60}=484\left(\frac{10}{6}\right)$

$I =\frac{220}

Vedclass · Accepted Answer

$14$

Vedclass · Answer

$R _{1}=\frac{ V ^{2}}{ P }=\frac{220^{2}}{100}=484$

$R _{2}=\frac{ V ^{2}}{ P }=\frac{220^{2}}{60}=484\left(\frac{10}{6}ight)$

$I =\frac{220}{484+484 	imes \frac{10}{6}}$

$P _{1}= I ^{2} R _{1}=14.06\,W$

An electrical bulb rated $220\,V , 100\,W$, is connected in series with another bulb rated $220\,V$, $60\,W$.If the voltage across combination is $220\,V$, the power consumed by the $100\,W$ bulb will be about $........... W$

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