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  • An electric bulb rated 220 v and 60 W is connected in series with another electric bulb rated 220 v and 40 W. The combination is connected across 220 volt source of e.m.f. Which bulb will glow more

An electric bulb rated 220 v and 60 W is connected in series with another electric bulb rated 220 v and 40 W. The combination is connected across 220 volt source of e.m.f. Which bulb will glow more?

Option: 1

P_1^{\prime}>P_2^{\prime}


Option: 2

\mathrm{P}_1^{\prime}<\mathrm{P}_2^{\prime}


Option: 3

P_1^{\prime}=P_2^{\prime}


Option: 4

P_1^{\prime} / P_2^{\prime}


Answers (1)

best_answer

R=\frac{V^2}{P}
\therefore resistance of first bulb is R_1=\frac{V^2}{P_1}
and resistance of the second bulb is \mathrm{R}_2=\frac{\mathrm{V}^2}{\mathrm{P}_2}
n series same current will pass through each bulb
\therefore Power developed across first is P_1^{\prime}=I^2 \frac{V^2}{P_1}
and that across second is P_2^{\prime}=I^2 \frac{V^2}{P_2}
\Rightarrow \quad \frac{\mathrm{P}_1^{\prime}}{\mathrm{P}_2^{\prime}}=\frac{\mathrm{P}_2}{\mathrm{P}_1}
\begin{array}{lll} \text { as } \mathrm{P}_2<\mathrm{P}_1 & \Rightarrow \quad \frac{\mathrm{P}_2}{\mathrm{P}_1}<1 \\ \Rightarrow \quad \frac{\mathrm{P}_1^{\prime}}{\mathrm{P}_2^{\prime}}<1 \quad & \Rightarrow \quad \mathrm{P}_1^{\prime}<\mathrm{P}_2^{\prime} \end{array}
The bulb rated 220 V & 40 W will glow more

Posted by

Gautam harsolia

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