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  • Two electric bulb rated at <img alt="25 \mathrm{~W}, 200 \mathrm{~V}$ and $100 \mathrm{~W}$, $200 \mathrm{~V}" src="https://entrancecorner.oncodecogs.com/gif.latex?25%20%5Cmathrm%7B%7EW%7D%2C%20200

Two electric bulb rated at 25 \mathrm{~W}, 200 \mathrm{~V}$ and $100 \mathrm{~W}$, $200 \mathrm{~V} are connected in series across a 200 \mathrm{~V} source. The 25 \mathrm{~W} and \mathrm{100W} bulb now draw \mathrm{P_1 \: and \: P_2} powers respectively, then -
 

Option: 1

\mathrm{P_1=4 \mathrm{~W}}
 


Option: 2

\mathrm{P_2=4 w}
 


Option: 3

\mathrm{P_r=18 w}
 


Option: 4

\mathrm{P_2=20 \mathrm{~W}}


Answers (1)

best_answer

Use, \mathrm{P=\frac{v^2}{R}}
then \mathrm{R_1=\frac{V^2}{R_1}=\frac{(200)^2}{25}=1600 \Omega}

\mathrm{ \text { \& } R_2=\frac{(200)^2}{100}=400 \Omega }

In series, \mathrm{ i=\frac{V}{R_1+R_2}=\frac{200}{1600+400}=\frac{A}{10} }

Now, \mathrm{ \quad P_1=i^2 R_1=\left(\frac{1}{10}\right)^2 \times 1600=16 \times watt }

and \mathrm{ \quad P_2=\left(\frac{1}{10}\right)^2 \times 400=4 \omega }

Hence option 2 is correct.

Posted by

mansi

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