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ADNAN MUHAMMED

Grade 12,

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I have posted my question as an attachement. Kindly refer to that .

     I have posted my question as an attachement.  Kindly refer to that .

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Grade:10

1 Answers

M H Rashid
35 Points
6 years ago
First we need to find out the resistance of the bulbs, then the equivalent resistance on series combination and finally the power consumption as required in the problem.
P=V I=\frac{V^{2}}{R}
\therefore R=\frac{V^{2}}{P}
Given: V1=V=200 V, P1= 60 W
           V2=V=200 V, P2=100 W
\therefore R_{1}=\frac{V^{2}}{P_{1}}
and
 R_{2}=\frac{V^{2}}{P_{2}}
Therefore the equivalent resistance is 
R=R_{1}+R_{2}=V^{2}\left \{ \frac{1}{P_{1}}+\frac{1}{P_{2}} \right \}=V^{2}\left \{ \frac{P_{1}+P_{2}}{P_{1}P_{2}} \right \}
Hence the Power consumption would be 
P=\frac{V^{2}}{R}=\frac{P_{1}P_{2}}{P_{1}+P_{2}} =\frac{60\times 100}{60+100}=\frac{75}{2}=37.5 W 

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