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CBSE Class 12 Physics 2019 Delhi Set 1 Paper

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Question : 6 of 27
Marks: +1, -0
SECTION -B
Two bulbs are rated (P1,V)(P_1, V) and (P2,V)(P_2, V). If they are connected (i) in series and (ii) in parallel across a supply VV, find the power dissipated in the two combinations in terms of P1P_1 and P2P_2.
Solution:  
Calculation of Power dissipation in two combinations
R1=V2P1,R2=V2P2,R_1 = \frac{V^2}{P_1}, R_2 = \frac{V^2}{P_2},
Ps=V2Rs=P1P2P1+P2P_s = \frac{V^2}{R_s} = \frac{P_1 P_2}{P_1+P_2}
1Ps=1P1+1P2\frac{1}{P_s} = \frac{1}{P_1} + \frac{1}{P_2}
1Rp=1R1+1R2=P1+P2V2\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{P_1+P_2}{V^2}
∴Pp=V2Rp=P1+P2\therefore P_p = \frac{V^2}{R_p} = P_1+P_2
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