CBSE Class 12 Physics 2019 Delhi Set 1 Paper

Question : 6 of 27

Marks: +1, -0

SECTION -B

Two bulbs are rated

(P_1, V)

and

(P_2, V)

. If they are connected (i) in series and (ii) in parallel across a supply

V

, find the power dissipated in the two combinations in terms of

P_1

and

P_2

Solution:

Calculation of Power dissipation in two combinations

R_1\;=\;{V^2}/{P_1}, R_2=\;{V^2}/{P_2},

P_s\;=\;{V^2}/{R_s}=\;{P_1 P_2}/{P_1+P_2}

\;{1}/{P_s}\;=\;{1}/{P_1}+\;{1}/{P_2}

\;{1}/{R_p}\;=\;{1}/{R_1}+\;{1}/{R_2}=\;{P_1+P_2}/{V^2}

∴ P_p\;=\;{V^2}/{R_p}=P_1+P_2

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