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Question
Answer the following question.
Two bulbs are rated (P1, V) and (P2, 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 P1 and P2.
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Solution
Given Bulbs are rated as (P1, V) and (P2, V) respectively
The resistance of 1st bulbs `R_1 = V^2/P_1`
The resistance of 2nd bulbs `R_2 = V^2/P_2`
(i) When both are connected in series with a power supply of voltage V. As both the bulbs are in series connection hence both will have the same amount of current flowing through them.
i = `V/(R_1 + R_2) = V/(V^2/P_1 + V^2/P_2) = 1/V((P_1P_2)/(P_1 + P_2))`
Power dissipated in the circuit
`P_d = i^2(R_1 + R_2) = 1/V^2((P_1P_2)/(P_1 + P_2))^2(V^2/P_1 + V^2/P_2)`
`P_d = (P_1P_2)/(P_1 + P_2)`
(ii) When both are connected in parallel In this case, both bulbs will get the same voltage supply. Hence, power dissipated
`P_d = V^2/R_1 + V^2/R_2 = V^2(P_1/V^2 + P_2/V^2)`
`P_d = P_1 + P_2`.
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