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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. - Physics


Answer in Brief

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.


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`.

Concept: Cells in Series and in Parallel
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