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.

Answer 1

Given Bulbs are rated as (P₁, V) and (P₂, V) respectively  
The resistance of 1^st bulbs `R_1 = V^2/P_1`

The resistance of 2^nd 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`.