#### Question

**Answer the following question.**

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

Given Bulbs are rated as (P_{1}, V) and (P_{2}, 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`.