Question

An electric heater can boil a certain amount of water in 10 minutes and another heater can do it in 15 minutes, both working at the same voltage. If the two heaters are connected parallel across the same voltage as before, how much time will they take to boil the same amount of water?

Answer 1

Let R1 and R2 be the resistances of the two heaters respectively. The amount of heat required is the same in all the three cases given in the problem as it is required to heat the same amount of water. Let t be the time required to heat the water with the parallel combination of heaters.

H₁ = V² x 15/ R₁
H₂ = V² x 15/ R₂
H₁ = H₂

Equating H₁ to H₂ we get the relation R₂ = 2R₁/3.
Parallel combination of heaters: H₃ = V² x (R₁ + R₂) x t / R₁R₂
Equating H₃ to H1 and substituting for R₂ in terms of R_1, we get

V² x (R₁ + R₂) x t / R₁R_{2 =} V² x 15/ R₁
(R₁ + R₂) x t / R_{2 =} 15

(R₁ + 2R₁/3) x t / (2R₁/3) ₌ 15
5t = 30 min

t = 6 min.

The time taken to boil the same amount of water when both are used in parallel is 6 min.