Find the total time:

Question

A heater boils 1 kg of water in time t1 and another heater boils the same in time t2. If both the heaters are connected in parallel, the combination will boil the water in how much time?

Answer 1

We know that electric heater contain coils which have some resistances. So 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 i.e 1 kg. Let t be the time required to heat the water with the combination of heaters connected in parallel.

We know from Joule’s Law of Heating

Heat (H) = V² x time(t) / R

So,
H1 = V² x t₁/ R₁
H2 = V² x t₂/ R₂
H1 = H2 

Equating  H1 to H2 we get the relation R₂ = t₂R₁ / t₁.
Parallel combination of heaters: H₃ = V² x (R₁ + R₂) x t / R₁R₂
Equating H₃ to H₁ and substituting for R₂ in terms of R₁, we get

=> V² x (R₁ + R₂) x t / R₁R₂ = V² x t₁/ R₁
=> (R₁ + R₂) x t / R₂ = t₁

=> (R₁ + t₂R₁/t₁) x t / (t₂R1 / t₁) = t₁
=> R₁(1 + t₂/t₁) x t t₁/ (t₂R1) =  t₁

Solving we get :

t = t₁ t₂ / (t₁ + t₂)

So, If both the heaters are connected in parallel, the combination will boil the water in time

t = t₁ t₂ / (t₁ + t₂)