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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?

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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 t/ (t1 + t2)

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

t = t1 t/ (t1 + t2)

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