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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?

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Solution

## 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 R1, we get V2 x (R1 + R2) x t / R1R2 = V2 x 15/ R1 (R1 + R2) x t / R2 = 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.

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