An electric kettle has two heating coils. When one of the coil is connected to an alternating current source the water in the kettle boils in 10 minutes. When the other coil is used, the same quantity of water takes 15 minutes in boil. How long will it take for the same quantity of water to boil, if the two coils are connected in parallel ? 

Concept:

We know that for parallel connection, the equivalent resistance is give n by:

\(\frac{1}{R}= \frac{1}{R_1}+\frac{1}{R_2}\)

Power (P) = VI = V2/R = I2R

Also

Heating Effect (Q) = Power × Time

Q = VIt = V2t/R = I2Rt

Since the coils are connected in parallel, therefore Voltage will be same, so we will use the following relation.

Q = V2t/R

\(R=\frac{V^2t}{Q}\)

Similarly

\(Q=\frac{V^2t_1}{R_1}=\frac{V^2t_2}{R_2}\)

\(R_1=\frac{V^2t_1}{Q}\)

\(R_2=\frac{V^2t_2}{Q}\)

Calculation:

Given:

t1 = 10 min, t2 = 15 min

\(\frac{1}{R}= \frac{1}{R_1}+\frac{1}{R_2}\)

\(\frac{Q}{V^2t}= \frac{Q}{V^2t_1}+\frac{Q}{V^2t_2}\)

\(\frac{1}{t}= \frac{1}{t_1}+\frac{1}{t_2}\)

\(\frac{1}{t}= \frac{1}{10}+\frac{1}{15}\)

\(\frac{1}{t}= \frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}\)

t = 6 min.