Question

An electric kettle has two heating coils. When one of the coils is connected to an A.C source, the water in the kettle boils in 10 minutes. When the other coil is used the water boils in 40 minutes. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be:
(A) 8 min
(B) 4 min
(C) 25 min
(D) 15 min

Answer 1

Hint: In order to solve this question we have to use the concept of heat produced since it will be common at the end so we will evaluate the R from that formula after which we will eliminate some of the terms and finally the equation will be in pure form of time.

Complete step by step solution:
For solving this question we need to apply the formula of heat produced in the first coil:
 $ Q = \dfrac{{{V^2}}}{{{R_1}}} \times {t_1} $
Substituting the values we get
 $ {R_1} = \dfrac{Q}{{{V^2}{t_1}}} $ ………………(1)
Where V is the voltage R is the resistance and t is the time taken to boil the water.
Since both the time we are boiling same amount of water and not the source was also changed so our heat produced and the voltage will be same for the second coil as for the first coil
So the heat produced for the second coil
 $ Q = \dfrac{{{V^2}}}{{{R_2}}} \times {t_2} $
On substituting the values we get
 $ {R_2} = \dfrac{Q}{{{V^2}{t_2}}} $ ……………………(2)
Now let us assume the equivalent resistance R when both the coils are connected in parallel to each other and as the amount of water and the A.C source are same so the heat produced and the voltage applied will be the same.
So the heat produced when the coils are connected in parallel:
 $ Q = \dfrac{{{V^2}}}{R} \times t $
On substituting the values we get
 $ R = \dfrac{Q}{{{V^2}t}} $ …………………….(3)
Now the resultant resistance will be:
 $ \dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} $
Putting the values of $ {R_1},{R_2}and{R_{}} $ we get:
 $ \dfrac{Q}{{{V^2}t}} = \dfrac{Q}{{{V^2}{t_1}}} + \dfrac{Q}{{{V^2}{t_2}}} $
On further solving we get:
 $ \dfrac{1}{t} = \dfrac{1}{{{t_1}}} + \dfrac{1}{{{t_2}}} $
On talking L.C.M and further solving we get:
 $ t = \dfrac{{{t_1}{t_2}}}{{{t_1} + {t_2}}} $
Putting the values of $ {t_1} $ and $ {t_2} $ on this equation:
 $ t = \dfrac{{10 \times 40}}{{10 + 40}} $
Now after further solving:
T = 8 min.

Note:
This is basically the question of calculating the equivalent resistance but we need to know the relation between heat produced and the resistance or else this question is not solvable so we need to be careful in our concept and should not mix the things.