Question

Heater of an electric kettle is made of a wire of length $ L $ and diameter $ d $. It takes 4 minutes to raise the temperature of $ 0.5kg $ water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of length $ L $ and diameter $ 2d $. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K.
(A) 4 if wires are in parallel
(B) 2 if wires are in series
(C) 1 if wires are in series
(D) $ 0.5 $ if wires are in parallel

Answer 1

Hint
The resistance of wires in series is greater than the individual resistance acting alone, while resistances of wires in parallel are less than the individual resistance acting alone. Electrical energy is inversely related to the resistance and directly related to the square of the voltage.
Formula used: $ Q = \dfrac{{{V^2}t}}{R} $ where $ Q $ is the electrical energy consumed by the heater, $ V $ is voltage, $ R $ is resistance and $ t $ is time.

Complete step by step answer
Let us begin with the formula for electric energy which is given by
 $\Rightarrow Q = \dfrac{{{V^2}t}}{R} $
This electrical energy is converted to heat energy and is used to raise a $ 0.5kg $ of water by 40 K.
The resistance is given by
 $\Rightarrow R = \rho \dfrac{L}{A} $ where $ \rho $ is the resistivity and $ L $ is length and $ A $ is the cross sectional area.
For wire of diameter $ d $ ,
 $\Rightarrow {R_1} = \rho \dfrac{L}{{\pi \dfrac{{{d^2}}}{4}}} = 4\rho \dfrac{L}{{\pi {d^2}}} $
For the wires with diameter $ 2d $ ,
 $\Rightarrow {R_{2,3}} = \rho \dfrac{L}{{\pi \dfrac{{{{(2d)}^2}}}{4}}} = \rho \dfrac{L}{{\pi {d^2}}} $
Compared to $ {R_1} $ we can see that
 $\Rightarrow {R_{2,3}} = \dfrac{{{R_1}}}{4} $
We calculate the equivalent resistance for the series and parallel combination of these wires
For series combination
 $\Rightarrow {R_{eqs}} = {R_2} + {R_3} $
By substituting the formula above, we have
 $\Rightarrow {R_{eqs}} = \dfrac{{{R_1}}}{4} + \dfrac{{{R_1}}}{4} = 2\dfrac{{{R_1}}}{4} $
 $ \therefore {R_{eqs}} = \dfrac{{{R_1}}}{2} $
 $ \therefore {R_{eqs}} = \dfrac{{{R_1}}}{2} $
For parallel combination
 $\Rightarrow \dfrac{1}{{{R_{eqp}}}} = \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} $.
 Substituting the formula again, we have
 $\Rightarrow \dfrac{1}{{{R_{eqp}}}} = \dfrac{4}{{{R_1}}} + \dfrac{4}{{{R_1}}} $
 $\Rightarrow \dfrac{1}{{{R_{eqp}}}} = \dfrac{8}{{{R_1}}} $
Inverting both sides, we get
 $\Rightarrow {R_{eqp}} = \dfrac{{{R_1}}}{8} $
Since the same amount of water was raised by the same amount of temperature with both first and second heater, we can say that
$\Rightarrow Q = \dfrac{{{V^2}t}}{{{R_1}}} = \dfrac{{{V^2}{t_s}}}{{{R_{eqs}}}} = \dfrac{{{V^2}{t_p}}}{{{R_{eqp}}}} $ where $ {t_s} $ and $ {t_p} $ are the time spent for the series and parallel configuration respectively.
Therefore,
$\Rightarrow \dfrac{{{V^2}{t_s}}}{{{R_{eqs}}}} = \dfrac{{{V^2}t}}{{{R_1}}} $
The potential differences are equal and can therefore cancel out.
We can then make $ {t_s} $ subject.
 $\Rightarrow {t_s} = \dfrac{{{R_{eqs}}}}{{{R_1}}}t $
Substituting the values for $ {R_{eqs}} $ and $ t $ we have that
 $\Rightarrow {t_s} = \dfrac{{{R_1}}}{2} \times \dfrac{1}{{{R_1}}} \times 4 $
 $\Rightarrow {t_s} = \dfrac{4}{2} = 2 $
 $ \therefore {t_s} = 2\min $
Similarly, for $ {t_p} $
 $\Rightarrow {t_p} = \dfrac{{{R_{eqp}}}}{{{R_1}}}t $
 $\Rightarrow {t_p} = \dfrac{{{R_1}}}{8} \times \dfrac{1}{{{R_1}}} \times 4 = 0.5
 $ \therefore {t_p} = 0.5\min $
Comparing the answers with the options, we see that the only matching option is D.
Hence, the correct option is (D).

Note
You may have wondered why we used the form $ Q = \dfrac{{{V^2}t}}{R} $ as the electrical energy when there are other forms we could have chosen from, say $ Q = IR $. The reason is because this is the form that will lead us to our solution because it allows $ V $ to be eliminated and leaves us with our important variables $ R $ and $ t $. $ V $ can be eliminated because the voltage across them are equal since the two heaters must have been plugged in the same spot or same A.C. main which doesn’t change with changes in resistance. However, current changes with resistance which would not have allowed us to eliminate it when equating.