A direct current of $2\;A$ and an alternating current having a maximum value of $2\;A$ flow through two identical resistances. The ratio of heat produced in the two resistances in the same time interval will be:
A. 1:1
B. 1:2
C. 2:1
D. 4:1

Hint: Think in terms of the difference between AC and DC currents and the behaviour of their amplitude with time. In other words, firstly, using ohm's law, and power in terms of voltage and current, determine the expression for the heat energy produced by the resistor in general. For the AC current, the equation will entail the rms current which flows through the resistor. Since we’ve been given the peak AC current, express the rms current in terms of this peak current and then you should be good to go with finding the ratio.

Formula used: Heat produced in a resistor by DC current $H_{DC} = I^2Rt$, and
Heat produced in a resistor by AC current $H_{AC} = I_{rms}^2 Rt$
Where I is the DC current, $I_{rms}$ is the AC current, R is the resistance value of the resistor and t is the time for which the current passes through.

Complete step by step answer:
Let us first look at how differently DC and AC currents behave.
DC current has a constant magnitude $I$ throughout its flow, whereas AC current has a periodically fluctuating current amplitude given by:
$I = I_{0}sin \omega t$ Thus, AC current gives us a sinusoidal supply.
In order to get a constant quantifiable value of AC current, we use the root mean square, which is a statistical measure of the magnitude of a periodically varying quantity. Heuristically, this is just the average of the AC current flowing through.
We are asked to find the heat produced in the resistances by a DC current $I$ and an AC current $I_{rms}$.
We know that heat is a form of energy, and Energy $H = P \times t$, where P is power and t is time.
Power $P = V \times I$, where V is voltage and I is current.
For a resistor, from Ohm’s law, $V = I \times R$, where R is the resistance.
Then, Power $P = I \times R \times I = I^2R$, and
Heat energy $H = I^2Rt$
Heat produced in a resistor by DC current $H_{DC} = I^2Rt$, and
Heat produced in a resistor by AC current $H_{AC} = I_{rms}^2 Rt$
In case of AC current, $I_{rms}$ can be expressed in terms of its peak (maximum current) $I_{0}$ as:
$I_{rms} =\dfrac{I_0}{\sqrt{2}}$
$\Rightarrow H_{AC} = I_{rms}^2 Rt =\left(\dfrac{I_0}{\sqrt{2}}\right)^2 Rt $
Therefore, the ratio of the heat produced in the resistor by the DC and AC currents is given as
$\dfrac{H_{DC}}{H_{AC}} = \dfrac{ I^2Rt }{ \left(\dfrac{I_0}{\sqrt{2}}\right)^2 Rt} = \dfrac{ I^2}{\left(\dfrac{I_0}{\sqrt{2}}\right)^2} $
Plugging in the values from the question:
$\dfrac{H_{DC}}{H_{AC}} = \dfrac{ 2^2}{\left(\dfrac{2}{\sqrt{2}}\right)^2} = \dfrac{4}{\dfrac{4}{2}} = \dfrac{2}{1} \Rightarrow H_{DC} : H_{AC} = 2:1 $

So, the correct answer is “Option C”.

Note: Remember that though we heuristically understood the rms current to be the average, it is not necessarily true when looked at from a statistical point of view. RMS stands for the root-mean-square of instantaneous current values, i.e., for a set of n instantaneous current values,
$I_{rms} = \sqrt{\dfrac{1}{n}(I_1^2 + I_2^2 + ….. + I_n^2)}$
Whereas, $I_{average} = \dfrac{1}{n}(I_1 + I_2 +….. +I_n)$