Question

A steady current of magnitude I and an AC current of peak value I is allowed to pass through the identical resistors for the same time. The ratio of heat production in the two resistors will be:
A. 2:1
B. 1:2
C. 1:1
D. None of these

Answer 1

Hint: In this question we first find the heat produce by a DC current passing through resistor R that is ${H_{DC}} = {I^2}R$ then we find the heat produce by an AC current passing through an identical resistor R that is ${H_{AC}} = {\left( {\dfrac{I}{{\sqrt 2 }}} \right)^2}R$ here we take the RMS value of current that is $\dfrac{I}{{\sqrt 2 }}$. After that we find the ratio as 2:1.

Complete Step-by-Step solution:
First, we consider the heat produced when DC current $I$ is flowing through a resistor $R$ in the circuit that is
$Heat = {I^2}R$
${H_{DC}} = {I^2}R$------------------------------- (1)
Similarly, the heat produces when an AC current with peak value $I$ is passing through a resistor $R$ in the circuit. Here the RMS value of current ${I_{RMS}}$ is responsible for the heat that is
\[Heat = {I_{RMS}}^2R\]
${H_{AC}} = {\left( {\dfrac{I}{{\sqrt 2 }}} \right)^2}R$
${H_{AC}} = \dfrac{{{I^2}R}}{2}$------------------------------- (2)
We now have the heat produced by the DC and AC current. In the question, we are asked to find the ratio of the heat produced by the DC circuit and the heat produced by the AC circuit.
For this we divide equation number (1) with equation number (2) and we will get
$\dfrac{{{H_{DC}}}}{{{H_{AC}}}} = \dfrac{{{I^2}R}}{{\dfrac{{{I^2}R}}{2}}}$
$ \Rightarrow \dfrac{{{H_{DC}}}}{{{H_{AC}}}} = \dfrac{{{{{I}}^2}{R} \times 2}}{{{{{I}}^2}{R}}}$
$ \Rightarrow \dfrac{{{H_{DC}}}}{{{H_{AC}}}} = \dfrac{2}{1} = 2:1$
Hence the required ratio of heat production in the two resistors is 2:1. Therefore option A is the correct one.

Note: To solve this kind of question we need to know how the current behaves in a DC and AC circuit. We need to know that in AC circuits we use RMS values of voltage that is $\dfrac{{{V_{Peak}}}}{{\sqrt 2 }}$ and current that is $\dfrac{{{I_{Peak}}}}{{\sqrt 2 }}$ for determining heat and power. We also use the average value of current that is $\dfrac{{{I_{Peak}}}}{\pi }$ and voltage that is $\dfrac{{{V_{Peak}}}}{\pi }$ in AC circuits. It is not the case in DC circuits.