Question

A DC of 5 produces the same heating effect as an AC of
A. \[50\,{\text{A}}\] rms current
B. \[5\,{\text{A}}\] peak current
C. \[5\,{\text{A}}\] rms current
D. None of these

Answer 1

Hint: Use the formula for the heat generated in a circuit. First determine the expression for the instantaneous current in the AC circuit. Also used the value of peak current in this expression. Then integrate this equation for heat produced in AC circuit for one time period. Equate this heat in AC circuit to the heat produced in the DC circuit and calculate the value of the rms current.

Formula used:
The heat \[H\] generated in the circuit is given by
\[H = {i^2}Rt\] …… (1)
Here, \[i\] is the current in the circuit, \[R\] is the resistance and \[t\] is the time.

Complete step by step answer:
We have given that the value of DC current is \[5\,{\text{A}}\].
\[{i_{DC}} = 5\,{\text{A}}\]
We have asked to calculate for which of the values of the AC current the heating effect is the same as that of the given DC current.Let \[{i_{rms}}\] be the root mean square current.The peak current in the circuit is given by
\[{i_0} = \sqrt 2 {i_{rms}}\]
We know that the instantaneous current \[i\] in the circuit is given by
\[i = {i_0}\sin \omega t\]
Here, \[{i_0}\] is the peak current in the circuit and \[\omega \] is the angular frequency.
Substitute \[\sqrt 2 {i_{rms}}\] for \[{i_0}\] in the above equation.
\[i = \sqrt 2 {i_{rms}}\sin \omega t\]

Let us now determine the heat produced by the AC circuit.The heat \[{H_{AC}}\] produced using the AC circuit is
\[{H_{AC}} = {i^2}Rdt\]
\[ \Rightarrow {H_{AC}} = {\left( {\sqrt 2 {i_{rms}}\sin \omega t} \right)^2}Rdt\]
\[ \Rightarrow {H_{AC}} = 2i_{rms}^2R{\sin ^2}\omega tdt\]
Let \[T\] be the time period of the AC cycle.Integrate the above equation for one time period.
\[{H_{AC}} = \int_0^T {2i_{rms}^2R{{\sin }^2}\omega tdt} \]
\[ \Rightarrow {H_{AC}} = i_{rms}^2R\int_0^T {{{\sin }^2}\omega tdt} \]
\[ \Rightarrow {H_{AC}} = 2i_{rms}^2R\left[ {\dfrac{t}{2}} \right]_0^T\]
\[ \Rightarrow {H_{AC}} = 2i_{rms}^2R\left[ {\dfrac{T}{2} - 0} \right]\]
\[ \Rightarrow {H_{AC}} = i_{rms}^2RT\]
This is the expression for the heat produced in the AC circuit.

The heat produced in the DC circuit is
\[{H_{DC}} = i_{DC}^2RT\]
Substitute \[5\,{\text{A}}\] for \[{i_{DC}}\] in the above equation.
\[{H_{DC}} = {\left( {5\,{\text{A}}} \right)^2}RT\]
\[ \Rightarrow {H_{DC}} = 25RT\]
We have given that the heat produced in the DC and AC circuit is the same.
\[{H_{DC}} = {H_{AC}}\]
Substitute \[25RT\] for \[{H_{DC}}\] and \[i_{rms}^2RT\] for \[{H_{AC}}\] in the above equation.
\[25RT = i_{rms}^2RT\]
\[ \Rightarrow i_{rms}^2 = 25\]
\[ \therefore {i_{rms}} = 5\,{\text{A}}\]
Therefore, the DC will produce the same heating effect as an AC for rms current \[5\,{\text{A}}\].

Hence, the correct option is C.

Note: The students should keep in mind that we can directly say that an AC circuit of an rms value of current same as the current in the DC circuit produces the same heating effect as that of the DC circuit if a pure resistor is used in the circuit. But we have not given in the question that the resistor is pure or not. Hence, we have done all the calculations to determine the value of the rms current in the AC circuit.