Question

Derive the relationship between the peak and the rms value of current in an a.c. circuit.

Answer 1

Hint: The current in the a.c. the circuit is known as alternating current and is defined as the current whose magnitude will change with time and also will reverse the direction periodically. Here, we will first consider a general equation of current. Also, we will consider the equation of rms value of current to find the relationship between rms value and peak value of current.

Complete step by step answer:
As we know that the general equation of current is given below
$$I = {I_0}\sin \omega t$$
Here, ${I_0}$ is known as the peak value of current.
Now, the heat produced in the circuit is given below
$dH = {I^2}Rdt$
Now, putting the value of $I$ in the above equation, we get
$dH = {\left( {{I_0}\sin \omega t} \right)^2}Rdt$
$ \Rightarrow \,dH = I_0^2R{\sin ^2}\omega tdt$
Now, the total heat produced in the circuit can be calculated by integrating the above circuit between the limits $0$ to $\dfrac{T}{2}$. Here, $\dfrac{T}{2}$ is the time taken by the circuit to produce the heat. Therefore, the total heat produced in the circuit in time $\dfrac{T}{2}$ is given below
$H = \int\limits_0^{\dfrac{T}{2}} {\left( {I_0^2R{{\sin }^2}\omega t} \right)dt} $
$ \Rightarrow \,H = I_0^2R\int\limits_0^{\dfrac{T}{2}} {{{\sin }^2}\omega tdt} $
$ \Rightarrow \,H = I_0^2R\int\limits_0^{\dfrac{T}{2}} {\left( {\dfrac{{1 - \cos 2\omega t}}{2}} \right)} $
$ \Rightarrow \,H = \dfrac{{I_0^2R}}{2}\int\limits_0^{\dfrac{T}{2}} {\left( {1 - \cos 2\omega t} \right)} $
$ \Rightarrow \,H = \dfrac{{I_0^2R}}{2}\left[ {\dfrac{T}{2} - 0} \right]$
$ \Rightarrow \,H = \dfrac{{I_0^2R}}{2}.\dfrac{T}{2}$
Now, as we know that the expression of rms value of current is given below
$H = I_{rms}^2R\dfrac{T}{2}$
Now, we will both the equation of rms value and peak value of current, we get
$I_{rms}^2R\dfrac{T}{2} = \dfrac{{I_0^2R}}{2}.\dfrac{T}{2}$
$ \Rightarrow \,I_{rms}^2 = \dfrac{{I_0^2}}{2}$
$ \Rightarrow \,{I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }}$
$ \therefore \,{I_{rms}} = 0.707{I_0}$

Therefore, the relationship between the peak and the rms value of current in an a.c. circuit is ${I_{rms}} = 0.707{I_0}$.

Note:Now, the peak value of current is defined as the maximum or highest value of current obtained in the one cycle. On the other hand, rms value of current is stated as the root mean square value of current that can be calculated when the dissipation in the resistor takes place as heat. Here, we have equated the equations of heat in case of rms value and peak value of current to find the relationship between the rms and peak value of current.