When do you use RMS values when calculating power? My physics textbook provides several equations for determining the power dissipated in AC circuits. For example, there are power equations using instantaneous values of current and voltage. Do you use RMS values when calculating only average power and the power factor?


The voltage drop and the current flowing through a circuit element are changing with time. For an AC source, there is no current through the circuit element at one point of time and at another time, the maximum current flows through it. Therefore, multiplying the peak values of voltage drop and the current will not give you the total energy consumed; rather, they give the power consumed at the instant when maximum current is flowing through the circuit element.

The instantaneous power for any circuit element in any state (DC or AC or anything) is given by:

$$P(t) = V(t)\times I(t) = [I(t)]^2 Z = \frac{[V(t)]^2}{Z} \tag{1}\label{(1)}$$

where $P(t)$ is the instantaneous power, $I(t)$ is the instantaneous current, $V(t)$ is the instantaneous voltage drop across the circuit element and $Z$ is the impedance(resistance) of the circuit element.

The aforementioned formula for instantaneous power is always true because the work done on a charge $q$ moving through a potential difference of $V$ (for that instant, $V$ is constant) is $qV$. The power is the rate of doing work, therefore, the instantaneous power is the product of $V$ and $\displaystyle \frac{dq}{dt}$ (or $I$).

The average of a function over an interval $[a, b]$ can be calculated using calculus as follows:

$$\langle f(x) \rangle = \frac{\displaystyle \int_a^b f(x) \space dx}{b -a} \tag{2}\label{(2)}$$

We can calculate the average power consumed using equation $\eqref{(1)}$ and concept $\eqref{(2)}$.

$$\langle P(t) \rangle = \frac{\displaystyle \int_0^T P(t) \space dt}{T}$$

$$\langle P(t) \rangle = \frac{\displaystyle \int_0^T \frac{[V(t)]^2}{R} \space dt}{T}$$

$$\langle P(t) \rangle = \frac{\displaystyle \frac{1}{R}\int_0^T [V(t)]^2 \space dt}{T}$$

$$\langle P(t) \rangle = \frac{1}{R} \langle [V(t)]^2 \rangle$$

If you had used instantaneous current to calculate the average power instead, you would have obtained:

$$\langle P(t) \rangle = R \langle [I(t)]^2 \rangle $$

Therefore, the average power consumed depends on the RMS(average value of the function squared) value of the voltage or current across it.

If you calculate the average power consumed per cycle for AC source, you will obtain the following:

$$P_{\text{avg}} = \frac{1}{2} \cdot \frac{V_{\text{peak}}^2}{R} = \frac{1}{2} \cdot v \cdot I_{\text{peak}}^2 R \tag{3} \label{(3)}$$

If you calculate the RMS values of current and voltage separately, you get:

$$V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}}, \quad I_{\text{rms}} = \frac{I_{\text{peak}}}{\sqrt{2}} \tag{4}\label{(4)}$$

Plugging the results of equation $\eqref{(4)}$ into equation $\eqref{(3)}$, you get:

$$P_{\text{avg}} = \frac{V_{\text{rms}}^2}{R} = I_{\text{rms}}^2 R \tag{5} \label{(5)}$$

There's something interesting here. You get the above two forms of average power using two different ways (integrating voltage squared expression and integrating current squared expression). Whatever mathematical method you prefer to use, the answer must be the same. Therefore:

$$\frac{1}{2} \cdot \frac{V_{\text{peak}}^2}{R} = \frac{1}{2} \cdot I_{\text{peak}}^2 R$$

$$V_{\text{rms}} = I_{\text{rms}} R \tag{6}\label{(6)}$$

Some of you might ask, what's so special about the above equation? You need to think more deeply. Is it obvious that the RMS values respect the Ohm's law relation? It might be for some who can think through the above equations but it isn't obvious for me.

Using equation $\eqref{(6)}$, you can write the average power in one more way.

$$P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} = \frac{V}{\sqrt{2}} \cdot \frac{I}{\sqrt{2}}$$

When to use RMS values of current and voltage while calculating power for AC circuits?

If you are calculating the average power over $n$ cycles ($n \in \mathbb{Z}$), you must use RMS values of voltage and current.

For instantenous power, you use instaneous voltage and current.

If you are calculating the average power over non-integer cycles (half of $0.67422$ or aything else), then you have to do the integration (equation \eqref{(2)}).


Short answer: use it when it makes physical sense.

Slightly longer answer: instantaneous power is the product of voltage and current. When there is a nice linear relationship, $V=I\times R $, then power is $P=VI=V^2 R$ and average power is calculated from the average voltage squared. If you use the RMS value of the voltage to compute the RMS value of the current you get the same result.

Does that help?

  • $\begingroup$ So just to summarize, the RMS values used in power equations should give you the same result as using any other power equation like P = IV = I^2 R = V^2 / R ? $\endgroup$ – Marie R Mar 25 '17 at 0:38
  • $\begingroup$ More to the point, if instantaneous power is proportional to $v^2(t)$, then the rms average of $v(t)$ is the correct average $V$ to use to get average power $P$ to have the same proportionality to $V^2$. $\endgroup$ – The Photon Mar 25 '17 at 5:26

If you have a purely resistive circuit then the power dissipated in the circuit can vary between zero and $V_{\rm peak}I_{\rm peak} = I_{\rm peak}^2 R = \dfrac{V_{\rm peak}^2}{R}$.

It is often useful to find out what the average power $<P>$ over a period and so what is required is $<VI> = <I^2> R = \dfrac{<V^2>}{R}$ where $<V^2>$ is the mean of the square of the voltage etc.

So a mean square value is defined as $V_{\rm rms}= \sqrt{<V^2>}$ for the voltage and $I_{\rm rms}= \sqrt{<I^2>}$ for the current.

For a sinusoidally varying voltage it is found that $V_{\rm peak} =\sqrt{2}\, V_{\rm rms}$.

Once this has been done it is often more convenient to use rms values rather than peak values because although the ratios of voltage and current stay the same, $\dfrac{V_{\rm peak}}{I_{\rm peak}}= \dfrac{V_{\rm rms}}{V_{\rm rms}}$, when average power is being calculated the use of peak voltages and currents would repeatedly introduce the (annoying) factor $\sqrt{2}$.

When circuits with inductance and capacitance as well as resistance are introduced there is still no problem with the ratios of voltage and current but again if average power is being considered the rms values are easier to use.

However because for such circuits the voltage and current are no longer necessarily going to be in pahse with one another one has to introduce a factor called the power factor $\cos \phi$ into the average power dissipated in the circuit equation $<P> = V_{\rm rms}I_{\rm rms} \cos \phi\, (= \frac 1 2 V_{\rm peak}I_{\rm peak} \cos \phi )$ where $\phi$ is the difference in phase between the voltage and the current.

So in general you will find that something with labels $V$ and $I$ in a textbook will very often refer to rms values.

One place where the peak values need to be used is when one considers the insulation placed around the conductors.
For example if you have a voltage supply of $230\,\rm V\, rms$ then the insulation must be able to withstand a voltage of $\sqrt{2}\times 230=325\,\rm V$.

This may not seem a lot different but when one is dealing with very high voltages eg $1.1\,\rm MV\, rms$ then the insulation has to withstand a peak voltage of $\sqrt{2}\times 1.1=1.55\,\rm MV$ which has lead to the development of a power super highway where dc rather than ac is used for transmitting electrical power..


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