Instantaneous, Peak, Mean, and RMS values of Alternating Current and Voltage

We generally know that there are two types of current which we use in our daily lives, viz. \((1)\) Direct Current \(\left( {{\rm{DC}}} \right)\) and \((2)\) Alternating Current \(\left( {{\rm{AC}}} \right).\) Direct current simply means a unidirectional flow of charge. In the circuit, the current simply keeps on flowing in the same direction based on a potential difference applied to it with the same magnitude unit the source or load is changed. In contrast, an Alternating current is an electric current that reverses its direction and changes its magnitude over fixed intervals of time. The graphs below show a clear difference in the pattern of direct current and alternating current.

As you can clearly see that the value of voltage and current across the load doesn’t change with time in the case of DC (\({{\rm{1}}^{{\rm{st}}}}\) picture), and it alternates with time in the case of AC (\({{\rm{2}}^{{\rm{nd}}}}\) picture). 

So, if values of voltage and current continuously change with time, then how do we calculate the amount of current at a particular time? How do we give ratings to the source or instruments? By reading this article, you will get to know about different types of values of current and voltage in the case of alternating current.

Study About Power In AC Circuits Here

Difference between Direct Current and Alternating Current

Instantaneous Value and Peak Value

As we know that the value of current and voltage in the case of alternating current changes with time. Hence, for any given time, there exist a different value of current and voltage according to the nature of the wave of alternating current. Normally, a source supplies an alternating voltage according to the equation

\(V = {V_0}\sin (\omega t)\)

Here, \({V_0}\) is the maximum value (peak value) of the voltage supplied, \(\omega \) is the angular frequency, and \(t\) is time. Hence, for any time \(t,\) we can obtain the instantaneous value of voltage in terms of the peak voltage.

For current, we consider various circuits because it functions in a different way for resistors, capacitors, and inductors. The phase difference between current and voltage when an AC source is applied across a resistor is zero. Hence, the instantaneous value of current in resistor will be \(i = {i_0}\sin (\omega t).\) For a similar case in the capacitor, current leads voltage by an angle \({\frac{\pi }{2}},\) hence the instantaneous value of current across a capacitor is \(i = {i_0} \sin \left( {\omega t + \frac{\pi }{2}} \right)\) and for the inductor, voltage leads current by the same angle. Therefore, the instantaneous value of current across the inductor will be \(i = {i_0}\sin \left( {\omega t – \frac{\pi }{2}} \right).\) 

But, for a combination of resistor, capacitor, and inductor, the instantaneous current, in general, will be written as \(i = {i_0} \sin (\omega t – \varphi )\). Where \(\varphi \) is the phase difference between current and voltage, which changes according to the value of the components connected and how they are connected.

Mean Value

Mean value is the average value of alternating current over time. Since an alternating current has equal values of both in the positive and negative direction, its means values always come out to be zero.

\(\langle V\rangle = \frac{1}{T}\int_0^T {{V_0}} \sin (\omega t)dt\)

\(\langle V\rangle = \frac{{{V_0}}}{T}[ – \cos (\omega t)]_0^T\)

\(\therefore \langle V\rangle = 0\)

Therefore, instead of going for the complete cycle, for mean value, we go for the half-cycle

\(\langle V\rangle = \frac{1}{{\frac{T}{2}}}\int_0^{\frac{T}{2}} {{V_0}} \sin (\omega t)dt\)

\(\langle V\rangle = \frac{{2{V_0}}}{T}[ – \cos (\omega t)]_0^{\frac{T}{2}}\)

\(\therefore \langle V\rangle = \frac{{2{V_0}}}{\pi }\)

\(\langle V\rangle = 0.637{V_0}\)

Hence, the mean value over a half cycle is \(0.637\) times its peak value. This was the average value for a sinusoidal waveform. Further in this article, we will have a look at the mean values of other waveforms too.

RMS Value

RMS stands for Root Mean Square, which means the value of steady current or voltage of DC, which would generate the same amount of heat in a given resistor in a given time as will be done by AC. As the name suggests, going step by step, first we need to square the quantity, then take its means, and final root it. 

Let us find the RMS value of a sinusoidal wave:

\(V = {V_0} \sin (\omega t)\)

Taking square on both sides: 

\({V^2} = V_0^2{\sin ^2}(\omega t)\)

Calculating mean over the time period:

\(\left\langle {{V^2}} \right\rangle = \frac{1}{T}\int_0^T {V_0^2} {\sin ^2}(\omega t)dt\)

\(\left\langle {{V^2}} \right\rangle = \frac{{V_0^2}}{T}\int_0^T {\left( {\frac{{1 – \cos (2\omega t)}}{2}} \right)} dt\)

\(\left\langle {{V^2}} \right\rangle = \frac{{V_0^2}}{2}\)

Finally, taking roots on both the sides

\(\sqrt {\left\langle {{V^2}} \right\rangle } = \sqrt {\frac{{V_0^2}}{2}} \)

\(\therefore {V_{rms}} = \frac{{{V_0}}}{{\sqrt 2 }}\)

The same will be the case of current if it’s a sinusoidal wave. Hence, the RMS value of voltage and current for a sinusoidal waveform is approximately \(0.71\) times its peak value.

Mean Value and RMS Value for Triangular and Square Waveforms

The diagram shows a triangular-shaped waveform. The equation of triangular waveform will be given by,

\(V = \frac{{4{V_0}}}{T}t\)

Taking the mean value over \({\rm{1/}}{{\rm{4}}^{{\rm{th}}}}\) of the cycle due to symmetry:

\(\langle V\rangle = \frac{1}{{\frac{T}{4}}}\int_0^{\frac{T}{4}} {\frac{{4{V_0}}}{T}} tdt\)

\(\langle V\rangle = \frac{{16{V_0}}}{{{T^2}}}\left[ {\frac{{{t^2}}}{2}} \right]_0^{\frac{T}{4}}\)

\(\therefore \langle V\rangle = \frac{{{V_0}}}{2}\)

\(\therefore \langle V\rangle = 0.5{V_0}\)

Hence, the mean value for a triangular waveform is \(0.5\) times its peak value:
Taking RMS value for the same:

\({V^2} = \frac{{16V_0^2}}{{{T^2}}}{t^2}\)

\(\left\langle {{V^2}} \right\rangle = \frac{1}{{\frac{T}{4}}}\int_0^{\frac{T}{4}} {\frac{{16V_0^2}}{{{T^2}}}} {t^2}dt\)

\(\left\langle {{V^2}} \right\rangle = \frac{{V_0^2}}{3}\)

\(\sqrt {\left\langle {{V^2}} \right\rangle } = \sqrt {\frac{{V_0^2}}{3}} \)

\(\therefore {V_{rms}} = \frac{{{v_0}}}{{\sqrt 3 }}\)

\(\therefore {V_{rms}} = 0.577{V_0}\)

Hence, the RMS value for a triangular waveform is \(0.577\) times its peak value.
The diagram here shows a square waveform. The equation of square waveform is given by,

\(V = {V_0}\)

The mean value over half cycle of the square wave due to symmetry

\(\langle V\rangle = \frac{1}{{\frac{T}{2}}}\int_0^{\frac{T}{2}} {{V_0}} dt\)

\(\therefore \langle V\rangle = {V_0}\)

Hence, the mean value of square wave is equal to its peak value:

Going ahead for RMS value:

\({V^2} = V_0^2\)

\(\left\langle {{V^2}} \right\rangle = \frac{1}{T}\int_0^T {V_0^2} dt\)

\(\left\langle {{V^2}} \right\rangle = V_0^2\)

\(\sqrt {\left\langle {{V^2}} \right\rangle } = \sqrt {V_0^2} \)

\(\therefore {V_{rms}} = {V_0}\)

Hence, the RMS value of the square wave is also equal to its peak value.

Summary

From this article, we got to know about instantaneous values for alternating current and voltage and also learned how to find mean values and RMS values for different types of waveforms.

For sinusoidal wave:
Mean value: \(\langle V\rangle = 0.637{V_0}\)
RMS value: \({V_{rms}} = \frac{{{V_0}}}{{\sqrt 2 }}\)

For triangular wave:
Mean value: \(:\langle V\rangle = 0.5{V_0}\)
RMS value: \({V_{rms}} = 0.577{V_0}\)
For square wave:
Mean value: \(:\langle V\rangle = {V_0}\)
RMS value: \({V_{rms}} = {V_0}\)

Frequently Asked Questions

Q.1. How do you find peak value from RMS value?
Ans: For every type of waveform, there are different relations between the peak value and RMS values. One can easily find peak values from that relation.

Q.2. What is the peak, average, and RMS value?
Ans: The peak value is the maximum value that an alternating current cycle can reach. The average value is the average of AC over a fixed time, whereas RMS is the root mean square value of the given equation.

Q.3. What is the difference between alternating current and alternating voltage?
Ans: Alternating current shows us the continuous change in magnitude and direction of the current, whereas alternating voltage shows us the continuous change in magnitude and direction of the voltage.

Q.4. How does alternating current affect voltage?
Ans: There is a phase difference between current and voltage in the case of AC supply. Sometimes it is along with the phase, sometimes it is lagging while sometimes it is leading, and due to this, the instantaneous value is affected.

Q.5. How many volts is AC?
Ans: Generally at home, the AC household supply is between \({\rm{220}}\,{\rm{V}}\) to \({\rm{240}}\,{\rm{V}}.\)

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We hope this detailed article on Instantaneous, Peak, Mean and RMS Values of Alternating Current and Voltage has helped you. If you have any queries, drop a comment below, and we will get back to you.