# Theory & Definitions

Last Modification: February 17, 2013

Connecting a measurement instrument in the right way only is one thing. But, to determine the correct parameter needs more considerations.
Very important is the question: For what purpose do I do this measurement?
Performing a current measurement to examine the heat development in a wire requires a different parameter than a current measurement to determinate the charge status of a capacitor or battery.

Parameters can be expressed as an average, RMS, instantaneous or peak value. Not only the type of load is important, but also whether this is an AC or DC source, and what the voltage and current shape looks like.
The closely related interaction between voltage and current, and power and energy on the other hand will be discussed on this page.

## Instantaneous values

The instantaneous voltage v, current i and power p has a value that corresponds to a specific time t. Every waveform has an infinity number of instantaneous values. Such a waveform is described as the parameter as a function of time. In the case of a voltage it will be written as v(t).
In the example below, the situation of a series circuit of a resistor and a coil connected to a sinusoidal voltage with a peak voltage of 3 V and a frequency f of 50 Hz.

The sinusoidal voltage as a function of the time is written as:
[equ. 1]
The current has a top value of 2 A and is shifted 60° in relation to the voltage.
[equ. 2]
The power as a function of time is the product of the corresponding instantaneous values of voltage and current:
[equ. 3]

Figure 1 shows the graphical representation of the voltage, current and power. As an example; the instantaneous values are shown for the time t = 4,2 ms marked with the gray line:
v(4,2 ms) = 2,906 V
i(4,2 ms) = 0,538 A
p(4,2 ms) = 1,563 W
At a certain time, the instantaneous voltage and current can always be multiplied to calculate the instantaneous power.

## Average values

The average value, also called the mean value, is the most commonly used parameter. If a multimeter is set for measuring DC-values, the average voltage or current is measured. Also the average value of an AC voltage or current is measured when the meter is set to DC. In case of a symmetrical AC voltage the multimeter will indicate 0 V, which is the correct value.

### Voltage and current

The average value is the sum of all products of the instantaneous values x multiplied by the infinitely small time dt divided by the period T wherein is measured. This summation with infinitely small time steps is called integrate. In general written as:
[equ. 4]
X for example can represent the voltage or current. Filled in for the voltage:
[equ. 5]

### Multimeter

As mentioned before, a multimeter selected into a DC-range measures the average value of the voltage or current. In digital meters, this average established by means of a RC-filter. This is the input signal continuously averaged over the RC-time. In formula form:
[equ. 6]

## Energy and power

Equation 3 shows that the product of the instantaneous voltage and current results in the instantaneous power p(t). If these instantaneous powers multiplied by the infinitely small time dt are continuously summed, it will return the energy in the system since t = 0 s:
[equ. 7]
Indeed, energy is the power multiplied by time: E = P·t, and energy packets may always added together to calculate the total energy.

Below are the signals showed again from the example coil-resistant series circuit as discussed in "Instantaneous values". In this figure represents the black line the energy development in time as calculated with equation 7.

As a result of the alternating polarity of the voltage and current, the power curve has also a periodic changing amplitude with a double frequency. Because energy is dissipated in the resistance, the gray coloured positive area of the power curve is greater than the negative area.
The value of the black energy line at any given time is equal to the previous area under the power curve. It is clear to see that the energy line periodically rises more than it falls as a result of the amplitude asymmetry of the power curve around the x-axis.

In figure 3 the time period T is indicated. The energy inside this time (0...T s) that is put into the system is indicated by Eper and will be calculated as follows:
[equ. 8]
The average power over a certain time period is equal to the total amount of energy within that time divided by time in wherein is measured:
[equ. 9]
If this division by the time is inserted in equ. 8, the average power can be calculated for any waveform:
[equ. 10]
This equation is consistent with the general equation for calculating an average value (equ. 4). The active power is always the average power.

This equation to calculate the average dissipated power is always valid because the calculation is based on instantaneous values. It does not matter whether this is the direct or alternating current, what the voltage and current shape looks like, or whether there is a phase shift between voltage and current exists.

The equation above to calculate the average power is the method by which the operation of a power meter is based on. An energy meter like a kilowatt-hour meter at houses and industries operates according to comparison 8. Or otherwise written as:
[equ. 11]
The upper limit T of the integral is the point of time where the energy meter is readout.

## RMS values

The RMS or effective value is a value for a voltage or current that an equally great power in a resistance dissipates as a DC voltage or current with the same value.
An alternating voltage with an effective value of 230 V will develop a same amount of heat in a resistor as a pure DC voltage of 230 V. The RMS value concerns only to the heat development in a resistive load. As an example: The RMS current is useful to monitor the load stress of a cable (= resistive), but not to measure the charging current from a battery or capacitor (= electron flow).

### Root Mean Square

RMS is an abbreviation for Root Mean Square. The voltage or current as a function of time will undergo successively three mathematical operations: quadrate, averaging and square root, to calculate the RMS value. Why these operations take place is explained below:

The power dissipated in a resistor that is connected to a voltage is calculated with:
[equ. 12]
For the instantaneous power and voltage this will be:
[equ. 13]
How to calculate the average power as function of time was shown in equation 10. p(t) can be filled in equation 13 above:
[equ. 14]
Because the resistance R is a constant, it can be brought forward:
[equ. 15]
When moving the voltage from equation 12 to the left side of the equal sign, the voltage can be calculated from the average power and resistance:
[equ. 16]
When the average power calculation from equ. 15 is filled in the equation 16 above:
[equ. 17]
Both resistor values R in the dividend and divisor eliminate each other and can be left out. This results in the equation that calculates the RMS value for any random voltage waveform:
[equ. 18]
It is clear to see that the equation consists of three parts: quadrate v(t)2, average, and square root.

The above described analysis is calculated with a voltage over a resistor. For currents trough a resistor a comparing evaluation can be made. The result for RMS currents will than be:
[equ. 19]

Most multimeters can not calculate the RMS value from the measured voltage. To know the RMS value usually a special instrument is needed.
The circuit in figure 4 shows how a RMS-meter the measured voltage computes. A RMS meter in practice will use a slightly different method of operation whereby just one multiplier is needed. Analog multipliers must have a very low temperature and offset drift, which makes these instruments expensive.
It is also possible to do the RMS calculation by software with the continues digitalize values of the measured voltages. This approach is commonly used with multimeters and digital oscilloscopes.

### Pseudo RMS

Most multimeters will not measure the RMS-value when the range selector is set to AC mode. Yet, they seem to give the effective value when measuring AC-voltages and currents. But the displayed values are only valid when a sinusoidal waveform is measured.
A simple AVO-meter rectifies the measured signal first. Then a following RC low-pass filter distilled the average value. This resulting mean value is than scaled so that the instrument will show the effective value. Written as an equation:
[equ. 20]
The consequence of this approach is that it's only usable for sinusoidal waveforms. Every other shaped waveform will give an erroneous effective value.

### RMS power?

Especially in audio communities there is a lavish use of the term "RMS power" or PRMS. This is by definition an erroneous term.

As in the chapter "Mean values" under the heading "Energy and power" is shown that the working power is calculated from the total amount of energy divided by the time wherein this energy is measured (equ. 9). The total energy is defined by summation of all instantaneous energy packets v(t)·i(t)·dt (equ. 11). This is the only correct way to calculate the active power.

As previously explained the RMS value is equivalent to a DC voltage or current which developed a same power in the same resistance. This is calculated by the square root from the average of the instantaneous voltage (or current) in quadrate. There is no reason to think why these three mathematical operations should be applied on the instantaneous power. This would be a nonsensical value.
[equ. 21]

To illustrate this, a calculation is made of a sinusoidal voltage with an amplitude of 2 Vtt and a frequency of 1 kHz.
Above the graph are the definitions: The load resistor R is 4 Ω. As a function of time are calculated: the sinusoidal voltage v(t), the current i(t) and the power p(t).

In the graph the voltage and current are displayed.

First the RMS voltage is calculated of the voltage as a function of time v(t). The result is equal to the well-known equation:

The second equation calculates the RMS current with the current as a function of time i(t). This is equal to:

Then, with three different methods the Active Power is calculated using the RMS voltage and current values: VRMS·IRMS, VRMS2/IRMS2·R. To check this, with a fourth calculation the average power is determined with the power as a function of time p(t). All these calculations result in the same value for the active or average power.
At last, at the bottom the calculation for the RMS power is done. The outcome of this (0,153 W) differs significantly from the four earlier calculations (0,125 W).

The above example is carried out using a sinusoidal voltage and current. But the shape of the voltage and current as well as the kind of load and possible phase shift are of subordinately importance.
The active power is always the average power. RMS Power is a nonsensical number.

###### At 19 okt 2013, 22:34:47 wrote George Daglish
Dear Freddy,

I am a retired Doctor of Applied Mathematics with a strong interest in Seismology.

In view of recent disasters and together with some colleagues, I am interested in working to automate the process of Location of Earthquake Epi/Hypocentres from the receipt of seismograms to the release of possible Tsunami warnings. This response must be as rapid as possible.

In this regard at least 300,000 lines of code have been designed, written and tested in prototype form.

I have attended two World Congresses for Engineering [London] in 2010 and 2011 and the 33rd General Assembly of the European Seismological Commission in Moscow [2012] and am likely to attend the 34th in Istamboul in 2014 to present some more of this work.

One of the many features that would appear to me to lend itself to purely hardware implementation is the pin-pointing of the initial onsets of the different forms of seismic energy [P-waves; S-waves; Love waves and Rayleigh waves, etc.] within the incoming seismogram at the point of receipt.

There are several methods for this, of which I have programmed 3. One, in particular involves two parallel FFT's and two parallel resultant PSD's [Power Spectral Density] evaluations and their subsequent comparison by ratio.

As stated, this would ideally act on an incoming seismogram in near real-time.

Would you be prepared to consider a design?

With Best Regards,

George Daglish

[BA PhD CMath FIMA]
Dear george,
###### At 24 feb 2014, 12:17:31 wrote Masello
Can some1 ple help me to get power and rms value of a signal: 10 cos ( 100t + pi/3)
###### At 24 feb 2014, 18:02:02 wrote George
I suspect the answer is 7.073282W; however, I will send an email to show the crucial part of the working.
###### At 24 feb 2014, 21:17:32 wrote Daglish
Gave wrong answer: should be RMS this signal = root (Pi). Have sent email to Freddy to show working. Need more data to calculate power - like \\\'R\\\'.
###### At 04 maa 2014, 10:08:31 wrote Ackon Nana Kobina
Pls,i am a student offering elec/electronics engineering, how can i get your book to study.
###### At 04 maa 2014, 23:32:16 wrote Freddy
Hello Ackon, I didn't write any book. All my articles are published on this website. Is there a specific subject you're missing?
###### At 01 aug 2014, 12:39:36 wrote Raghavendra H R
sir, during braking mode i got power curve in that curve how to calculate energy I knew formula E=P*t . my power curve is between 1s to 1.0097s, that power curve is delay and its peak value is -14*10^6. then reply in that situatiobn how to calculate energy.
###### At 02 aug 2014, 01:49:03 wrote Freddy
I assume that you're speaking about breaking an electric motor. Normally the breaking energy is calculated by integrating the product of the measured voltage and current during the breaking action. Is it an AC or DC type and can you describe the curve shape?
###### At 02 aug 2014, 10:25:35 wrote Raghavendra H R
Good afternoon Sir , just I red your reply. I used BLDC motor which is AC type of motor. now let i explain the voltage graph of BLDC motor. During 0s to 1s the bldc motor is running i.e called motoring mode of BLDC motor in that time period power is zero. After 1s to 1.0097s means between fraction of seconds i got a POWER CURVE in braking mode of BLDC Motor . see At t=1s, power is peak value =-14*10^6 in negative direction of a graph ( the power curve coming downwards at 1s which is the peak value ) then the power curve is dacaying up to the time period t= 1.0097s . Finally the time period t= 1.0097s the power bacame zero means coming back to original zero position. in this curve i dont know how to calculate enery please solve out my problem quickly. if its graph not understand clear to you then write your email ID in this massage site i will take one photo of that graph and then mailling to you today. since the power curve is exponentially decal i.e at t= 1 p=-14*10^6 which is high value and at t=1.0097 p=0 which is low vaue of power. hence power is decaying from the period 1s to 1.0097s . THANK YOU SIR
###### At 02 aug 2014, 14:02:58 wrote Freddy
Simply calculate the area (P*t) under the curve. Because the curve is exponential it is hard to do. You can ease this by dividing the total time in smaller time chunks, and calculate the the energy in each chunk (Pav*dt). After that sum all the chunks (Etotal = A+B+C+D...).

The example above shows how to do this. (only the first five chunks are calculated.)
A = -11.3*10^6 W * 0.001 s = -11300 J
B = -6.25*10^6 W * 0.001 s = -6250 J
C = -4.0*10^6 W * 0.001 s = -4000 J
D = -2.5*10^6 W * 0.001 s = -2500 J
E = -0.5*10^6 W * 0.001 s = -500 J
Total energy Etot = A+B+C+D+E = -24550 J

If you want a more accurate calculation divide the total time in more chunks.
###### At 03 aug 2014, 20:59:38 wrote Raghavendra H R
1)In the Graph There is negative Power but you have taken positive for the calculation.

2)Regarding the time that you have taken for each chunks is 0.001s, can you explain why is it so?
###### At 03 aug 2014, 21:18:39 wrote Freddy
Off course the numbers must be negative, I've corrected it. I was thinking positive ;)

The width of the time chunks is a choice. I choose 1 ms to reduce the number of calculations. But for more accurate calculations you can choose for example 0.2 ms intervals, but the number of calculations is than five times as high.
###### At 04 aug 2014, 08:40:47 wrote Raghavendra H R
Thank you sir .. you are thinking is right. it was more helpfull to me for doing M.Tech project. it was simple method and I got accurate answer. thank you lot.
###### At 29 apr 2015, 18:21:04 wrote Carl000
Pav1=Vrms*Irms=Pav4
Thanks
###### At 03 mei 2015, 13:42:46 wrote Freddy
P=Vrms*Irms is only valid for sinusoidal signals with no phase shift. In the Pav1 equation, replace Vrms and Irms by the RMS equations above, and replace v(t) and i(t) by a sinusoidal function sin(2*pi*t). Because p(t)=v(t)*i(t), you can simplfy the equation to the one given as Pav4.
###### At 21 jan 2016, 15:56:08 wrote abel george
Sir i have a doubt.a mobile charger when not connected but on.what kind of losses will happen. It will be great if u can tell with values.plz sir
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