Average value of current or voltage [closed]

I had read that average value of current must be calculated for half time period. For example if we apply it for a sine value full time period then its value becomes zero. But for any arbritrary graph which doesn't have shape similar above and below x axis. I mean that they have different kind of shape for half timperiod and for another half, (Periodic function/graph) Then by calculating avg. Value for current (calculation for half timeperiod) don't we lose the real meaning of average. Should we include that negative part of waveform to our average I , by taking modulus?

• I think what you mean is better called "effective" not "average". E.g. effective voltage means the voltage a DC source would have that caused the same average power in a resistor.
– Curd
Sep 8, 2017 at 9:38
• What are you trying to do? You may want the mean value the rms value or something else depending on the application. Sep 8, 2017 at 10:03
• I want to understand the concept of average current . What i have learnt that it had been introduced to have that ac current average and we find it by only considering for half timeperiod (sine waveform), the reason for doing so if we consider negative part of waveform we would have Iavg.=0. But i want to take a waveform with different shapes up and below x axis but is periodic. Then,too we will calculate I avg. by considering for half timeperiod. Sep 8, 2017 at 10:21

Generally, for AC currents we want to know the effect of the current. For example, how much power or heating does it provide. The standard method for doing this is to calculate or measure the true RMS (root mean square) of the current waveform. If you like, the resultant value gives you the equivalent DC current that would have the same effect.

simulate this circuit – Schematic created using CircuitLab

Figure 1. A strange AC waveform.

A simple example may help. In Figure 1 we have an AC waveform whose geometric average or integral will be zero. Clearly the power delivered is non-zero so let's calculate the effective current.

• Power is proportional to $I^2$.
• For the first second power is proportional to $10^2 = 100$.
• For the second second(!) power is proportional to $5^2 = 25$.
• For the third second power is proportional to $0^2 = 0$.

That's the squared part of RMS done. Now get the mean.

• $Mean \; of \; squares = \frac {sum\;of\;squares}{periods} = \frac {100 + 25 + 0}{3} = \frac {125}{3} = 42$.

Now get the root.

• $RMS = \sqrt {Mean \; of \; squares} = \sqrt {42} = 6.5$. So the effective current is 6.5 A.

For the negative half-cycle the result will be the same due to the squaring.

Note that if we just calculated the "average" current for one half-cycle (the the positive one, for example) we would have got $I_{AVG} = \frac {10 + 5 + 0}{3} = 5 \; A$. The RMS value is much higher because the $10^2$ term has a large effect.

• Good description of what is RMS, but please take into account in last parragraph that the average of the periodic signal in the draw, $\lim_{A->\infty} \frac{1}{2A} \int_{-A}^{A} i(t) dt$, is 0, it is not 5A. Sep 8, 2017 at 12:11
• I meant for one half-cycle. I've clarified that, thanks. I deliberately tried to avoid the calculus. Sep 8, 2017 at 13:59

The real meaning of average is whatever it is specified to mean.

An average is not well defined, you have to define it. If you mean over some certain time period, or some particular waveform feature, then you have to measure it that way.

Reading between the lines of your question, and taking a guess, it sounds like you are referring to the diode current in a rectifier, as specified in its data sheet. As a diode tends to have essentially constant voltage across it while conducting, the average power dissipated in the diode (which heats it up) is going to be closely approximated by the average current (when conducting) (a relevant specification). The same will not be true of other devices, or other situations.

I had read that average value of current must be calculated for half time period.

If a text says that, it is an error.

For example if we apply it for a sine value full time period then its value becomes zero.

Because the average (arithmetic mean) of a sine is zero: $$\lim_{A->\infty} \frac{1}{2A} \int_{-A}^{A} \sin(t) dt = \frac{1}{T} \int_{0}^{T} \sin(t) dt = 0$$

But for any arbritrary graph which doesn't have shape similar above and below x axis. I mean that they have different kind of shape for half timperiod and for another half, (Periodic function/graph) Then by calculating avg. Value for current (calculation for half timeperiod) don't we lose the real meaning of average.

Absolutely true, you must considerer the full period. As an extreme example, think about a periodic sequence of delta signal: $x(t)=\sum_{n}\delta(t-nT)$. This signal has a non-zero average but you can easily find a T/2 period with all signal equal to zero.

Should we include that negative part of waveform to our average I , by taking modulus?

No, if you want obtain the average. Average is all signal with its original sign.

If you want to obtain something else, as the energy, you should square it (that made positive all signal ) and evaluate again over all the period.

Only if you know some additional property of the periodic signal, as being odd or similar, you could optimize in some cases the calculus and use a part of the period.