# Is current calculation in copper wire for AC simple or not?

I'm trying to learn electronics by myself, and I'm starting with simple current calculations. Let's assume we have a straight wire of length L connecting a source and a sink. I know the voltage (V), and for finding the current I need to apply Ohm's law:

$$I = \frac{V}{R}$$

This formula works fine for a DC source, and it's an easy one. What happens when one has an AC system? I saw that the formula changes to:

$$I = \frac{V}{Z}$$ where Z is the impedance, which is a complex number. This formula is a new universe for me because of this complexity. Or at least that's how it looks when trying to learn it.

From this website I saw that one could calculate the impedance at different frequencies, but AC also has to do with magnetic fields as well, right? There are two issues I don't understand yet:

1. Why is the impedance a complex number and what does the imaginary number suppose to reveal?
2. Is the medium around that wire changing the current calculations due to the magnetic field? If so, how does it get into OHM's law?

I know it might sound stupid to most of you, but for me is not easy to understand.

• You should watch out before confusing yourself ! That website also includes skin effects etc and that only comes into play at high frequencies. This is also EM field related. Electronics is very broad and diving into things like this will confuse you. I suggest that as a lead, you get yourself a book which treats the subjects that you like. If you've just started with Ohm's law then I would not bother with Impedance of wires just yet. – Bimpelrekkie Jul 5 '17 at 14:48
• Yes every wire has self inductance – sstobbe Jul 5 '17 at 14:55
• It would be useful to add some context here. In some circumstances you need to go into a lot of detail and solve field equations particularly in RF circuits or when designing wound components but I suspect this is a lot more complex than you need – Warren Hill Jul 5 '17 at 16:12
• for many simple calcs, esp on wire, you can just use rms and pretend it's DC – dandavis Jul 5 '17 at 20:07

It's got nothing to do with magnetic fields in a simple situation (skin effect does come into this later but for now ignore it and just learn the basics of impedances). Walk first, then run.

If the load is a resistor then the load impedance = R (or Z = R). So you get a sinewave current with a sinewave voltage and the two waveforms are in sync: -

However, in AC circuits there are capacitors and inductors and these numerically are represented by complex numbers. Simple reason: the voltage current relationship is at 90 degrees. See this for a capacitor: -

And for an inductor: -

So, if you have any inclination about complex numbers this should make sense. If you are a bit rusty on complex numbers then you probably need to do some more research on the topic.

Pictures taken from here and this might be a useful learning resource.

• CIVIL: In a Capacitor I leads V. V leads I in an inductor (L). Darn! The mnemonic broke down. – Transistor Jul 5 '17 at 16:08
• @Transistor it works for me and has done for too many bloody years (sigh). – Andy aka Jul 5 '17 at 16:17

At low frequencies (hertz), and for short wires (length WAY less than a wavelength) it's simple. As the frequency increases, you will need to consider "Skin effect", maybe dielectric loss into the insulator.

If the frequency is high enough, the wire can be 1/4 wavelength long, and if it is an unterminated wire, the reflection can make its impedance look near infinite at the driven end.

Things get complicated pretty quickly.

No need to feel foolish; this quickly becomes complicated. This site isn't suited to explain such a broad topic, but here's somewhere to start:

First off, impedance is a combination of resistance and reactance. You already know resistance. Reactance describes capacitive or inductive qualities of the load. Putting resistance and reactance together gives you the total impedance of the circuit.

The resistive element is the "real" part of the complex impedance, where the reactive part is the "imaginary" part. If you don't know complex number math, the examples you find probably won't make sense. One more thing to learn!

With an AC circuit whose load is purely resistive, that is, with no reactive element, the impedance is simply a real number. In this case Ohm's Law works exactly the way you would expect from DC.

Once you add a reactive component, however, things get to be less intuitive. @AndyAka's answer shows one of the primary issues. That is, that voltage and current are no longer in phase. When you want to dig into this, look up "power factor".

And when you get more advanced, there are even more considerations. For example, there aren't any "purely resistive" circuits in reality. But the calcs are often close enough. Also, there is "skin effect" at high frequencies, etc... I wouldn't about these things until much later.

Finally, something interesting: When you hear about AC voltages, the number is the RMS ("root mean square") value of the AC voltage. In other words, the number (e.g. 120 VAC) is the average of the either half of the sine wave. The peak voltage is actually much higher.

In the case of 120 VAC, the peak voltage is 170 Volts:

(source)