Current Dividers

[Nowhere-Man]

So are you asking for help to understand this?

If so, I'd say start with Ohm's law. Yes, there is an equation for current division, just like there is for voltage division, but these equations are generally just two steps of Ohm's law combined into one equation.

For 2 resistors in parallel, you need to first find the voltage dropped across the pair. Then, knowing that, it's a simple application of Ohm's law (I=V/R) to find the currents. The first step - finding the voltage seen by the pair of resistors - might require a bit more thinking, but not much.

Before attempting any maths, it's useful to have an approximate idea in your head of what might be happening. Then, when the numbers come back, you can ask yourself if they seem to be plausible. Perhaps an obvious thing to say, but at work we find that many people suffer from blindly believing what their calculators tell them, forgetting that garbage-in equals garbage-out. Indeed, we all make typos at times.

So, look at the two resistors. If they are equal in value, they will share the current equally. If one is 10 times the value, then it offers 10 times the resistance to the flow of electricity, so it will have a correspondingly small flow of current compared to the lower value resistor that offers the easiest route. Having an intuitive "feeling" for what is going to happen is more important than being able to precisely quantify it in most practical cases - and if nothing else, this intuition needs to come before the application of mathematics. Well, that's my opinion (as someone who teaches electronics for a living) - no-doubt some people might disagree.

Although the plumbing analogy isn't always 100% applicable to electrical currents, this is a case where it might help. Imagine two water pipes connected in parallel. One is a 4" soil pipe. The other is a half-inch copper water pipe. Which offers the least resistance to the flow of water - which pipe has the greatest current flow?

I don't think I was having any major difficulty with it as such although I have been known to have late nights if I get stuck on something. Lately, I discovered I was also a bit rusty on capacitive reactance and power factors. So, all in all, I did a bit of revising of some of my theory. I learned that impedance can be calculated using square roots after Ohms resistance and capacitive reactance has been taken into account. However, I was using an old HAM theory book to do all of this revision.