Current Dividers

[Nowhere-Man]

For some reason, not a lot has been said in my electronics books about current division. They do include plenty of detail over resistors in parallel but the simple current divider chart was absent. I winded up working some of it out by my own formula, taking the model of 2 resistors in parallel of the same value being equal to half the original Ohms value and then dividing current from there.

Later I found the simpler system.

I find most books I came across make a far bigger deal of voltage dividers instead. In radio books the most practical thing you may be advised to do is to drop a certain voltage in the supply in cases where a mains dropper cable can't be repaired. My own radio book doesn't go into much detail over this and tends to leave out any distinction between a half-wave rectifier and full wave.

I sometimes find I may have learned more advanced stuff such as AVC systems but often some very basic theory has passed me by. So now it may be easier to think of both voltage dividers (series resistors) and current dividers (parallel resistors).

[Mark Hennessy]

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?

[Ed_Dinning]

Hi,
The basic electrical network theory is covered by Kirchoff's laws and the Norton and Thevinin theorems.
A quick Google will bring them up as will reference to Art of Electronics.

Ed

[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.