Using Admittance or Conductance (G) instead of Resistance (R) in circuit calculations

When you have parallel circuits, it is often handy to use G instead of R to calculate resistors in parallel and current dividers.

Now let's talk about admittance and
how to use it in current dividers.

Remember that resistance is
a value that's given in ohms.

It's inverse,
1 over R is called admittance.

We usually use G to represent that and
it's given in inverse ohms or

ohms to the -1, or 1 over ohms,
and sometimes we call that Mhos.

Kinda see it's ohms backwards.

So admittance,
right here is one over the resistance.

We use that in a lot of good math.

For example, remember that if we have
resistors in series, R1, R2, and

R3, that if we want to
have the total resistance

over this region, R = R1 + R2 + R3.

If we had values that were in parallel,
then we had to handle that differently.

One very easy way of handling that
is to put admittances in parallel.

If we use values G1, G2,
and G3, then we can say

that the entire admittance G,
from this point a to this point b,

so G from a to b = G1 + G2 + G3.

So it's very handy to add up resistors in
parallel by treating them as admittances

instead of as resistors.

Now let's do a current
divider using resistance.

Remember that when we've got
a current divider like this,

we first have to combine two
of the resistors in parallel.

Let's say that what I want to find is I1,
so I would combine these in parallel.

This would be equal to R2
times R3 over R2 + R3.

Then if I wanted to find I1,

that would be equal to I0 times
the resistors that it's not going through.

Which is this R2 times R3 over R2 + R3

divided by all the resistors,

which is R1 + R2R3 over R2 + R3.

That's how we handle a current
divider using resistance.

But if we wanted to use
admittance instead,

here's the way that we can do that.

We don't even need to combine
this into series and parallel,

we're just going to be able to say
that I1 = Io times the admittance

that it is going through,
divided by G1 + G2 + G3.

That's really a lot easier than what
we did on the previous page, so

I'm gonna say this is a really cool
way of doing current dividers.

I'm going to show you how similar
that is to doing voltage dividers.

If we have, so this is a voltage divider,
if we have resistors in series,

R1, R2, and R3,
let's say that I wanted to find V1.

Then V1 would be equal to,
this is the total voltage across here,

V0 times R1, divided by R1 + R2 + R3.

You can see the form of these two
equations is equivalent except,

that one is handling parallel admittances,
the other is handling series resistances.

Now let's do an example just with numbers,
let's do a resistive current divider

like this with 1 ohm, a 2 ohm,
and a 3 ohm resistor.

And let's say that I want to find
I1 given that I have 10 amps

of current going in, that is I0.

So we will first of all need to
convert this two admittances and

what we would have Is 1 divided by 3 mhos,

1 divided by 2 mhos and
1 divided by 1 mhos.

And again, we're looking for
I1 with an I0 of 10 amps going in.

So what we would say is, I1 = 10 amps,

that's the I0 value times the admittance
that it is going through,

1 mho, divided by the sum of
all of the other admittances.

And that's our I1 value.

That's really very simple, whole lot
simpler than doing it with resistances.