# Voltage Divider Rule using Admittance

Discussion in 'Electronic Basics' started by golson, Oct 14, 2004.

1. ### golsonGuest

I have a question about the Voltage Divider rule. I am looking at a
example in a textbook to calculate a transfer function for this
circuit.

----inductor------|------------|
| |
| |
C R
VIN a E VOUT
p S
| |
| |
------------------|------------|

Transfer Function is

Vout(s) (1/sL)
------ = --------------------------------
Vin(s) (1/sL) + [sC + (1/R)]

looking at the transfer function it appears that the voltage divider

------ g1 ----------|-------|
| |
Vin g g Vout
2 3
| |
--------------------|-------|

g1
Vout = Vin * ( 1/(g1 + g2 + g3) ) * g1 ) = Vin *
---------------------
g1 + g2 + g3

Assume above are resistors where : g1 = .5 mhos, g2 = .25 mhos, and
g3 = .1666 mhos or r1=2 ohms r2=4 ohms r3=6 ohms

then using admittances and my theory above:

.5
Vout = Vin * ------------ = 0.545 Vin
.5 + .25 + .1666

Using Voltage Division for Resistance:

Vout = Vin * (r2 || r3) / ( r1 + ( r2 || r3) ) = 2.4 / 4.4 = .545 *
Vin

Is the approach I took for using admittances correct for all cases.
It seems to be working for the example at the end.

I was looking at page 11 of RC Active Filter Design Handbook by F.W.
Stephenson.

It makes the equation much easier using admittances over impedences.

I can not find admittance example in my textbooks.