# Phasor Domain Circuit

1. Mar 6, 2014

### slh3410

1. The problem statement, all variables and given/known data

Where Vs = 120 ∠0° V

2. Relevant equations

All values are represented in ohms, so they are treated as resistors.

3. The attempt at a solution

Series inductor + resistor

30+j10 ohms = 31.62∠18.435°

Parallel resistor and inductor+resistor

(1/50)= 0.02∠0°, 1/(31.62∠18.435°)= 0.0316∠-18.435°
0.02∠0° + 0.0316∠-18.435° = .050968∠-11.31°
1/(050968∠-11.31°) = 19.62∠11.31°

Series [j10+parallel section]

j10 + 19.62∠11.31° = 23.704∠35.746°

Now the current

120∠0° / 23.704∠35.746° = 5.062∠-35.746°

Not sure how to set up the Voltage Division equation. I think I'm doing all the impedance calculations correctly though.

Last edited: Mar 6, 2014
2. Mar 6, 2014

### donpacino

assign symbols to the passive elements, then solve using the symbols. It makes the work easier to follow and easier to correct a mistake.

L1=j10
L2=J10
R1-50 ohm
R2=30 ohm

series inductor + resistor
R2+L2

parallel resistor and inductor+resistor

z1=1/(1/R1 + 1/(R2+L2))

current= vs / (L1 + Z1 )
now plug in #s
Its much easier

now lets get to solving the problem...
Why do you need the current at this point in time?

To set up the voltage division equation you use the same method that you would for all resistors
Vo=Vin*R1/(R1+R2)

so in this case Vn= Vin* Z1 / (L1+Z1)

3. Mar 6, 2014

### slh3410

I thought I needed the current to eventually use V=IZ, but I guess not.

I followed you up until you got to current. Why is Vs only over L1+Z1? Because you are calculating voltage drops over that outermost loop?

Where did Vn come from as well?

Last edited: Mar 6, 2014
4. Mar 6, 2014

### slh3410

So for multiplication of phasors I use a cross product? What do I do for division?

Last edited: Mar 6, 2014
5. Mar 6, 2014

### slh3410

I have no idea how to solve what I think is the proper voltage division setup:

(120∠0) [(10∠90) / (23.7∠35.75)]

6. Mar 7, 2014

### donpacino

vn is the node between the 2 inductors.
z1 is the parallel combo of the resistor and inductor+resistor. that plus the first inducotor is the total impedance of the circuit.

for division you divide the amplitude and subtract the phase

that is incorrect voltage division.

Vn= Vin* Z1 / (L1+Z1), with vn being that middle node, and z1 being the parallel combo of the resistor and inductor+resistor