# Determine source frequency and impedance

1. Apr 6, 2015

### TheRedDevil18

1. The problem statement, all variables and given/known data
The current in a series circuit of R=5 ohms and L=30 mH lags the applied voltage by 80 degrees. Determine the source frequency and impedance.
(Ans: Z=5+j28.4 , f=150.4 Hz)

2. Relevant equations

3. The attempt at a solution

Current in the inductor,
IL = Vm/wL*sin(wt-80)

Current in the resistor,
Ir = Vm/R * sin(wt-80)

Current in the resistor equals current in conductor
Equating them you get,
1/wL = 1/R
1/w(30*10-3) = 1/5

w = 2*pi*f
f = w/2*pi
= 166.67/2*pi
= 26.53Hz

But they get a frequency of 150.4Hz

2. Apr 6, 2015

### donpacino

have you learned phasors yet? if you have there is an easy way to do this problem.

That being said, your problem is voltage Vm is equel to the voltage drop across the inductor AND the resistor. They are is series. Therefore the voltage drop across the inductor is not vm. You also wrote your time domain equation wrong.

3. Apr 6, 2015

### TheRedDevil18

Yes, I have done phasors

I think I found a way using phasors

if z = x∠80

Then, x*cos(80) = 5 (The real part, which is the value of the resistor)

Therefore x = 28.79

So, z = 28.79∠80

Therefore the impedance of the inductor is equal to 28.79sin(80) = j28.4

Angular frequency for inductor
z = jWL
W = 28.4/30*10-3

Therefore the frequency is
w/2*pi
= 946.67/2*pi
= 150.67Hz

I think that's correct, yes ?

Thanks

4. Apr 6, 2015

### donpacino

yes! here is how i did it

i=v/z
z=R+LS

we know I=mag(I)∠-80
so z=mag(z)∠80=sqrt(R2+(Lw)2)∠tan-1(Lw/R)

80=tan-1(Lw/R)
solve for w

in other words, the same thing, but used algebra until the end. I recommend you do that, with more complicated problems avoiding calculations until the end can prevent mistakes.

Last edited: Apr 6, 2015
5. Apr 6, 2015

### TheRedDevil18

Yeah, I see my final answer was a bit off there

Thanks for you help