Determine source frequency and impedance

In summary, the conversation discussed a series circuit with R=5 ohms and L=30 mH, where the current lags the applied voltage by 80 degrees. The source frequency was determined to be 150.4 Hz and the impedance was found to be Z=5+j28.4. The conversation also mentioned using phasors to simplify the problem and the importance of avoiding calculations until the end to prevent mistakes.
  • #1
TheRedDevil18
408
1

Homework Statement


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)

Homework Equations

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 = 166.67 rad/s

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

But they get a frequency of 150.4Hz
 
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  • #2
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
donpacino said:
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.

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
= 946.67 rad/s

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

I think that's correct, yes ?

Thanks
 
  • #4
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:
  • #5
donpacino said:
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.

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

Thanks for you help :smile:
 

FAQ: Determine source frequency and impedance

1. What is source frequency and why is it important to determine?

Source frequency refers to the rate at which a wave or signal is generated by a source. It is important to determine because it can affect the behavior and properties of the wave, such as its wavelength, amplitude, and phase.

2. How do you measure source frequency?

Source frequency can be measured using a frequency meter or oscilloscope. The frequency can be calculated by counting the number of cycles within a specific time period.

3. What is impedance and how does it relate to source frequency?

Impedance is the measure of opposition to the flow of an electrical current. It is important to determine the impedance of a source because it affects the amount of energy that can be transferred from the source to the load, and it can also impact the frequency response of a circuit.

4. What factors can affect the source frequency and impedance?

The properties of the source, such as its material, size, and design, can affect the source frequency and impedance. External factors such as temperature, humidity, and interference can also impact the source frequency and impedance.

5. How can the source frequency and impedance be controlled or adjusted?

The source frequency and impedance can be controlled or adjusted by changing the properties of the source or by using external components such as capacitors, inductors, and resistors to modify the impedance. Additionally, using filters and equalizers can help adjust the frequency response of a circuit to meet specific requirements.

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