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Circuit Problem With Phasors

  • #1

Homework Statement


When the circuit shown in the figure is at steady state, the mesh current is

i(t) = 0.3255 cos(10t + 133.3°) A

Determine the values of L and R.

Zc is the impedance of the Capacitor
ZL is the impedance of the Inductor
I1 is the current in the top loop
I2 is the current in the bottom loop

Homework Equations


Z= V/I
A∠θ = a+bi
KCL and/or KVL
ZL = iωL
Zc = -i(1/ωC)

The Attempt at a Solution


I'm attaching the image to the problem.

It seemed to be hinting that I should use mesh analysis. Which I did, and here is what I got.

For the top loop (I1):
Vs1 + [(I1 -i(t))*Zc]=0
33∠-20° + [(I1 -i(t))*(-25i)]=0
Solving for I1 gives me... I1 = -.675 - i

For the bottom loop (I2):
Vs2 + 22(I2-i(t))
-7∠208° + 22(I2 - i(t))=0
Solving for I2 gives me... I2 = -.0504 + .088i

Well. As I went to solve the middle loop to find R and L, I realized that I would end up with one equation with 2 variables. As so:
22(i(t) - I2) + (i(t)*ZL) + (i(t) - I1)(Zc) + R*i(t) = 0
In it's full form:
http://www4a.wolframalpha.com/Calculate/MSP/MSP1141gi2i269116eaa1h000037b5gfa436511hfc?MSPStoreType=image/gif&s=5&w=362.&h=38. [Broken]
Where x is L and y is R.

This is where I get stumped. What am I supposed to do?
 

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Answers and Replies

  • #2
gneill
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You can simplify things greatly if you recognize that the 4mF capacitor and 22 Ω resistor are both paralleled by fixed voltage sources. Those components cannot change the voltages delivered by those sources, no matter what currents they draw from them...
 
  • #3
You can simplify things greatly if you recognize that the 4mF capacitor and 22 Ω resistor are both paralleled by fixed voltage sources. Those components cannot change the voltages delivered by those sources, no matter what currents they draw from them...
Are you telling me that the capacitor and 22Ω resistor might as well be non existent when solving for the voltage across R and L?

If so.. Does that mean I can do KVL while taking only R, L, and the 2 voltage sources into consideration? Giving me my 2nd equation?
 
  • #4
gneill
Mentor
20,781
2,759
Are you telling me that the capacitor and 22Ω resistor might as well be non existent when solving for the voltage across R and L?

If so.. Does that mean I can do KVL while taking only R, L, and the 2 voltage sources into consideration? Giving me my 2nd equation?
I am, and you can :smile:
 
  • #5
I am, and you can :smile:
Hmm.. Okay this took me a hell of a long time to calculate between those two equations. Was I supposed to use the equation in my first post and the equation found doing the KVL across L,R, and the two voltages to find R and L?

If so, I got these monstrous values (10^16) that have complex numbers for R and L. If I am supposed to find R and L between those two, then I must have done the math wrong.
 
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  • #6
gneill
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20,781
2,759
Okay, you've probably suffered enough :smile: Here's the easy way. Write KVL around the loop and solve for I (symbolically). You'll end up with something of the form
$$I = \frac{E}{R + j\omega L}$$
But you're given an expression for I. Convert it to complex form phasor. Do the same for 'E'. Do what's required to equate coefficients.
 
  • #7
Okay, you've probably suffered enough :smile: Here's the easy way. Write KVL around the loop and solve for I (symbolically). You'll end up with something of the form
$$I = \frac{E}{R + j\omega L}$$
But you're given an expression for I. Convert it to complex form phasor. Do the same for 'E'. Do what's required to equate coefficients.
I'm not sure what "E" is meant to be, but I can meet you halfway I think :biggrin:

I = E /(R + jwL)

.3255cos(10t+133.3°) = E / [R + j(10)(L)]

-.223 + .237j = E / (R + 10jL)

If I were to take a guess at what 'E' is, it's the combination of the voltage sources, correct?

E = -7∠208° + 33∠-20° ?
 
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  • #8
HA YES I DID IT. HAHAHAHAHA :rofl:

I assumed E was the combined voltage sources and got -37.19 + 8i as the complex form

R + jwL = I/E

R +10iL = 96.2 + 66.38i

I had two variables and one equation, but I saw that there was real and nonreal numbers. So..

i terms -> 10iL = 66.38i
L = 6.638

Real terms - > 96.2187 = R

Thank you again gneill! I made things way harder than they really were.
 
  • #9
gneill
Mentor
20,781
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You're welcome. Cheers.
 

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