Current in RLC Series Circuit

  • Thread starter lylos
  • Start date
  • #1
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Homework Statement



I need to solve the following equation to get I = V/((R^2+(Xl-Xc)^2)^1/2).

Homework Equations



-I[(1/c - w^2L)sin(Φ)-Rwcos(Φ)]-wV=0
we know that tan(Φ) = (Xl-Xc)/R
Xc = 1/wc
Xl = wL

The Attempt at a Solution



I[(1/c - w^2 L)sin(Φ)-Rwcos(Φ)]=wV
I[sin(Φ)/c - sin(Φ)w^2L - Rwcos(Φ)] = wV
I[sin(Φ)/wc - sin(Φ)wL - Rcos(Φ)] = V

By using a helping right hand triangle, I know that sin(Φ)=(Xl-Xc)/((R^2+(Xl-Xc)^2)^1/2) and cos(Φ)=R/((R^2+(Xl-Xc)^2)^1/2)...

So I won't have to type this over and over, I'll assume the hypotnuse of the triangle is h = ((R^2+(Xl-Xc)^2)^1/2). Now when I plug in the values I have...

I[(Xl-Xc)/h (Xc) - (Xl-Xc)/h (Xl) - R^2/h]=V
I[(XcXl-Xc^2)/h - (Xl^2-XcXl)/h - R^2/h]=V
I[(XcXl-Xc^2-Xl^2+XcXl-R^2)/h]=V
I = (Vh)/(XcXl-Xc^2-Xl^2+XcXl-R^2) plug in for h now from above...
I = [V ((R^2+(Xl-Xc)^2)^1/2)] / (XcXl-Xc^2-Xl^2+XcXl-R^2)

This is where I get stuck... I don't know what to do here... This could be better to post in the Mathematics forum. If someone could help me out, it would be greatly appreciated...
 

Answers and Replies

  • #2
SGT
Are you familiar with complex numbers?
The complex impedance is [tex]Z = R + j\omega L - \frac{j}{\omega C}[/tex]
The current is [tex]I = \frac {V}{Z}[/tex]
So, [tex]|I| = \frac{|V|}{|Z|}[/tex]
and phase(I) = phase(V) - phase(Z)
 
  • #3
79
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I know that the current is equal to V/Z... We just have to show this using Kirchoff's Laws for circuits. It's part of the assignment that we do it this way.
 
  • #4
SGT
I know that the current is equal to V/Z... We just have to show this using Kirchoff's Laws for circuits. It's part of the assignment that we do it this way.
Use KVL. [tex]V = R I +j X_L I - j X_C I[/tex]
 
  • #5
960
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hey thats cheating, i thought complex variables were out and this was to be done the ole fashioned way:yuck:
JS
 
  • #6
79
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It does have to be done the "ole fashioned" way... I have it, I believe... Going to hand it in today, when I get time today I'll post what I found.
 
  • #7
79
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Solve the following equation to get I0 = V0 / ( (XL-XC)^2 + R^2 )^1/2.

I0 [(( 1/C – ω2L ) Sin (φ)) + Rω Cos (φ)] = ωV0

Known
Tan (φ) = (XL-XC) / R
XC = 1 / ωC
XL = ωL
Sin (φ) = (XL-XC) / ( [(XL-XC)^2+R2]^1/2)
Cos(φ) = R / ([(XL-XC)^2+R^2]^1/2)

Solution
I0 [Sin(φ)/C - Sin(φ)ω^2L) + RωCos(φ)] = ωV0
Divide both sides by I0 and ω.

[Sin(φ)/ ωC - Sin(φ)ωL) + RCos(φ)] = V0/ I0
Put in XC for 1/ωC and XL for ωL and factor out Sin(φ).

Sin(φ) (XC-XL) + Cos(φ)R = V0/ I0
Put in the values found for Sin(φ) and Cos(φ) from right triangle.

[(XL-XC)^2 / ( [(XL-XC)^2+R^2]^1/2)] + [R^2 / ([(XL-XC)^2+R^2]^1/2)] = V0/ I0
Combine numerators with common denominator.

[(XL-XC)^2 + R^2] / ([(XL-XC)^2+R^2]^1/2) = V0/ I0
Divide numerator into denominator.

([(XL-XC)^2+R^2]^1/2) = V0/ I0
Cross multiply to get I0 equal to an expression.

I0 = V0 / ([(XL-XC)^2+R^2]^1/2)
 
  • #8
960
0
nice, and for having done so, you'll be all the more appreciative of Laplace transforms and complex variables. Both are very powerful tools for this sort of problem and many others.
 

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