# Homework Help: Current in RLC Series Circuit

1. Mar 27, 2007

### lylos

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

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

2. Relevant 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

3. 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...

2. Mar 27, 2007

### SGT

Are you familiar with complex numbers?
The complex impedance is $$Z = R + j\omega L - \frac{j}{\omega C}$$
The current is $$I = \frac {V}{Z}$$
So, $$|I| = \frac{|V|}{|Z|}$$
and phase(I) = phase(V) - phase(Z)

3. Mar 27, 2007

### lylos

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. Mar 27, 2007

### SGT

Use KVL. $$V = R I +j X_L I - j X_C I$$

5. Mar 27, 2007

### denverdoc

hey thats cheating, i thought complex variables were out and this was to be done the ole fashioned way:yuck:
JS

6. Mar 28, 2007

### lylos

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. Mar 28, 2007

### lylos

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. Mar 28, 2007

### denverdoc

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.