Series RLC circuit

Thread Starter

jetpac

Joined Feb 5, 2010
7
Hello all, I'm having some difficulty doing this problem. I feel like I'm nearly there so if anyone has a minute to help me out with the last few steps I'd really appreciate it.

I know the formatting's not great - if requested I will have another go at formatting it a bit better..


=========================

Question:

In a series RLC circuit the value of the inductance is 0.02H
Find the values of the resistance and capacitance where the voltage and current are given by:

v(t) = 353.5 cos(3000t -10) V
i(t) = 12.5 cos(3000t -55) A


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Cos(x) = Sin(x + 90)

=> v(t) = 353.5 sin(3000t, -10 + 90)
= 353.5 sin(3000t, +80)

& i(t) = 12.5 sin(3000t, -55 + 90)
= 12.5 sin(3000t, +35)

=>
ω = 3000
=> Xl = 3000 * .02 = 60 Ω
=> Zl = 60 ∠ 90

Vpeak = 353.5
=> V = 353.5 * .707 = 249.9 ∠ 80
= 249.9 * cos(80) + j 249.9 * sin(80)
= 43.4 + j246.1

Ipeak = 12.5
=> I = 12.5 * .707 = 8.8 ∠ 35
= 12.5 * cos(35) + j 12.5 * sin(35)
= 7.2 + j 5.0


Ir = Il = Ic = Itotal = (7.2 + j 5.0) OR (8.8 ∠ 35)

Ztotal = Etotal / Itotal
= (249.9 ∠ 80) / (8.8 ∠ 35)
= (249.9/8.8) ∠ (80 - 35)
= 28.4 ∠ 45
= 28.4 * cos(45) + j 28.4 * sin(45)
= 20.1 + j 20.1

El = Il * Zl
= (8.8 ∠ 35) * (60 ∠ 90)
= (8.8 * 60) ∠ (35 + 90)
= 528 ∠ 125
= -302 + j 432.5

So I now have values for El, Etotal, Ir, Il, Ic, Itotal, Zl, Ztotal - basically everything except Er, Ec, Zr and Zc. I'm just not sure how to go about the last portion of splitting the impedance between the inductor, capacitor and resistor.

Also - would anyone like to hazard a guess as to what percentage of the marks getting this far would get me in an exam situation?
 
Last edited:

kingdano

Joined Apr 14, 2010
377
heres a hint -

if i read this right you were GIVEN the inductance value - and you know that inductance and capactiance account for all of the imaginary component of the impedance.

therefore resistance is the "real" part of the impedance.

...does that help you at all?
 

Thread Starter

jetpac

Joined Feb 5, 2010
7
Thanks for the response!!

Would this look something like the right answer then?

Ztotal = 20.1 + j 20.1 (from above)
Xl = 60Ω (from above)
=>
R = 20.1Ω
Xc = 60 - 20.1 = 39.9Ω


Also - could you just verify that I'm handling the cosine functions correctly above (ie. adding 90 to the phase angle and dealing with as a sine)?

Thanks!
 

kingdano

Joined Apr 14, 2010
377
to be honest i havent done that stuff in about 6 or 7 years - so i wont be much help checking your work - i only remember the basic ideas and theories.

heres a bump though.
 
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