# Series RLC circuit

#### 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..

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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.

#### 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.

#### jetpac

Joined Feb 5, 2010
7
OK thanks anyway, useful tip - solved the problem I think!

#### t_n_k

Joined Mar 6, 2009
5,455
You haven't quite finished - you've found Xc but not C.

#### jetpac

Joined Feb 5, 2010
7
Ah good point.

Xc = 1/ωC

=> 39.9 = 1/3000 * C
=> C = 1/3000 * 39.9
=> C = 8.35421888 μF

Cheers!