# R L C circuit

Discussion in 'Math' started by lil coob, Nov 12, 2009.

1. ### lil coob Thread Starter New Member

Nov 12, 2009
7
0
Hi all,
i am stuck on how to find the resistor and coil values on this circuit.
i have been given the following:

Voltage = 100v phase angle 0° at 20hz
Current = 1.5A phase angle -30°
Capacitor = 50μF

any help would be much appreciated
many thanks

Last edited: Nov 12, 2009
2. ### count_volta Active Member

Feb 4, 2009
435
24
Very easy. Phasor analysis.

But what is the voltage 100V across and current 1.5A through? I assume the voltage 100V is the input voltage source since it has the frequency term. But what is 1.5A through?

After you figure that out, convert everything to impedances and the voltage source to a phasor.

ZL = jwL Ω and ZC = 1/jωC are the inductor and capacitor impedances respectively.

where ω= 20hz *2π = 125.6 rad/s

Do you know what I'm talking about? Not sure if you learned this yet.

If you do phasor analysis on your circuit you can easily find the resistance and inductance values. It would also help if you gave us the schematic of your circuit.

Last edited: Nov 13, 2009
3. ### lil coob Thread Starter New Member

Nov 12, 2009
7
0
ye i can sort of see what you mean
all it says is the current flowing in the circuit is 1.5A phase angle -30

i have made a quick sketch from the data for you

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4. ### lil coob Thread Starter New Member

Nov 12, 2009
7
0
i expect the coil to be of of a greater size than the capacitor due to the -30° angle of the current, am i right in thinking that?

5. ### neonstrobe Member

May 15, 2009
14
1
Hi

By size I think you mean impedance: yes the coil will have a greater impedance than the capacitor to give you a current lagging the voltage.

It should be easy to calculate. The reactive parts of your circuit are as previous contributor said j(ωL-1/(ωC)). The phase angle is obtained from the imaginary part of the ac divided by the real which gives you the ratio -1/sqrt(3) in this case (remember triangles?). You know C so you should be able to find L and R.

Hope this is a bit more of a clue.

6. ### count_volta Active Member

Feb 4, 2009
435
24
Ok so its a series circuit.

All you need to do is KVL using phasor analysis. Or even a voltage divider. I will go ahead and do it for you this time, but I want you to learn from this. Pay attention to what I do. Step by step.

7. ### neonstrobe Member

May 15, 2009
14
1
OK a bit more help.
I mentioned that the phase angle is tan-1(Im /Re) which we know is -30 deg.
SInce the total imaginary is jwL-j/(wC) we can set

[wL-1/(wC)]/R=-1/sqrt(3)

We know that the magnitude of the current is 100/1.5 = 66.67 ohms
but this is made up from the reactive and resistive bits.

The mag of Z=sqrt(R**2 + (wL-1/wC)**2) =66.67 ohms

Substitute for wL-1/(wC) from the phase ....

cheers ...mine's a pint

8. ### mojojerry New Member

Oct 27, 2009
2
0
I am only a 2 yr electronic student, 30 years ago and never studied since. We only studied RCL circuits 1 quarter. I was never to smart anyway. This discussion sounds so intresting that I just have to make a fool of myself. I though that you had two have 2 of any 3 in order to find the third in a regular current, voltage, resistance circuit. If this is true in and RCL circuit as well? Would you not have to test for an actual voltage across the resistor and or inductor to see what the volt drop is and then R=V/I? If the volt drop on R1 is 50 v then R= 33.333 ohm. God only knows what phasor analysis is? never heard of this. It do sound intresting though. mojojerry

Last edited: Nov 24, 2009
9. ### zgozvrm Member

Oct 24, 2009
115
2
For a series RLC circuit, we know the following:

$E_T = E_R \: + \: E_L \: + \: E_C$

$I_T = I_R = I_L = I_C$

$Z_T = Z_R \: + \: Z_L \: + \: Z_C$

$E_L$ leads $E_R$ by 90° which, in turn leads $E_C$ by 90°

$X_L$ leads $R$ by 90° which, in turn leads $X_C$ by 90°

$I_T$ and $E_R$ are in phase with each other.

So, first we have:

$

I_T = I_R = I_L = I_C = 1.5 \angle -30^\circ A = 1.299 - j0.75 A

$

and we know

$X_C = \frac{1}{2 \pi fC} \: = \: \frac{1}{2 \pi 20 \times 50 \mu F} \: = \: 159.155 \Omega$

Which gives us

$Z_C = 159.155 \angle -90^\circ \Omega \: = \: 0 - j159.155 \Omega$

Now, we can find

$Z_T =\frac{E_T}{I_T}$ and $\: E_C = I_C \: \times \: Z_C$

$Z_T = \frac{100 \angle 0^\circ V}{1.5 \angle -30^\circ A} \: = \: 66.667 \angle 30^\circ \Omega \: = \: 57.735 \: + \: j33.333 \Omega$

and

$E_C = 1.5 \angle -30^\circ A \: \times \: 159.155 \angle -90^\circ \Omega \: = \: 238.732 \angle -120^\circ V \: = \: -119.366 \: - \: j206.748 V$

Now, for the tricky part:

Since we know the angular relationship between $E_R$ and $E_C$, we know that $E_R$ is 90° ahead of $E_C$ which turns out to be (-120°) + 90° = (-30°)

Also, we know that $E_L$ is 90° ahead of $E_R$, which is (-30°) + 90° = 60°

Now, all we have to do is find the magnitude of those values: Since $E_L$ and $E_C$ are 180° apart, the sum of their angles must be either 60° or -120°, therefore it must be 60° so that, when added to [E_R] (which has an angle of -30°), we get a vector resulting with an angle of 0°.

Also, the x-components of the sum of $E_L$, $E_C$, and $E_R$ must be 100 and the sum of their y-components must be zero.

This gives us:

$E_{L+C} \: \times \: \cos{60} \: + \: E_R \: \times \: \cos(-30) =100$

$\frac{E_{L+C}}{2} \: + \: \frac{E_R \: \times \: \sqrt{3}}{2} \: = \: 100$

$E_{L+C} \: + \: (E_R \: \times \: \sqrt{3}) \: = \: 200$

and

$E_{L+C} \: \times \: \sin{60} \: + \: E_R \: \times \: \sin(-30) \: = \: 0$

$\frac{E_{L+C} \: \times \: \sqrt{3}}{2} \: - \: \frac{E_R}{2} \: = \: 0$

$\frac{E_{L+C} \: \times \: \sqrt{3}}{2} \: = \: \frac{E_R}{2}$

$E_{L+C} \: \times \: \sqrt{3} \: = \: E_R$

$E_{L+C} \: = \: \frac{E_R}{\sqrt{3}}$

substituting, we get

$\frac{E_R}{\sqrt{3}} \: + \: (E_R \: \times \: \sqrt{3}) \: = \: 200$

$E_R \: + \: 3E_R \: = \: 200 \sqrt{3}$

$4E_R \: = \: 200 \sqrt{3}$

$E_R \: = \: 50 \sqrt{3} = 86.603$

or, more correctly, $E_R \: = \: 86.603 \angle -30^\circ \: V \: = \: 75 \: - \: j43.301 \: V$

From here, you can divide $E_R$ by $I_R$ to get $Z_R$ which gives you $57.735 \angle 0^\circ \: \Omega$

and, you can calculate $E_L$ by:

$E_L \: = \: E_T \: - \: E_R \: - \: E_C \: = \: (100 \: - \: 75 \: - \: (-119.366)) + j(0 \: - \: (-43.301) \: - \: (-206.748)) \: V$

or

$E_L \: = \: 144.366 \: + \: j250.049 \: V \: = \: 288.732 \angle 60^\circ \: V$

Divide $E_L \: \div \: I_L$ to get $Z_L$

$Z_L \: = \: 288.732 \angle 60^\circ \: V \: \div \: 1.5 \angle -30^\circ \: A = 192.488 \angle 90^\circ \: \Omega$

This gives us R = 57.735Ω and XL = 192.488Ω
Since XL=2∏fL, and f=20Hz, we can solve for L:

$L = \frac{X_L}{2 \pi f} \: = \: \frac{192.488 \Omega}{2 \pi \: \times \: 20Hz} \: = \: \frac{192.488}{125.664} \: = \: 1.532H$

Last edited: Dec 14, 2009
10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784

R=57.735 Ω and L=1.532H

Solution attached as pdf

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11. ### zgozvrm Member

Oct 24, 2009
115
2

Yes, you're right. I inadvertently used the magnitude of $E_R$ (86.603) as the value of the resistance, forgetting to divide by the current first.

I made the appropriate corrections to my earlier post.