# RLC problem

Discussion in 'General Electronics Chat' started by dileepchacko, Jul 25, 2008.

1. ### dileepchacko Thread Starter Active Member

May 13, 2008
102
1
Hi All

Consider an RCL series circuit connected to a sin wave which is having 12V RMS and 20KHz frequency. The value of R1, C1, L1 is 100ohm, 10uF and 10mH respectively. please refer fig.1 in the attachment. The circuit will switch on by on/off switch. please calculate voltage drop across R1, voltage drop across C1 and voltage drop across L1 after 10ms.

Note: Sine wave is not having any phase shift.
Capacitor is not charged before connecting to the circuit.

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2. ### theamber AAC Fanatic!

Jun 13, 2008
321
0
If the sine wave is not having phase shift you do not need complex numbers. Just add the reactances to get the total resistance plus the resistor= (Zt) and use ohms law from there. However the Time constant of the capacitor is not going to be dropping full voltage you have to plug the voltage value in correctly. This differential charge equates to a storage of energy in the capacitor at that given time.

3. ### Norfindel AAC Fanatic!

Mar 6, 2008
325
25
Had you taken a look here?: http://www.allaboutcircuits.com/vol_2/chpt_5/2.html
It explains an series RLC circuit much better than can be explained here.

As the circuit is pure AC, i think that it would be steady at 10 ms, and the normal RLC calculations, with complex numbers, can be used. I'm assuming you want the RMS values, not the instantaneous values, right?

4. ### silvrstring Active Member

Mar 27, 2008
159
0
dileepchacko,

Do you need to solve this using a particular method? Are you trying to solve it by using polar and rectangular notation, or are you trying to solve with a differential equation?

Can you post your work?

5. ### theamber AAC Fanatic!

Jun 13, 2008
321
0
I think is an imaginary question if there is no phase shift then there are not vectors. Current and Voltage are at 0 phase angle. He will need the RMS values and use a regular ohms Law.

6. ### silvrstring Active Member

Mar 27, 2008
159
0

But there will be a measurable/calculable phase shift on the measured loads (you can see when using polar and rectangular notation)--the circuit is inductive. The differential equation, though, offers a more accurate calculation of the instantaneous value of the voltage drop across the loads.

If you (dileepchacko) are practicing AC analysis with polar and rectangular notation, then the phase angle is important. If you are studying differential equation applications, then the instantaneous value is important. Well, I guess they're both important. The answers with each method are similar, but which method are you working with right now?

Last edited: Jul 26, 2008
7. ### theamber AAC Fanatic!

Jun 13, 2008
321
0
It is asumed you have a source sine wave with no phase shift to start with.
Under normal circumstances the capacitive or inductive properties will retard the waves. But the problem said there is NO phase shift so you cannot ignore that. That is the thing with this silly questions instead of making the student think, it creates more confusion, not only you have to tackle the question but also the puzzled text.

8. ### Norfindel AAC Fanatic!

Mar 6, 2008
325
25
I think they mean that the source itself has no phase shift, like it's a 12v @ 0 deg source, instead of a 12v @ 30 deg source, or some thing like that. The voltage drops on the components are going to have different phase shift, there's no way around that.

9. ### Ratch New Member

Mar 20, 2007
1,068
4
To the Ineffable All,

You are spot on, Norfindel. Only at resonance will there be no phase shift between the source voltage and series current. I wonder if the RLC values given are correct. At that frequency, the inductive reactance complete swamps out the resistance and capacitive reactance, thereby making the voltage across the resistor and capacitor negligible. 10ms is 200 cycles of 20kHz, so I think it can be assumed that the circuit has entered into the steady state. So the voltages across each circuit element is a trivial calculation.

Ratch

10. ### dileepchacko Thread Starter Active Member

May 13, 2008
102
1
As usual take the KVL equation for the network.

The equation become A sin (wt + ∂) = R I (t) + 1/C Int (I dt) + L diff (I dt) ---------------- (1)

Since ∂ = 0

equation (1) become A sin (wt ) = R I (t) + 1/C Int (I dt) + L diff (I dt) --------------------- (2)

Take the Laplace transform of the equation (2)

The equation (2) becomes A * wt / ( (wt)2 + S2 ) = R I (s) + (1/C)(1/S) I (s)  I(0) + L S I(s)  I(0) - (3)

Laplace of A sin (wt) is A * a / (a2 + s2) and I(0) is the initial condition.

Int : Integral
diff: differential

assuming all intial current in this circuit are zero.

Rearange the equation (3) for getting the equation I (s) in left hand side and all other tems is in right hand side.

Take the inverse laplace transform of I (s) and we will get the equation in time domain , after that sustitute the value of R, C, L and time in that equation.

11. ### silvrstring Active Member

Mar 27, 2008
159
0
Here, dileeepchacko.

It's a bit of a mess, but hey--I'm tired. If you can't make sense of it, let me know; I can try to put a 'neat' one on for you tomorrow. Gotta go to bed now.

I'm assuming this is not homework. Though, even if it is, it is important that you see the process. MATLab matches up, so it's good.

Take care,
silvrstring

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