RLC series-parallel

Discussion in 'Homework Help' started by caique221, Nov 17, 2013.

1. caique221 Thread Starter New Member

Nov 10, 2013
7
0
Hello,

I've already asked questions relative to a RL series circuit in another post, but somehow I'm no longer able to post on it. I'd like to thank the people who helped me: I managed to solve the problems and got the correct answers.

Unfortunately, I'm having a hard time again with another circuit, this time a RLC series-parallel.

I'm attaching the circuit sketch.

a) Equivalent impedance of the circuit;
b) The instantaneous current and the RMS current in the circuit;
c) Phasor diagram for RMS V1, VA and VB;
d) Phasor diagram for RMS I1, I2 and I3;
e) The true, reactive and apparent powers used by the circuit.

a) I found the total impedance of the circuit by using the basic formulas, such as XL=wL and XC=1/(wC). By solving the circuit, I found it to be 667,52+j544,22 (ohm) or 861,25ohm with an phase angle of 39,19 degrees.

b) I'm really not sure if I got it right, but since the problem gave the maximum voltage on the source (169,7 volts), I simply divided it by the square root of two and found the RMS voltage to be approximately 120 volts.

To find the instantaneous current (max.) of the source, I applied Ohm's Law using the equivalent impedance and the max voltage. I found it to be I(t)=0.197sen(377t-0,684). The minus 0,684 is the phase angle of the current in radians. I'm not sure if its correct.

To find the RMS current, I divided the max. current magnitude(0,197) by the square root of two, and got I(rms)=0,139 with a phase angle of -39,19 degrees.

c,d) "Drawings". Since I'm not sure that I got the (b) values correctly, maybe they are wrong:

- I1=0,139A with an angle of -39,19 degrees;
- I2=0,33A with an angle of -70,51 degrees;
- I3=0,25A with an angle of approximately 90 degrees.

Since I1 are in series with the source, they are receiving all the RMS current sent by the source; to find I2, I found the voltage drop across the branch and its equivalent impedance. For I3, I simply used the voltage drop I found on the I2 branch, since they are in parallel.

e) Thats where I started to think I had messed up something. First, my apparent power, on its rectangular form, gave some crazy values. I got 12,92+j10,54 VA using the source voltage with 0 degrees multiplied by the conjugate complex of the current with -39,19 degrees.
Since it gave some pretty low value, I went with another method to try to find the apparent power. I got the individual power on each component. The true power component I calculated was 3355,6W, totally off from the 12,92 I found on the rectangular calculation. On the reactive power component, I subtracted the total capacitive reactive power from the total inductive power, and got 33,49 VAR. Using the pythagorean theorem, I found the apparent power to be 3355,77 VA. Thats still dumb. Since the phase difference between RMS voltage and RMS current is 39,19 degrees, which gives a power factor of 0,78 approximately, I can tell that there would be at least some intermediate contribution of the reactive power to the apparent power. By comparing the two power values, the reactive power is pretty much insignificant, which would lead to a much better power factor value, I think. I'm pretty sure I messed something along the way, and would appreciate some help.

File size:
8.1 KB
Views:
83
2. WBahn Moderator

Mar 31, 2012
23,576
7,214
You appear to be going along pretty well until you say that you found the individual power on each component. At that point you say you got over 3kW of average power but since you show no work there is no way for us to even begin to guess where you went off the rails.

3. caique221 Thread Starter New Member

Nov 10, 2013
7
0
I'm trying to redo my steps. Starting by the currents, I'm guessing something on them went wrong.

The sum of the current in the branches should be equal to the total current sent by the source (I guess). Since the RMS current is 0,139A with an angle of -39,19 degrees, then the sum of I2 and I3, which are both 0,33A with an angle -70,51 degrees, and 0,25A with an angle of 90 degrees, should equal the RMS current. Instead of getting around 0,139A with an phase of -39,19 degrees, I'm getting 0,126A with an angle of -29,02 degrees. I did something wrong on the currents or are they really supposed to be off? I started to study electricity this year, so I don't dominate most concepts.

4. WBahn Moderator

Mar 31, 2012
23,576
7,214
Walk through your work in detail. Preferably presented in a format that is easy to follow as opposed to written in prose. Show us HOW you arrived at each of these currents.

5. caique221 Thread Starter New Member

Nov 10, 2013
7
0
By going through some of my steps to give you exactly what I did, I found that I somehow put a wrong phase angle at the impedance of the RL in series with the source. Redoing the calculation, the correct angle is 81,84 degrees (converted from 20+j139,49). Maybe thats the problem. I'm going to sleep now and I'll be trying to do all the calculations again with this new angle tomorrow. I'll also check to see if I missed some other points due to lack of attention. After that, I'll post here what I get.

Thank you.

6. WBahn Moderator

Mar 31, 2012
23,576
7,214
It is very often the case that if you carefully and systematically try to walk someone else through what you've done, that you will catch the places where you went wrong. At times, I have been known to literally speak to an empty chair as though there were someone sitting there and walk them through my work. If my dog is around I'll walk him through it. It's a win-win: I get to find my problem and he enjoys the attention.