# RLC circuit question

Discussion in 'Electronics Homework Help' started by Ali_Law, May 8, 2014.

1. ### Ali_Law

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May 8, 2014
Hi there,

I was wandering if anyone could help with this question:

"An alternating voltage v = 120sin(1000t + (pi/3) ) is maintained across a series circuit containing a 1Ω resistor, 400μF capacitor and 4.5mH inductor.

Calculate the output Voltage in Cartesian notation, taken from the resistor"

So far i have calculate Xc = 2.5 Ω and Xl = 4.5Ω but am unsure of the next step..

Any help would be greatly appreciated!

2. ### duke37

5,364
772
Jan 9, 2011
Subtract these two figures to get the nett reactance.
Add the resistance in the correct way (not simple) to get the total impedance.

3. ### LvW

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146
Apr 12, 2014
In case the question asks for the steady-state response (that means: not for the switch-on transient) we have a just a complex voltage divider.

4. ### Ali_Law

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May 8, 2014
The solution i was given was Vo = (53.6- j3.2). I managed to calculate the total impedance using |Z| = Srt(R^2 + (Xl - Xc)) This gave |Z| = Srt(5). How would i go from this to the solution?

One formula i have found is V0/Vin = Zr / (Zr + Zl + Zc) but this does not give me the solution i was provided

5. ### LvW

604
146
Apr 12, 2014
Did you consider that Zl and Zc are imaginary quantitties? Thus, the whole denominator is complex (magnitude and phase).

6. ### duke37

5,364
772
Jan 9, 2011
In the equation for Z, you have an R^2 term, the reactance should also have a ^2 term.

7. ### Ali_Law

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May 8, 2014
yeah i did actually square the reactance term as well to get sqrt(5) that was just a typo.

8. ### Ali_Law

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May 8, 2014
I know that they are imaginary quantities i am just unsure how you would apply Vin in that formula to obtain the solution

9. ### Ratch

1,094
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Mar 10, 2013
You should be able to get help on RLC circuits on the web, such as http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rlcser.html . That should be your first attempt at help. Your next step is to calculate the impedance and divide that into the voltage (120 volts) to get the current. Since the current in a series circuit is the same through each component, you can get the voltage with phase across each component by multiplying the current by the impedance of the components.

Although mathematicians and physicists call reactance voltages by the misnomer "imaginary", those voltages are really othogonal quantities with respect to the current. They can, however, become very "real" if you get your body parts across a highly energized "imaginary" coil or capacitor.

Ratch

Ali_Law likes this.
10. ### Ali_Law

5
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May 8, 2014
Thank you very much this was very helpful!

I = (120 |60 ) / (srt(5) |63.4 )
I = 120 / srt(5) + j (60 - 63.4)
I = 53.66 - j 3.4
R = 1
Vr = I x R
Vr = (53.66 - j 3.4) x 1
Vr = 53.66 - j 3.4

11. ### Ratch

1,094
334
Mar 10, 2013
Why did you put 60° into the excitation voltage? The excitation phase is not going to change the impedance or the current values with respect to the voltage. You want to figure the current with the max voltage of 120 volts, not at some lessor value like 60°. Why do you change those figures into polar? Can't your calculator do rectangular coordinates? This time figure the voltage across R, L, and C and see if those three voltages add up to 120 volts. If they don't, you got it wrong.

Ratch  