RL, RC and RC circuits

Thread Starter

SilverKing

Joined Feb 2, 2014
72
Hi everyone,

I have multiple questions and I need to be sure of my answers to them.

1. For the following circuit, if Vrms= 40V, write the current and the voltage equations.


My answer:
Vmax=Vrms/0.707=56.57 V
ZT=sqrt(R^2+XL^2)=12.72 ohms
Imax=Vmax/Z=4.44 A

So the equations would be:
i=Imax sin(wt) = 4.44 sin (wt)
v=Vmax sin(wt) = 56.57 sin(wt)

Should I write -90 or +90 since I'm dealing with inductor?

2. For the following circuit, if i=0.8 sin(1000t-30), write the voltage equation and determine the average value of current.


My answer:
Since Imax=0.8 A, Vmax=Imax*ZT =0.8*5=4 (ZT=sqrt(R^2+(XL-XC)^2) =5)
v=4 sin(1000t-30)

Iavg=2*Imax/pi = 0.5 A (Or is it zero since we are dealing with sine wave?)

Is that right?
 
Last edited:

WBahn

Joined Mar 31, 2012
29,976
When you asked, "Should I write -90 or +90 since I'm dealing with inductor?" you were asking it about the current and voltage of the source. But the 90 degrees is the phase relationship between the voltage and current ONLY for the voltage across the inductor relative to the current through the inductor. The phase difference between the voltage and current of the source will be dictated by the impedance angle of the entire load, which is the combination of both the resistor and the inductor. Do you know how to calculate that angle?

As for the second question, I can't make heads or tails of what you are doing. The question asks for "the voltage equation." What voltage equation? The KVL equation around the loop? The voltage of the source? I'm guessing the latter. It then asks for the "average" current. Well (as you noted), this is a sinusoidal current, so the average is zero. They probably meant to ask for the effective (or RMS) current, but that's not what it says and it might well be a trick question aimed at seeing if you know the difference. I'd recommend providing both answers and making it clear which answers the question that was actually asked and which answers the question that might have been intended.

You then proceed to say that Iavg=2*Imax/pi. Where does this come from? Then you claim that this equation, with Imax=0.8A, works out to be 0.5A. Does that make any sense at all? At the very least you are rounding too excessively and not keeping a reasonable number of sig figs in your answer (accepted convention is three sig figs unless indicated otherwise).
 

Thread Starter

SilverKing

Joined Feb 2, 2014
72
WBahn
When you asked, "Should I write -90 or +90 since I'm dealing with inductor?" you were asking it about the current and voltage of the source. But the 90 degrees is the phase relationship between the voltage and current ONLY for the voltage across the inductor relative to the current through the inductor. The phase difference between the voltage and current of the source will be dictated by the impedance angle of the entire load, which is the combination of both the resistor and the inductor. Do you know how to calculate that angle?

If you mean the angle in the impedance triangle:
http://www.datwiki.net/images2/Impedence-triangle.jpg
This yes. ( theta=Arctan (Xc/R) )

As for the second question, I can't make heads or tails of what you are doing. The question asks for "the voltage equation." What voltage equation? The KVL equation around the loop? The voltage of the source? I'm guessing the latter.

This is might a trick question, but I'm also guessing the latter. But anyway, I hope you can help with both cases.
This is what I came up with:

KVL: V=VC+VR+VL
=(XcI)+(RI)+(XL I) ----> (The I (current) here should be the peak value or the rms?)
Voltage source: Vmax=Imax*Z
v=4 sin(1000t-30) (I'm not sure about the angle 30)


You then proceed to say that Iavg=2*Imax/pi. Where does this come from? Then you claim that this equation, with Imax=0.8A, works out to be 0.5A. Does that make any sense at all? At the very least you are rounding too excessively and not keeping a reasonable number of sig figs in your answer (accepted convention is three sig figs unless indicated otherwise).

I found in the following site that Vavg=2*Vmax/pi. I though the same can be applied on current.
http://www.electronics-tutorials.ws/accircuits/average-voltage.html
 

MrAl

Joined Jun 17, 2014
11,389
Hello,

The voltage is simply the voltage that the source would need to be in order to produce a current of the said amplitude and phase.

Your voltage amplitude looks correct being equal to 4.0, but the voltage phase shift is relative to the phase of the forcing function (I) and also the phase shift of the network itself, so the phase of 30 degrees for the result is not correct. If there was no phase shift, then the voltage would simply be out of phase by the amount due to the network alone, but because both the current and the network have a phase shift, the voltage phase shift will be either the sum or the difference of the two (you should try to determine this yourself). To start you should try to calculate the phase shift of the network with a current source that has no phase shift (ie zero degrees). You will then easily see that this phase shift must be considered as well as the source phase shift. Luckily, phase shifts like this either add or subtract, but much care has to go into this because one may be positive and the other negative, or any other combination of signs depending on the network and forcing function.

I am not sure if you know how to calculate the phase angle of the network or not. If you use complex math it's straightforward and the same for every network, but you could look up the technique to do this with a series RLC circuit if you dont want to use complex math and that will tell you right away how to get the phase angle. For the general network however you'll have to use complex math.
 
Last edited:

WBahn

Joined Mar 31, 2012
29,976
You then proceed to say that Iavg=2*Imax/pi. Where does this come from? Then you claim that this equation, with Imax=0.8A, works out to be 0.5A. Does that make any sense at all? At the very least you are rounding too excessively and not keeping a reasonable number of sig figs in your answer (accepted convention is three sig figs unless indicated otherwise).

I found in the following site that Vavg=2*Vmax/pi. I though the same can be applied on current.
http://www.electronics-tutorials.ws/accircuits/average-voltage.html
I think I would stay away from that site. Maybe that's a rush to judgment, but they give absolutely no basis for choosing to define the "average" voltage over just one half of the waveform. That has no physical meaning that I am aware of and it certainly has no physical meaning applied to this circuit.
 

Thread Starter

SilverKing

Joined Feb 2, 2014
72
OK. Another question:If I want to find the voltage across the resistor or the inductor, should I use VR=R*I (or VL=XL*I)? And what is this ''I''? Is it Imax or Irms?
 

WBahn

Joined Mar 31, 2012
29,976
OK. Another question:If I want to find the voltage across the resistor or the inductor, should I use VR=R*I (or VL=XL*I)? And what is this ''I''? Is it Imax or Irms?
Since I is the series current that passes through each of the components, yes, you multiply I by the resistance/reactance of THAT component to find the voltage across THAT component.

As to whether you use Imax or Irms, that depends on whether you want to find Vmax or Vrms.
 

MrAl

Joined Jun 17, 2014
11,389
Hi,

You do have to be careful though.
For example, we have a resistor 4 ohms and capacitor 1/3000 Farad and AC voltage source 5vac with angular frequency (w) of 1000, all in series.
The current in the circuit is 1 amp AC, and the reactance of the capacitor with w=1000 comes out to 3 ohms. The voltage across the cap measures 3vac and the voltage across the resistor measures 4vac, but the sum of 3+4 does not equal the source voltage which is 5vac.
 
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