Complex power and circuits

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  • #1
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1. Homework Statement [/
1) Find the average and relative power for the voltage source and for each impedance branch. Are they absorbing or delivering average power/magnetizing VARS?

Homework Equations


Those derived and
average power=P=((Vm*Im)/2)*(cos(theta(v)-theta(i))
reactive power=Q=((Vm*Im)/2)*(sin(theta(v)-theta(i))

The Attempt at a Solution


These are the equations that I derived from the schematic:

for the first loop on the left, there really is no need for big equations since: R1=50 ohms
Then V=IR => I = V/R => I= (340<0 degrees/50 ohms)= 17/125 <0 degrees

such that for the second loop we want to find i2, and I derived this equation:
(-j100)*i2-(i2-i1)(-j100)+80*i2+(j60)*i2=0

substitute for i1 into the equation above and solve for i2 but I am not sure if this is correct.
Also, how would you simplify complex numbers, or do I need to use phasor notation (how do I do that)?

To calculate the average power, we simply use the formula mentioned above. But what do we input for the power factor angle?So, if it is positive power, it is an absorbing component otherwise it delivers the power if negative result.

Then to obtain the magnetizing VARS is only for the capacitor, so we use the same formulas but only the name changes right?

Thank you for your time!
 

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Answers and Replies

  • #2
Zryn
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for the first loop on the left, there really is no need for big equations since: R1=50 ohms
Then V=IR => I = V/R => I= (340<0 degrees/50 ohms)= 17/125 <0 degrees

Assuming that you were given that Vg = 340V, and you are correct in that I = V/R, can you explain what 17/125 < 0 degrees means.

such that for the second loop we want to find i2, and I derived this equation:
(-j100)*i2-(i2-i1)(-j100)+80*i2+(j60)*i2=0

I believe there's something wrong with the part that takes into account the current through the capacitor. You should count both the i1 current and the i2 current, though in opposite directions so what remains is the difference of the two i.e. the part of the current that will actually travel along the capacitor branch. At the moment you are counting the i2 current and the 'difference between i2 and i1' current. If we rearrange/factorise the math in that section, (-j100)*i2-(i2-i1)(-j100), we get (-j100)*(i2-i2+i1) = (-j100)*i1, which is not the [STRIKE]droids[/STRIKE] current we are looking for. Otherwise, the Kirchoffs equation around the loop looks good.

substitute for i1 into the equation above and solve for i2 but I am not sure if this is correct.

The process sounds good since you have only one equation and only one unknown, so it should work.

Also, how would you simplify complex numbers, or do I need to use phasor notation (how do I do that)?

Given (X +/- iY) +/- (A +/- iB) = (X +/- A) +/- i(Y +/- B)
Given (X +/- iY) * (A +/- iB) expand as per normal algebra albeit keeping in mind i^2 = -1
Given (X +/- iY) / (A +/- iB) convert from rectangular to polar

Given (X < Y) +/- (A < B) convert from polar to rectangular
Given (X < Y) *// (A < B) = (X *// A) < [Y + B for *] OR [Y - B for /]

For Polar --> Rectangular: X < Y = Xe^(iY) = X{cos(Y) + i*Sin(Y)}
For Rectangular --> Polar: (X +/- iY) = sqrt(X^2 + Y^2) < tan^(-1)(Y/X)
Or a number of calculators do this bit for you nowadays.

These are the rules I use (from memory, you should double check them).
 
  • #3
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I think that the <0 degrees just means that the sources are in phase but I think you are right in that they may be irrelevant in the calculations.

so, is this correct for Kirchoff's equation?
-j100(i2-i1)+81*i2+(j60)*i2=0
I think that the <0 degrees just means that the sources are in phase but I think you are right in that they may be irrelevant in the calculations.

so, is this correct for Kirchoff's equation?
-j100(i2-i1)+80*i2+(j60)*i2=0

if so,
=> -j100*i2-(340/50)*(-j100) +80i2+j60*i2=0
=> -j40*i2+680j+80i2=0
=> solving for i2=(-j680/(-j40+80))=(-j68/(-j4+8))=17j/(j-2) A

so i have i1=6.8A and i2=17j/(j-2)

To find the average power: P=(Vm*Im/2)*cos (theta(v)-theta(i))
since theta is zero we ignore this and substitute cos (...)=1?

Psource=(340*i1)*1=(340*6.8)=-2312 W delivering power and no magnetizing vars?
P50ohm=2312 W absorbing power
Pc=Vm*(I2-I1)= Vm*(17j/(j-2) -6.8)?
PL=(i2/R)*(i2) = ((i2)^2)/R =(j17340)/(j^2-j4=4)
where c stands for capacitor, L for inductor

and the reactive power: P=(Vm*Im/2)*sin(theta(v)-theta(i))
let sin(0)=0

Psource=0 W
P50ohm=0W
Pc=0 W??
PL=0W??
 
Last edited:
  • #4
Zryn
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I think that the <0 degrees just means that the sources are in phase but I think you are right in that they may be irrelevant in the calculations.

When you have an angle of 0, it usually means that that component is the reference i.e. everything else is leading or lagging relative to that specific component.

The bit I was curious about was how (340 < 0) / 50 = 17 / (125 < 0). You solved i1 = 6.8A, but what math / explanation leads to the bold part of the equation.

so, is this correct for Kirchoff's equation?
-j100(i2-i1)+80*i2+(j60)*i2=0

This equation says to me that you expect a voltage rise due to i2 across the capacitor, a voltage drop due to i1 across the capacitor, a voltage rise due to i2 across the resistor, and a voltage rise due to i2 across the inductor. If you label these voltage polarities on your diagram, are the voltages due to i2 arranged so they have the same polarity around the loop?

if so,
=> -j100*i2-(340/50)*(-j100) +80i2+j60*i2=0
=> -j40*i2+680j+80i2=0
=> solving for i2=(-j680/(-j40+80))=(-j68/(-j4+8))=17j/(j-2) A

so i have i1=6.8A and i2=17j/(j-2)

That math looks good, assuming that you have the previous equation correct.
 
  • #5
Zryn
Gold Member
310
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Also, after you look into the KVL, if i1 = 6.8A and V = IR, what is the voltage drop across the 50R resistance and what does that mean for the rest of the circuit?
 

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