# Homework Help: Complex power S

1. Apr 8, 2016

### Ivan Antunovic

1. The problem statement, all variables and given/known data
In sinusoidal circuit shown in Figure 13 is known : w =10^6 1/s, R = 100 Ohm ,
L = 300μH , C1 = 10nF and C2= 5nF . Reactive power of coil inductance L is QL = 3kVAr ,
RMS value of the voltage receiver impedance Z is UZ = 100 V , and the voltage UZ phase delaying behind
electricity generator Ig for pi / 2. Calculate the complex apparent power S of that generator.

2. Relevant equations

3. The attempt at a solution
I set current source at angle 0. Ig = Ig exp( j*0) = Ig
By compensation theorem , I replaced impendance Z with ideal voltage source Ek =Uz exp(-j*pi/2) = -j100 V
From QL = IL^2 * XL , I get IL = sqrt(QL/wL) = sqrt(10) A.
Using node voltage method
I get result from the picture.
The problem is that there are too many unknowns (2 equations and 3 unknowns) , well I got RMS value of IL but I still don't know it's angle.
I am pretty stuck at this point.

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2. Apr 8, 2016

### Staff: Mentor

If I may make a suggestion? Try mesh analysis. The loop with the current source $I_g$ is already "solved" since the mesh current will equal $I_g$. So remove the loop (remove $I_g$) and replace it with a suitable voltage source in each branch that that loop current passes through. Solve for the loop current through L. It will be in complex form. That will take of your missing angle for the current through the inductor.

3. Apr 9, 2016

### Ivan Antunovic

Mesh analysis I got
Uz-Ig = I2*(ZC1 + R) - I1 *ZC1
Ig*ZC2 = -I1(ZL+ZC+ZC1)+I2*ZC2

and IL=I1=(UZ-ZC2*Ig)/ZL

After solving these equations I get that Ig = -j
and IL = UZL (-90) / ZL ( 90) => gives angle of -180 degrees therefore I got IL = -sqrt(10) in complex form
But as you can see from the picture I get really big I2 I am doing something wrong .

And UIg = Ig*ZC2 + I1 * ZC2 + Ig*R + I2 *R ?

4. Apr 9, 2016

### Staff: Mentor

In your first mesh equation on the left had side Ig should be multiplied by R. I'm not seeing how you arrived at your equation for IL.

Note that since Ig provides the phase angle reference, it should end up being a purely real value.

Why don't you drop in the actual impedance values for the components? The mesh equations should simplify quite a bit that way.

(Note that the polarity of the source in loop (1) has been reversed rather than write its value as -j200 Ig)

5. Apr 9, 2016

### Ivan Antunovic

But I still don't know the current I1 ,and therefore I can't calculate Ig.I know that | I1 | = | IL | = sqrt(10) , but that seems kinda useless when those are complex equations.

screen cap

6. Apr 9, 2016

### Staff: Mentor

Taking your equation (2) and dividing though by -100:

$I_g + j = -j I_1 + (j - 1) I_2~~~~~~~~~~~$..(2')

Substitute (1) into (2'), solve for $I_1$:

$I_g + j = -j I_1 + 2(j - 1) I_g$
$I_1 = -j(-3 I_g + 2j I_g - j)$
$I_1 = 2 I_g - 1 + j 3 I_g~~~~~~~~~~~~~$..(3)

That's a bit easier to work with.

The complex power associated with the inductor will be j3000 (units: VA). It will be given by the inductor voltage ($I_L Z_L$) multiplied by the complex conjugate of the inductor current. The inductor current is identical to $I_1$. See if you can find $I_g$ by that route.

7. Apr 9, 2016

### Ivan Antunovic

If I understood you well

take a screenshot

Still I am getting wrong result , seems like I messed up again.

8. Apr 9, 2016

### Ivan Antunovic

Okay it should be (EXPRESSION for current complex current IL) , IL * exp(j*fi) * IL * exp(-j*fi) ... not angle (180 - fi) , but that 's gives zero exp(j*fi -j*fi) , and I get sqrt(10) * 1 * sqrt(10) * 1 ---> IL = 10 and that doesn't make any sense .

9. Apr 9, 2016

### Staff: Mentor

$j 3000 = (I_1 Z_L)(I_1^*)$

So inserting (3) from post #6:

$j 3000 = \left[ (2 I_g - 1 + j 3 I_g)(j 300) \right] (2 I_g - 1 - j 3 I_g)$

Solve for $I_g$. There should be two candidate values.

10. Apr 9, 2016

### Ivan Antunovic

Thank you I got the right answer now

image sharing

By the way what angle should I put for theta at red equation.
And by what theorem did you replace current source Ig with Ig*ZC2 and Ig*ZC1 ?

11. Apr 9, 2016

### Staff: Mentor

For a complex conjugate you just negate the angle. That is, $\left( I e^{j θ} \right)^* = I e^{-j θ}$

I'm not aware of any particular named theorem that prompts what I did to distribute the $I_g$ current supply as voltage sources. But it certainly obeys KVL and mesh analysis rules. If you left the current source loop in place and wrote the mesh equations for loops (1) and (2) you would be able to pick out the terms that correspond to the mesh current $I_g$ passing through the common branch impedances. These are the potential changes imposed by $I_g$ flowing through those impedances. Since $I_g$ is a given current quantity they correspond to voltage sources. So I suppose this is a version of the "compensation theorem" that you invoked in your first post..

12. Apr 10, 2016

### Ivan Antunovic

Not sure if it's compensation theorem http://www.electrical4u.com/compensation-theorem/
, more like Norton - Thevenin ideal current / voltage source equivalent , will try to check it on some other networks and see if it works. I am really trying to remember all these rules from 'fundamentals of EE' , since I am not at 1st year anymore.
Also is there limitation for posting threads ? Because I am stuck with some other problems .

13. Apr 10, 2016

### Staff: Mentor

Nope, there are no restrictions on the number of threads you can start. If you have more than one or two going simultaneously make sure that you have enough time to pay attention to all of them at once.