Calculate the effective values in this 3-Phase Circuit

• Engineering
• diredragon
In summary, the student was trying to solve the homework problem, but was having difficulty understanding what was needed to do so. They found that the current in the middle LC network was not affected by one of the inductors and that this allowed them to calculate the value of the active power.
diredragon

Homework Statement

Given the known quantities ##R, C, L##, ##k##, and the effective values of the voltages ##E_1=E_2=E_3=E## , ##w = \frac{1}{\sqrt{LC}}## of the direct-symmetrical 3-Phase system find the expressions for the effective values of ##U_{12}##, ##U## and ##I## and calculate the active power ##P## of the 3-Phase generator.

Homework Equations

3. The Attempt at a Solution [/B]
Let me write down the solutions to this problem so we know what to aim for when solving it:

I feel kinda overwhelmed to not even know where to start from. I think can express the current ##I_2## of the generator ##E_2## in terms of the total impedance of the middle part and the voltage ##U##.

##U=jwI_1L+jwkLI'_2 => I'_2 = \frac{U}{jwkL}## since ##I_1=0##. The ##Z_e## in the diagram equals ##Z_e = jwL||\frac{1}{2jwC} = -jwL##.
The current ##I'_2 = I_2\frac{Z_e}{Z_e+L} => I_2 = \frac{U(k-jw)}{kLw^2}## and this is all i can get out of this first part. I could also make the left part like this:
But i don't see how that would help. What am i missing? Is there a course of action to take that i don’t see here?

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First look at the middle LC network. One of the coupled inductors is open-circuited, so it won't have any effect on the main circuit. That leaves you with 2 Ls and 2 Cs, all in parallel. What would be their effective reactance?

cnh1995 said:
First look at the middle LC network. One of the coupled inductors is open-circuited, so it won't have any effect on the main circuit. That leaves you with 2 Ls and 2 Cs, all in parallel. What would be their effective reactance?
I get ##\frac{jwL}{2}||\frac{1}{2jwC} = \frac{-jwL}{2(1-wL)} = \frac{-jL}{2(\sqrt{LC}-L)}##

No.

What can you say about Xc and XL given that ω=1/√LC? What would be the equivalent reactance of the LC network?

diredragon
cnh1995 said:
No.

What can you say about Xc and XL given that ω=1/√LC? What would be the equivalent reactance of the LC network?
Oh, ##X_L = X_C##. So at the points where ##Z_e## should be is open circuit, no current ##I_2## in that branch and since the only resistive element is in that branch that explains ##P = P_R = 0##. The voltage on the two capacitors must be the same as the same current goes through them and that's how i got ##U_{12}##. One thing about ##I##, is there an easier way than to turn this into a 3 inductors on the right side as drawn on the picture in the original post?

1. How do you calculate the effective values in a 3-Phase Circuit?

The effective values in a 3-Phase Circuit can be calculated by finding the root mean square (RMS) value of the voltage or current in each phase. This can be done by squaring each value, finding the average, and then taking the square root of the average.

2. Can the effective values in a 3-Phase Circuit be different for each phase?

Yes, the effective values can vary for each phase in a 3-Phase Circuit. This is because the voltage and current in each phase may have different magnitudes and/or phase angles, leading to different RMS values.

3. What is the purpose of calculating the effective values in a 3-Phase Circuit?

The effective values in a 3-Phase Circuit are important because they represent the average power that is being delivered to a load. This allows for accurate measurement and analysis of the circuit's performance.

4. Are there any other methods for calculating the effective values in a 3-Phase Circuit?

Yes, there are other methods such as using Fourier analysis or using a power analyzer. However, the most common method is to calculate the RMS values as described in question 1.

5. How can the effective values in a 3-Phase Circuit be improved?

The effective values in a 3-Phase Circuit can be improved by reducing any harmonics or distortions in the voltage and current waveforms. This can be achieved by using filters or other power quality improvement techniques.

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