Phase voltage and Line-Line voltage relation

In summary, the conversation is about finding the phase and line-line voltage relation in a circuit using KVL. The circuit contains a wye transformer, ideal DC sources, and switches. The main question is what the circuit is supposed to do and how the switches are controlled. The conversation also includes attempts at using KVL to calculate the voltages Ua, Ub, and Uc in different switch states. The conversation ends with uncertainty about the correctness of the approach and a reference to a user guide for more information.
  • #1
PhysicsTest
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Homework Statement
Find the phase voltage and Line-Line voltage relation
Relevant Equations
Kirchhoff voltage and current laws
The below is the diagram i want to find the phase and line-line voltage relation
1676305359954.png

I am finding difficulty in identifying the loop and applying the KVL.
One attempt is
U_{VA} - U_R - U_L -U_O - U_L - U_R = U_VB
U_{VA} - 2(U_R+U_L) = U_VB -> eq1

Is my attempt correct? i am not confident please help.
 
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  • #2
UVA and UVB should actually be equal if this is a schematic of a wye transformer. They should both be 120V. The line to line voltage would be UAB.
 
  • #3
Probably a 3Φ motor drive or inverter, since there are ideal DC sources in there. But it could be a rectifier, I guess. Much depends on the switches. Normally KVL would only be applied for a specific switch state since the network changes when switches change.

What is this circuit supposed to do? How are the switches controlled?
 
  • #4
Yes it is inverter
SAtSBtSCtU_VAU_VBU_VCU_ABU_BCU_CAUaUbUc
100udcbus/2-udcbus/2-udcbus/2udcbus0-udcbus
010-udcbus/2udcbus/2-udcbus/2-udcbusudcbus0
001-udcbus/2-udcbus/2udcbus/20-udcbusudcbus
110udcbus/2udcbus/2-udcbus/20udcbus-udcbus
101udcbus/2-udcbus/2udcbus/2udcbus-udcbus0
011-udcbus/2udcbus/2udcbus/2-udcbus0udcbus
I want to calculate the voltages Ua, Ub, Uc
For the 1st case if i consider
1 0 0 the circuit will be as below
1676390674623.png


If i write KVL
a. 1 0 0
VDCBus/2 + Ua -Ub + VDCBus/2 = 0;
Ua - Ub + VDCBus=0 -> 1

b. 0 1 0
VDCBus/2 + Ub -Ua + VDCBus/2=0
-Ua + Ub + VDCBus = 0 -> 2

c. 0 0 1
VDCBus/2 + Uc - Ub + VDCBus/2 = 0 -> 3

Very confusing am i in correct direction? The document is user guide page 51.
 

FAQ: Phase voltage and Line-Line voltage relation

What is the difference between phase voltage and line-line voltage?

Phase voltage is the voltage measured across a single component in a three-phase system, typically between a phase and the neutral point. Line-line voltage, on the other hand, is the voltage measured between any two of the three phases. In a balanced system, the line-line voltage is √3 times the phase voltage.

How do you calculate line-line voltage from phase voltage?

To calculate line-line voltage (VLL) from phase voltage (VPH), you use the formula: VLL = VPH × √3. This relationship holds true for a balanced three-phase system.

Why is the line-line voltage higher than the phase voltage?

The line-line voltage is higher than the phase voltage because it is the vector sum of two phase voltages that are 120 degrees apart. This results in a line-line voltage that is √3 times the phase voltage, due to the geometry of the vector addition in a three-phase system.

What is the significance of the √3 factor in three-phase systems?

The √3 factor arises from the 120-degree phase difference between any two phases in a three-phase system. When you calculate the voltage difference between two phases, you are essentially dealing with the sides of an equilateral triangle, where the side length (line-line voltage) is √3 times the height (phase voltage).

Can you measure phase voltage directly in a three-phase system?

Yes, you can measure phase voltage directly by connecting a voltmeter between a phase and the neutral point in a three-phase system. However, in many industrial settings, it is more common to measure line-line voltage because the neutral point may not be accessible or the system may be designed without a neutral.

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