Kirchoff's Voltage Law in RLC circuit

In summary, the conversation discusses setting up KVL circuit equations for a circuit with alternating voltage sources. The conversation includes mention of Kirchoff's Voltage Law, Ohm's Law, and Steinmetz notation. The final solution involves representing the voltages as complex quantities and using them in the equations to solve for the currents.
  • #1
majin_andrew
20
0

Homework Statement



I am required to write KVL circuit equations for the following circuit

KirchoffRCL.jpg


I don't need to finish solving for I1, I2, I3 and I4 at the moment. I just need a bit of help setting up the equations. Thanks.

Homework Equations



Kirchoff's Voltage Law, which states that the sum of voltage drops in a loop must be equal to the sum of the voltage rises.

Ohm's Law, V = IR

The Attempt at a Solution



There are a few things I am unsure of.

The first is that I don't know what the 30 degrees and the 45 degrees refers to beside the voltage sources. My first guess was that it means the phase that the alternating voltage source was up to at that particular moment in time, but I'm not sure if EA is even alternating, because it doesn't have the squigly line in the circle. In my attempted solution I guessed that both EA and Eab are alternating, and that the angle given refers to where in the phase it is at that point in time (with 0 degrees at the peak).

Another thing I'm unsure of is is the notation of Eab = 10. Does this mean the the potential at a is 10 higher than the potential at b? Or the other way around? I am unsure of the direction of the voltage. For my attempted solution I guessed that 'a' had the higher potential.

Anyway, here is my attempt at a solution:
Eab= 10 cos(30 degrees) = -20*I1

Eab= 10 cos (30 degrees) = 60*j*I2

Eab + EA= 10 cos (30 degrees) + 20 cos (45 degrees) = -30*-j*I3 = 30*j*I3

...
and from here I would be able to perform simultaneous equations to find the currents.

Any help would be greatly appreciated!
Andrew
 
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  • #2
majin_andrew said:

The Attempt at a Solution



There are a few things I am unsure of.

The first is that I don't know what the 30 degrees and the 45 degrees refers to beside the voltage sources. My first guess was that it means the phase that the alternating voltage source was up to at that particular moment in time, but I'm not sure if EA is even alternating, because it doesn't have the squigly line in the circle. In my attempted solution I guessed that both EA and Eab are alternating, and that the angle given refers to where in the phase it is at that point in time (with 0 degrees at the peak).
It's called Steinmetz notation. It just tells you the amplitude and relative phase of the signal. In your problem, [itex]E_{ab}=10 \cos(\omega t+30^\circ)[/itex] and [itex]E_A = 20 \cos(\omega t+45^\circ)[/itex].
Another thing I'm unsure of is is the notation of Eab = 10. Does this mean the the potential at a is 10 higher than the potential at b? Or the other way around? I am unsure of the direction of the voltage. For my attempted solution I guessed that 'a' had the higher potential.
"Higher" probably isn't the best word to use since it's an alternating source. The subscript "ab" usually means that you're measuring the voltage of point a relative to point b.
 
  • #3
Both the things you said make sense. Thanks vela!
 
  • #4
So, with this new understanding, I think the KVL equations are:

[itex] E_{ab}=10 \cos(\omega t+30^\circ) [/itex] = -20*I1

[itex] E_{ab}=10 \cos(\omega t+30^\circ) [/itex] = 60*j*I2

[itex] E_{ab} + E_A= 10 \cos(\omega t+30^\circ) + 20 \cos(\omega t+45^\circ) [/itex] = -30*-j*I3 = 30*j*I3

Which can then be solved to obtain the currents. Is this correct?
 
  • #5
Essentially, but you have a slight problem. The lefthand sides are real while the righthand sides are complex. You want to represent the voltages as a complex quantity and use that on the LHS of the equations.
 
  • #6
Ok thanks. Representing the voltage as a complex quantity:

[tex]E_{ab}=10e^{\frac{\pi}{6}j}e^{j\omega t}=-20I_{1}[/tex]

[tex]E_{ab}=10e^{\frac{\pi}{6}j}e^{j\omega t}=60jI_{2}[/tex]

[tex]E_{ab} + E_A=10e^{\frac{\pi}{6}j}e^{j\omega t}+20e^{\frac{\pi}{4}j}e^{j\omega t}=-30jI_{3}[/tex]
 
  • #7
That's good. The notation is for representing phasors, so you don't actually need to write in the frequency factor explicitly. It's assumed to be in there.
 
  • #8
Ok thanks a lot for your help vela!
 

1. What is Kirchoff's Voltage Law (KVL) in RLC circuits?

Kirchoff's Voltage Law states that the sum of all voltage drops in a closed loop in a circuit is equal to the sum of all voltage gains in that same loop. In other words, the total voltage in a closed loop circuit must equal zero.

2. How is KVL applied in RLC circuits?

In RLC (resistor-inductor-capacitor) circuits, KVL is used to analyze the voltages across each component in the circuit. By applying KVL to each individual loop in the circuit, we can determine the relationships between voltages and currents in the circuit.

3. Why is KVL important in RLC circuit analysis?

KVL is important in RLC circuit analysis because it allows us to determine the voltages across each component in the circuit, which is essential in understanding how the circuit functions and how to design and troubleshoot it. It also helps us to check for errors or inconsistencies in our circuit calculations.

4. What are the limitations of KVL in RLC circuits?

One limitation of KVL in RLC circuits is that it assumes ideal components, meaning no resistance in wires and no internal resistance in components. In real-life circuits, there will always be some resistance, so KVL may not always be completely accurate. Additionally, KVL is only applicable to series circuits, so it cannot be used in parallel RLC circuits.

5. How can KVL be verified in RLC circuits?

KVL can be verified in RLC circuits by performing a voltage loop analysis and comparing the calculated voltages to the measured voltages in the circuit. If the calculated and measured voltages are within a reasonable range of each other, KVL is verified to be accurate in that circuit.

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