KVL for a RLC circuit given current direction and polarity o

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Homework Statement



KVL for a RLC circuit given current direction and polarity of a resistor?

Here is the diagram of the circuit:

2.JPG


Homework Equations


KVL

Sum of voltages equal zero

The Attempt at a Solution



I am actually looking for the voltage across the inductor (VL), so I decided the best approach in finding it is through Kirchoff's Voltage Law.
Being given the polarity of the resistor, should I follow the passive sign convention for the other components in the circuit while doing KVL...if I decide to do a loop in the counterclockwise direction?

I understand that current conventionally flows through the positive terminal to the negative.
So would it be correct if I label the circuit like this, if I did a KVL mesh counterclockwise?:

RLC.JPG



Any help would be appreciated, thank you. :)
 

Answers and Replies

  • #2
cnh1995
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I am actually looking for the voltage across the inductor (VL), so I decided the best approach in finding it is through Kirchoff's Voltage Law.
What about voltage source? Where is it placed in the diagram? If RL branch and RC branch are in parallel to the voltage source, you can analyze them independently.
 
  • #3
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What about voltage source?
Well, this is the original problem...I've done part a) already, but I'm at part b) trying to find the general equation for i_x(t)...and to do that I need to find the inductor voltage when the switch is open--since I'm solving for the initial conditions di(0+)/dt and that equals VL/L.

Was just wondering if my polarities/KVL is correct in the diagram above? :frown:


Capture.JPG
 
  • #4
cnh1995
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Well, this is the original problem...I've done part a) already, but I'm at part b) trying to find the general equation for i_x(t)...and to do that I need to find the inductor voltage when the switch is open--since I'm solving for the initial conditions di(0+)/dt and that equals VL/L.

Was just wondering if my polarities/KVL is correct in the diagram above? :frown:


View attachment 95472
If the switch was closed for a long time, the capacitor must be charged to 36V. Is the direction of Ix given in the problem or you assumed it?
Edit:sorry I didn't see the diagram, it is given in the problem itself.
 
  • #5
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If the switch was closed for a long time, the capacitor must be charged to 36V. Is the direction of Ix given in the problem or you assumed it?
Ix is given. I don't think the capacitor will be charged to 36V. t=0- , capacitor should be 16V.
 
  • #6
cnh1995
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You should reverse the direction of Ix. It is not possible for given polarity of the resistor.
 
  • #7
cnh1995
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  • #8
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You should reverse the direction of Ix. It is not possible for given polarity of the resistor.
Oh why is that...if you don't mind me asking? The resistor polarity was given as positive on top and negative on bottom. I figured if I did KVL and followed the polarity of that resistor, then I would be on the right track.
 
  • #9
cnh1995
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After opening the switch, the capacitor will discharge. So, the current direction should be from +ve plate to -ve plate. Polarity of the resistor is correctly shown. You can change the direction of Ix or consider the magnitude of Ix as -Ix.
 
  • #10
gneill
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You should reverse the direction of Ix. It is not possible for given polarity of the resistor.
Actually, since it's an LC circuit it's likely going to oscillate (you'll have to check the amount of damping provided by the resistances). So Ix will probably take on either direction over time. It should simply be seen as an instruction as to how to interpret the current direction.

The initial current in the inductor is going to force the initial direction of current. Inductors are stubborn that way :smile:
 
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  • #11
cnh1995
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Right. But wouldn't the initial current direction be opposite to the given direction of Ix? I thought it would be easier to write KVL with the actual initial current direction. Hence I suggested to reverse the direction. After solving the equation, the response will be damped oscillations, but in the beginning, I thought it would be better to go with the actual direction of initial current.:smile:
 
  • #12
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Right. But wouldn't the initial current direction be opposite to the given direction of Ix? I thought it would be easier to write KVL with the actual initial current direction. Hence I suggested to reverse the direction. After solving the equation, the response will be damped oscillations, but in the beginning, I thought it would be better to go with the actual direction of initial current.:smile:
This was the solutions to the problem...but I just wanted to understsand the polarity of the circuit itself. I'm guessing as you mentioned, since capacitor is discharing, I_X goes the other direction ? Sorry I'm still a little unsure about this..

sol.JPG
 
  • #13
gneill
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Right. But wouldn't the initial current direction be opposite to the given direction of Ix? I thought it would be easier to write KVL with the actual initial current direction. Hence I suggested to reverse the direction. After solving the equation, the response will be damped oscillations, but in the beginning, I thought it would be better to go with the actual direction of initial current.:smile:
Yes, the initial current must be clockwise as driven by the inductor's initial current. But that just makes Ix negative initially.

It's usually not a good idea to tamper with predefined interpretation labels. It's too easy to forget what you did when it comes time to writing the result that's going to be submitted. But you can feel free to introduce your own currents, solve the problem, and then interpret the result in light of the defined currents/potentials.
 
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  • #14
cnh1995
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This was the solutions to the problem...but I just wanted to understsand the polarity of the circuit itself. I'm guessing as you mentioned, since capacitor is discharing, I_X goes the other direction ? Sorry I'm still a little unsure about this..

View attachment 95481
Right. Initial current goes in the other direction.
 
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  • #15
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Yes, the initial current must be clockwise as driven by the inductor's initial current. But that just makes Ix negative initially.

It's usually not a good idea to tamper with predefined interpretation labels. It's too easy to forget what you did when it comes time to writing the result that's going to be submitted. But you can feel free to introduce your own currents, solve the problem, and then interpret the result in light of the defined currents/potentials.
Yes, the initial current must be clockwise as driven by the inductor's initial current. But that just makes Ix negative initially.

It's usually not a good idea to tamper with predefined interpretation labels. It's too easy to forget what you did when it comes time to writing the result that's going to be submitted. But you can feel free to introduce your own currents, solve the problem, and then interpret the result in light of the defined currents/potentials.
So you're saying that we can completely ignore the defined polarity of the resistor that was given? And do KVL from there?
 
  • #16
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Right. Initial current goes in the other direction.
So if this was the case...shouldn't the capacitor's polarity be (+) on the bottom and (-) on the top? Since current is flowing clockwise this time, and should flow from positive terminal of capacitor to negative. The solutions seem to have it the other way around.
 
  • #17
cnh1995
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Since current is flowing clockwise this time, and should flow from positive terminal of capacitor to negative
It is flowing from +ve terminal to -ve terminal.
 
  • #18
cnh1995
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And the polarity of inductor voltage should be reversed.
 
  • #19
gneill
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So you're saying that we can completely ignore the defined polarity of the resistor that was given? And do KVL from there?
It's an interpretive label. At some point you may be asked to state what VR is at some particular time, or over time in general. You'd then use that label to interpret the result that you present. But you don't need that label or its implied polarity to solve the circuit for the current that flows.
 
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  • #20
gneill
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The potential and current labels on a diagram are merely there for interpretation purposes. They tell you how to (hypothetically) connect a meter to read a value for reporting a result. When you're solving the circuit you can introduce all your own currents and potentials and ignore the labels if you wish. When you report results that are to be interpreted in light of those labels, that's where they come into play.

Just leave the interpretive labels alone and solve the circuit. Then interpret the result in terms of those definitions.
 
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  • #21
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The potential and current labels on a diagram are merely there for interpretation purposes. They tell you how to (hypothetically) connect a meter to read a value for reporting a result. When you're solving the circuit you can introduce all your own currents and potentials and ignore the labels if you wish. When you report results that are to be interpreted in light of those labels, that's where they come into play.

Just leave the interpretive labels alone and solve the circuit. Then interpret the result in terms of those definitions.
I appreciate your explanation-- I guess the sign convention has got me overthinking it. One question that still ponders me is if i_x was originally going downwards to the 12 ohm and capacitor, but now i_x is going the other direction due to the discharge of the capacitor, does this mean the capacitor voltage is now (-16V) and the 12 ohm's voltage is now (12*(-2)) ? Aside from not labeling the signs of the elements, should we also consider the signs of the values itself?
 
  • #22
cnh1995
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I appreciate your explanation-- I guess the sign convention has got me overthinking it. One question that still ponders me is if i_x was originally going downwards to the 12 ohm and capacitor, but now i_x is going the other direction due to the discharge of the capacitor, does this mean the capacitor voltage is now (-16V) and the 12 ohm's voltage is now (12*(-2)) ?
At t=0+, inductor and capacitor are acting as sources. So, if you model them as voltage sources, you'll see that potential gain takes place in them along the direction of current and potential drop takes place in the resistors. So, given the polarity of resistor, I believe VL and Vc should be negative if Vr is positive.
 
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  • #23
gneill
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I appreciate your explanation-- I guess the sign convention has got me overthinking it. One question that still ponders me is if i_x was originally going downwards to the 12 ohm and capacitor, but now i_x is going the other direction due to the discharge of the capacitor, does this mean the capacitor voltage is now (-16V) and the 12 ohm's voltage is now (12*(-2)) ?
The polarities of potential changes that you write down depend on your direction of "KVL walk" around the circuit. The actual potential changes across components does not depend upon this; they are what they are. You can choose to "walk" around the loop in either direction, and you'd "see" the changes happen with reversed signs.

Capacitor's generally don't change their charges instantaneously. What has happened in this circuit is that the capacitor charged up to some value and then current through the capacitor stopped. When the switch opened current once again began to flow, this time in the opposite direction of the current that originally charged the capacitor. The capacitor begins with the same physical potential across it. How you interpret it is up to the the direction of your "KVL walk" when you write your KVL equation.
 
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  • #24
cnh1995
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The polarities of potential changes that you write down depend on your direction of "KVL walk" around the circuit. The actual potential changes across components does not depend upon this; they are what they are. You can choose to "walk" around the loop in either direction, and you'd "see" the changes happen with reversed signs.

Capacitor's generally don't change their charges instantaneously. What has happened in this circuit is that the capacitor charged up to some value and then current through the capacitor stopped. When the switch opened current once again began to flow, this time in the opposite direction of the current that originally charged the capacitor. The capacitor begins with the same physical potential across it. How you interpret it is up to the the direction of your "KVL walk" when you write your KVL equation.
I wish I had even 10% of your fluency!
 
  • #25
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The polarities of potential changes that you write down depend on your direction of "KVL walk" around the circuit. The actual potential changes across components does not depend upon this; they are what they are. You can choose to "walk" around the loop in either direction, and you'd "see" the changes happen with reversed signs.

Capacitor's generally don't change their charges instantaneously. What has happened in this circuit is that the capacitor charged up to some value and then current through the capacitor stopped. When the switch opened current once again began to flow, this time in the opposite direction of the current that originally charged the capacitor. The capacitor begins with the same physical potential across it. How you interpret it is up to the the direction of your "KVL walk" when you write your KVL equation.
Interesting...that is somewhat starting to make sense to me now.. if I understand you correctly, "previously" when I_x was flowing down the 12 ohm resistor and capacitor, the capacitor's polarity remains (+) on top and (-) on bottom since current normally flows in the positive terminal. But while keeping in mind that the capacitor's voltage and polarity does not change instantaneously, the polarity remains the same. Right?
 

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