Understanding Kirchhoff's Law in LC Circuits: Exploring the Confusion

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  • #1
Youngwoo Cho
2
0
I am confused with explanation about Kirchhoff's law in LC circuit.
Please refer to the file I attached for the LC circuit.
First, the switch was connected to the emf and the capacitor was charged to its full capacitance.
Then the switch is connected to the capacitor, the capacitor will start to discharge clockwise. Let me trace the closed loop clockwise and sum all of the voltage differences. Then I get (q/c) - L(dI/dt) = 0 for the Kirchhoff's second law, because I trace the capacitor from its cathode to anode and the inductor will opposite the current. But common textbooks says that (q/c) + L(dI/dt) = 0 with no thorough explanation. What am I missing for this issue?
 

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  • #2
Welcome to PF!

Both equations are right if you have the correct relationship between q and I. What is it in your case?

ehild:
 
  • #3
ehild said:
Welcome to PF!

Both equations are right if you have the correct relationship between q and I. What is it in your case?

ehild:
Thank you ehild.
What do you mean by the correct relationship between q and I?
 
  • #4
Q (the charge on the capacitor) and the current I are related. Taking this relation into account you get a differential equation for Q that you can solve and you get the time dependence of the charge and from that, the time dependence of the current.
How are Q and I related?
 
  • #5
Hi!

I have been agonizing over the exact same problem. I spent several hours today thinking about this and trying to figure out the stupid sign. I slightly edited the image posted by OP to add the capacitor charges and to specify my starting point. I am considering a situation where at time t = 0 the capacitor has +q on the top plate and -q on the bottom. Current will thus flow in the direction of the arrow, initially. If I start at point a in the attached image and go around clockwise, then, just as the OP wrote, I get:
q/C - L(di/dt) = 0

The inductor will experience a voltage drop when following the direction of current; thus I get the negative sign in front of the L term since it's a "drop". Qualitatively, the inductor will initially oppose the increasing current, so it creates a potential opposite in direction to the potential across the capacitor, similarly to how it would be if this were an RC circuit.

To answer ehild's question, i = dq/dt. But I still don't see how this allows me to get the correct differential equation with the correct signs.

I read what Young & Freedman wrote in their book about this, but it made no sense (to me, at least). I would really really appreciate help here. This has been driving me absolutely crazy.
 

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  • #6
fr0zensphere said:
To answer ehild's question, i = dq/dt.
Hmmmm when you close the switch does q increase or decrease?
 
  • #7
After the switch is closed, dq/dt < 0 for sure. I am treating q(t) as the charge on the positive plate of the capacitor. So, as the switch is closed, charge begins draining away from the top plate as the capacitor discharges -- i.e. q is decreasing.

I believe what Young & Freedman wrote in their book has to do with this, but I don't see why it matters. How does the fact that dq/dt is initially negative have to do with d2q/dt2? That's the term that really matters.

In the analogous mechanical system, 1D oscillator on a spring, if I stretch the mass to the right of the origin (positive x), then the initial dx/dt is also < 0 because the mass moves to the left. But that doesn't change change the sign you put in front of the m*dx2/dt2 term in the differential equation. If I instead were to stretch the spring to the left of the origin, then dx/dt would be > 0, but I would still have the same differential equation. Put another way, dx2/dt2 doesn't care whether dx/dt < 0 or > 0.
 
  • #8
Bump. No one knows how to resolve this?
 
  • #9
dq/dt is the rate of increase of the charge on the upper plate and your diagram defines "i" as charge leaving it.
So the equation relating the two is . . . . .
 
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1. What is Kirchhoff's Law in an LC circuit?

Kirchhoff's Law in an LC circuit is a set of rules that describe the behavior of electrical currents and voltages in a closed circuit. It states that the sum of the voltages in a closed loop must be equal to zero, and the sum of the currents entering a node must be equal to the sum of the currents leaving that node.

2. How does Kirchhoff's Law apply to an LC circuit?

In an LC circuit, Kirchhoff's Law applies to the sum of voltages and currents in the loop that includes the inductor and capacitor. The voltage across the inductor is equal to the voltage across the capacitor, and the sum of these two voltages is equal to the applied voltage source.

3. What is the significance of Kirchhoff's Law in an LC circuit?

Kirchhoff's Law is significant in an LC circuit because it helps us understand the behavior of electrical currents and voltages in a circuit, and allows us to predict and analyze the circuit's response to different inputs. It also helps us to ensure that the circuit is functioning correctly and efficiently.

4. How is Kirchhoff's Law used to solve problems in an LC circuit?

Kirchhoff's Law is used to solve problems in an LC circuit by setting up and solving a system of equations based on the sum of voltages and currents in the circuit. This allows us to determine the unknown values of currents and voltages in the circuit, and analyze the behavior of the circuit under different conditions.

5. Are there any limitations to Kirchhoff's Law in an LC circuit?

One limitation of Kirchhoff's Law in an LC circuit is that it assumes ideal conditions, such as zero resistance and no external interference. In real-world circuits, these conditions are not always met, which can lead to inaccuracies in the predictions and calculations using Kirchhoff's Law. Additionally, Kirchhoff's Law only applies to circuits in steady-state conditions, and may not accurately predict the behavior of the circuit during transient states.

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