Solving LC-Circuit Problem: Q & I as Func. of Time

  • Thread starter gralla55
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In summary, the conversation is about an LC-circuit with given values for L and C, and initial conditions for charge on the capacitor (Q) and current through the inductor (I). The question is to write an expression for Q and I as a function of time, and the attempted solution involves using a standard solution form and solving a system of equations with two unknowns. However, there is a discrepancy with the answer in the book, and upon further examination, it is found that the book's answer is incorrect. It is suggested to approach the problem by solving the differential equation directly.
  • #1
gralla55
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Homework Statement




In an LC-circuit, with:

L = 64 mH
C = 121 nF

where at t=0,

Q = 10*10^-6 C , (the charge on the capasitor)
I = 0,3 A, (the current through the inductor)

Write an expression for the charge on the capacitor and the current through the inductor as a function of time.

The Attempt at a Solution



I've tried this several times from stratch now, and while I do get an answer, according to the book I'm wrong. The standard form of a solution for Q, is Q = Qm*cos(wt + phi), where phi is the phase displacement. My strategy is to write both the expressions for Q and I, both containing the unknown parameters phi and Qm, and solving the system of two equations with two unknowns.

However, I always get phi = -1,20, whereas the answer in the book is phi = -0,922. I've attached my latest handwritten attempt at a solution, thanks for any help!
 

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  • #2
Update:

I decided to check the answer in the book. The answer the book gives for the current function is:

I = 0,642*cos(Wt + pi/2 - 0,922), where w = 1,14*10^4

However, when setting t=0, this function gives an output I of 0,511. That is not equal to the problem's initial condition of I = 0,3!

The answer I got after substituting the variables back is:

I = 0,322*cos(wt + pi/2 - 1,2), where w = 11364

which coincidentally checks out as I = 0,3 when t=0.

I therefore conclude that the book is wrong, but I still would very much like a second opinion on this one. This is my first LC-circuit problem, so it's really hard to be too confident in my calculations.

Would greatly appreciate any help! :)
 
  • #3
Your differential equation is wrong. The solution of LQ"-Q/C=0 is not periodic at all. There should be "+" instead of "-". ehild
 
  • #4
Try solving it directly using differential equation rather than using the derived formula. It is always easier and better to understand and solve the equations rather than memorizing formulae.

This will help. http://en.wikipedia.org/wiki/LC_circuit
 
  • #5
You are correct, but it's just a typo, I solved it as if it were a plus. Anyway, I do know how to solve this differential equation. What I thought gave me trouble was solving for the initial conditions, but seeing how the answer in the book didn't check out when I tested it, whereas mine did, I actually do believe I got the right answer from the beginning.

Thanks for replying!
 

1. How do I solve an LC-circuit problem?

Solving an LC-circuit problem involves using the equations for charge (Q) and current (I) as functions of time. The first step is to identify the values for inductance (L) and capacitance (C) in the circuit. Then, you can use the differential equations Q=Qmaxcos(ωt) and I=Imaxsin(ωt) to find the values of Q and I at any given time.

2. What is the significance of Q and I in an LC-circuit?

Q represents the charge on the capacitor and I represents the current in the inductor. These values are constantly changing as the circuit oscillates between storing energy in the electric field of the capacitor and storing energy in the magnetic field of the inductor.

3. How do I calculate the maximum charge and current in an LC-circuit?

The maximum charge on the capacitor (Qmax) can be found using the equation Qmax=CVmax, where C is the capacitance of the circuit and Vmax is the maximum voltage across the capacitor. The maximum current in the inductor (Imax) can be found using the equation Imax=Vmax/ωL, where Vmax is the maximum voltage across the inductor and ω is the angular frequency of the circuit.

4. How do I determine the frequency of an LC-circuit?

The frequency of an LC-circuit is determined by the values of inductance and capacitance in the circuit. The formula for calculating frequency is ω=1/√(LC). This means that the smaller the values of L and C, the higher the frequency of the circuit will be.

5. What are some common applications of LC-circuits?

LC-circuits are commonly used in electronic devices such as radios, televisions, and computers. They are also used in power distribution systems to regulate and stabilize the flow of electricity. Additionally, LC-circuits are used in wireless communication systems, such as cell phones and Wi-Fi routers.

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