Find Inductance and Capacitance

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To construct an AM radio's RC circuit resonating at 580 Hz with a maximum current of 0.2 A and a capacitor's maximum potential energy of 2 x 10^-5, the relationship between inductance (L) and capacitance (C) must be established using the resonance frequency formula f = 1/(2π√(LC)). The maximum energy stored in the capacitor can be expressed as U = 1/2 * C * Vmax^2, while the maximum energy in the inductor is U = 1/2 * L * I^2. The Q-factor is crucial for understanding energy dissipation, and since there are no resistors in a pure LC circuit, the energy is conserved. By rearranging these equations, L can be solved first, followed by substituting it back into the resonance equation to find C. This method should yield the correct values for L and C in the circuit design.
MulatuOyarta
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Hello,

I need to construct an AM radio's RC circuit that:

1. resonates to frequency f= 580 Hz
2. has a maximum current of I = 0.2 A
3. the capacitor has a maximum potential energy U = 2 x 10^-5

I need to find: The inductance and capacitance (L & C) of this RC circuit.

Please help?

I keep trying different things and end up with really odd answers.
 
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Impedance from inductor is equal to impedance from capacitor at resonance. That should give you value of LC product. Have you gotten that far?
 
Well, I already did the f= w/2(pie) = 1/2(pie)((LC)^(1/2)).

What puzzles me is how you find L & C after you do this. You see, I thought that somehow if I use i = -wQsin(wt) & q = Qcos(wt), I'd be able to do it. Because I'm told that the maximum current must be a certain value "I", I can say:

I = -wQsin(90) ----> I = -wQ

now the problem is that I don't know Q. How do I find Q? I could plug in w = 2(pie)f for w. So I really need Q in terms of L & C.

Once I have a system of equations, I can solve it. But... What do I do to get there?

EDIT: Thankyou Yungman, I will have a look at the link.
 
I have a new question. After reading the article, I am wondering whether the Q-factor they talk about here is the same as the Q in

q = Qcos(wt)
i = -wQsin(wt)

because I thought that "Q" represented the maximum charge of the capacitor. Additionally, if this Q-factor is the same as the Q in those equations, I run into the problem that the amount of energy dissipated per cycle is 0. LC circuits don't have resistors, meaning they don't consume energy.

Q = U / 0
 
first use the current equation in the circuit.
if resistance put is R and (ifthe circuit is in series) and if the applied voltage is V(rms) then
V=I^2*R (include the phase also, because at resonance XL cancels XC ).
also from the energy equation we have energy stored in capacitor and inductor and resistor (if distributed equally)
then U=(.5*LI^2)+(.5*CV^2)+(R*I^2)

and from the frequency equation u get another relation of L and C.

then just solve it.
you may get better result.

an attempt .
 
The problem with that is I don't know R or V. In fact, a pure LC circuit doesn't have an R... because it has no resistors.
 
is that a series or parallel LC circuit?
 
It's a series circuit.
 
  • #10
Okay guys, I think I might understand it. Because I'm given the maximum potential energy of the capacitor, I can use what shankar gave me:

U of inductor = 1/2 [ L I(t)2 ]
U of capacitor = 1/2 [ Q(t)2 / C]
U total = 1/2 [ L I(t)2 ] + 1/2 [ Q(t)2 / C ] = the sum of the individual potential energies of the components

There is no resistor, so I don't have to worry about resistance. Now, because I know that when the charge on the capacitor = 0, the current in the circuit is at its maximum, and I happen to know that the maximum current is qual to "I", then I can say:

U given to me = 1/2 [ L I(t)2 ]

The "q" term gets zeroed out. So, now I can just rearrange and solve for L. Then, I can plug L into the resonance equation "f" and get C.

Does this sound liek the right way to solve it?
 
  • #11
MulatuOyarta said:
I need to construct an AM radio's RC circuit that:

1. resonates to frequency f= 580 Hz
2. has a maximum current of I = 0.2 A
3. the capacitor has a maximum potential energy U = 2 x 10^-5.
a). Do you mean 580 Hz or 580 kiloHertz?

b). For a capacitor, the charge Q = CV.
the maximum current is dQ/dt = I = C dV/dt = ω0C Vmax
The maximum stored (potential?) energy is ½CVmax2

Bob S
 
  • #12
hope you are on the right way.
and things are looking fine now.try that one out. and if you don't get absurd result and on application if your set doesn't flop then your guess should be right.
 
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