Find Inductance and Capacitance

In summary, the maximum potential energy of the capacitor is 2 x 10^-5 volts, and the maximum current is 0.2 amps. The inductance and capacitance of the RC circuit are unknown, but can be calculated if the impedance and frequency of the circuit are known.
  • #1
MulatuOyarta
6
0
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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  • #2
Impedance from inductor is equal to impedance from capacitor at resonance. That should give you value of LC product. Have you gotten that far?
 
  • #3
  • #4
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.
 
  • #5
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
 
  • #6
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 .
 
  • #7
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.
 
  • #8
is that a series or parallel LC circuit?
 
  • #9
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.
 

What is the difference between inductance and capacitance?

Inductance and capacitance are both properties of electrical circuits. Inductance is the ability of a circuit to store energy in the form of a magnetic field, while capacitance is the ability to store energy in the form of an electric field. Inductance is measured in henries (H), while capacitance is measured in farads (F).

How do you calculate inductance and capacitance?

Inductance can be calculated using the formula L = (N^2 * μ * A)/l, where N is the number of turns in the coil, μ is the permeability of the material, A is the cross-sectional area of the coil, and l is the length of the coil. Capacitance can be calculated using the formula C = ε * A/d, where ε is the permittivity of the material, A is the area of the plates, and d is the distance between the plates.

What is the unit of measurement for inductance and capacitance?

Inductance is measured in henries (H), while capacitance is measured in farads (F).

How does inductance and capacitance affect an electrical circuit?

Inductance and capacitance can affect an electrical circuit in several ways. Inductance can cause a delay in the flow of current, leading to a phase shift in the circuit. Capacitance can store energy, which can affect the overall energy consumption of the circuit. Both inductance and capacitance can also affect the frequency response of a circuit, leading to changes in the behavior of the circuit at different frequencies.

What are some practical applications of inductance and capacitance?

Inductance and capacitance are essential components in many electronic devices and circuits. They are used in filters, oscillators, and amplifiers, among other applications. Inductors and capacitors are also used in power supplies to regulate voltage and filter out unwanted noise. In addition, inductance and capacitance are crucial in the design of wireless communication systems, such as antennas and RF circuits.

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