Finding the Maximum Current in an LC-only Circuit

In summary, the maximum current in an LC-only circuit can be calculated using the equation I_max = V/R, where V is the initial voltage and R is the resistance. This equation takes into account the natural frequency of the circuit, which is determined by the values of the inductor and capacitor. By finding the maximum current, one can determine the peak energy stored in the circuit, which is useful for designing and optimizing LC circuits for specific applications. Additionally, the maximum current can also be found graphically by plotting the current as a function of time and identifying the peak value. Overall, understanding and calculating the maximum current in an LC-only circuit is essential for analyzing and optimizing the performance of such circuits.
  • #1
Underhill
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Homework Statement



Pages 6 and 7 http://apcentral.collegeboard.com/apc/members/exam/exam_information/2008.html

I am struggling with Question (2), Part (b), Sections (ii.) and (iii.)

Homework Equations



RL circuit with battery: I = V/R(1 - e-Rt/L), Imax=V/R

RC circuit with battery: I = V/R(e-t/RC)

The Attempt at a Solution



While studying for the AP Physics C E&M exam, I found this LC problem. I don't recall learning about LC circuits in my high school course, and am stumped on how to proceed. Even the internet seems to provide scarce resources on this topic.

I've solved RL circuits as well as RC circuits, but can't quite figure out how to make the leap to LC circuits.
 
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  • #3
Sorry, this is the 2011 test, E&M
 
  • #4
In order to save a lot of people the trouble of hunting down the question,...

Essentially the circuit for this part of the question looks like this:

attachment.php?attachmentid=45536&stc=1&d=1332774134.gif


and you're asked to find:
i) The initial energy stored in the capacitor
ii) The maximum current
iii) The time rate of change of the current when Qc = 50 mC.

Now, what are the expressions for the energy stored in capacitors and inductors? How much energy do you have to play with? What then are the maximum V's and I's?

The circuit will produce oscillations. What will be the natural frequency? What might you do with that information?
 

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  • #5
Thanks, gneill.

(i.) is easy. The energy stored in a partially charged capacitor is [itex]\stackrel{1}{2}[/itex]QV

In our case, we get [itex]\stackrel{1}{2}[/itex](.105)(9.0) = 0.4725 Joules

(ii.) is harder. The total energy we have to work (or play) with is the amount of energy stored in the capacitor at time t2, which is 0.4725 Joules. The energy stored in the magnetic field of an inductor is -[itex]\stackrel{1}{2}[/itex]LI2. (By the way, I'm not sure how relevant the minus sign is here. I think it has to do with Lenz's Law, but don't know if it makes sense in this context. Can you explain?)

So setting this expression equal to 0.4725 Joules, the total energy in the circuit, and substituting in L = 5.0 Henri's, we have

0.4725 = [itex]\stackrel{1}{2}[/itex](5.0)(I2)

I ≈ 0.4347 Amps

(iii.) I'm not sure what to do here. It looks like I need to take the derivative of something, but I'm not sure where to begin. What do you mean by natural frequency? I didn't study this in high school. Could you quickly brief me on the natural frequency of oscillating LC circuits?

Maximum voltage or electric potential appears to be the same 9.0V as was present before the switch was flipped. Does this make sense?

[EDIT]:

I've just discovered that I can use the voltage of an inductor, Vi = -L[itex]\stackrel{dI}{dt}[/itex] to help calculate the change in current at a given instant. Again, however, I'm confused. How relevant is this minus sign from Lenz's Law?

L[itex]\stackrel{dI}{dt}[/itex] = [itex]\stackrel{Q}{C}[/itex]

(iii.) Q = 0.050 C, therefore [itex]\stackrel{dI}{dt}[/itex] = -0.40 A/s
 
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  • #6
Underhill said:
Thanks, gneill.

(i.) is easy. The energy stored in a partially charged capacitor is [itex]\stackrel{1}{2}[/itex]QV

In our case, we get [itex]\stackrel{1}{2}[/itex](.105)(9.0) = 0.4725 Joules
Um, how did you decide that the voltage on the capacitor is 9.0V when it has a charge of 105 mC? Remember that even though the original circuit had a 9V source, the switch was operated before charging completed, giving the capacitor just 105 mC of charge.

So what's the actual starting voltage on the capacitor for times t2+?
(ii.) is harder. The total energy we have to work (or play) with is the amount of energy stored in the capacitor at time t2, which is 0.4725 Joules. The energy stored in the magnetic field of an inductor is -[itex]\stackrel{1}{2}[/itex]LI2. (By the way, I'm not sure how relevant the minus sign is here. I think it has to do with Lenz's Law, but don't know if it makes sense in this context. Can you explain?)
The negative sign won't matter here as we only need to deal with the magnitude of the energy.

When a circuit oscillates the energy is shuttled back and forth between storage components. In an undamped circuit, at certain instants of time ALL of the energy will be in one component, at other instants it will all be stored in the other component.

Our story begins with all of the energy crammed into the capacitor; No current is flowing in the inductor so there is no magnetic field (inductors store energy in their magnetic fields).
A half cycle of oscillation later and we can expect all the energy stored in the capacitor to have been shuttled over to the inductor. It will then have ALL the energy, stored in its magnetic field tied to the current flowing through it. Current will be maximum at that instant.
So setting this expression equal to 0.4725 Joules, the total energy in the circuit, and substituting in L = 5.0 Henri's, we have

0.4725 = [itex]\stackrel{1}{2}[/itex](5.0)(I2)

I ≈ 0.4347 Amps
Fix your energy calculation to reflect the actual starting voltage on the capacitor and you'll be able to calculate the correct maximum current.
(iii.) I'm not sure what to do here. It looks like I need to take the derivative of something, but I'm not sure where to begin. What do you mean by natural frequency? I didn't study this in high school. Could you quickly brief me on the natural frequency of oscillating LC circuits?
If you have the maximum current and voltage and charge values and you know that the circuit oscillates. You can therefore write equations for the current, voltage, charge, with respect to time. All you need is the natural frequency of the oscillations. Look up "LC circuit" "natural frequency".
Maximum voltage or electric potential appears to be the same 9.0V as was present before the switch was flipped. Does this make sense?
Nope, as I explained above :smile:
 
  • #7
Um, how did you decide that the voltage on the capacitor is 9.0V when it has a charge of 105 mC? Remember that even though the original circuit had a 9V source, the switch was operated before charging completed, giving the capacitor just 105 mC of charge.

So what's the actual starting voltage on the capacitor for times t2+?

I didn't mean voltage on the capacitor - I meant the voltage source in the circuit. If 1/2QV can be used to calculate energy stored in the capacitor at any point in time BEFORE the switch is flipped in a normal RC circuit, then can't I just transfer that value over if I know the switch was flipped when the capacitor holds a certain known charge?

Wait - doesn't the V in U = 1/2QV refer to the battery in the circuit?? Are you saying that V is entirely dependent on the state of the capacitor?
 
  • #8
Underhill said:
Wait - doesn't the V in U = 1/2QV refer to the battery in the circuit?? Are you saying that V is entirely dependent on the state of the capacitor?

The energy stored in the capacitor is entirely due to the state of the capcitor. For any Q on the capacitor it will have a corresponding voltage V and energy E. Remember that capacitors charge over time -- its voltage is not instantly equal to the supply voltage when it begins charging.
 
  • #9
Thanks, gneill.

I see now. I always thought that the V in 1/2QV referred to the battery voltage, not the capacitor voltage. Thanks!

Would you mind explaining when the minus sign due to Lenz's Law is relevant? Do I use it when calculating the time rate of change of the current? -LdI/dt + Q/C = 0, for example?

[EDIT]

One more question. Part iii. asks the time rate of change of the current when the charge on the capacitor plate is 50mC. But how do I know whether charge is building up or leaving the capacitor? I don't know whether the answer should be negative or postive.
 
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  • #10
Underhill said:
Would you mind explaining when the minus sign due to Lenz's Law is relevant? Do I use it when calculating the time rate of change of the current? -LdI/dt + Q/C = 0, for example?
Yes, it's relevant when you use it to determine the direction of the induced emf due to a changing flux in a circuit.

Most of the time for circuit analysis we can rely on the definition of inductance: ##V = L\frac{dI}{dt}## where V is the potential difference that appears across the inductor due to a change in current through the inductor.
 
  • #11
gneill said:
Yes, it's relevant when you use it to determine the direction of the induced emf due to a changing flux in a circuit.

Most of the time for circuit analysis we can rely on the definition of inductance: ##V = L\frac{dI}{dt}## where V is the potential difference that appears across the inductor due to a change in current through the inductor.

So, my answer for part (iii.) is 0.4 A/s, but I'm not sure whether there should be a minus sign in front, because I don't know whether the capacitor is charging or discharging. What should I do?
 
  • #12
Underhill said:
So, my answer for part (iii.) is 0.4 A/s, but I'm not sure whether there should be a minus sign in front, because I don't know whether the capacitor is charging or discharging. What should I do?

Write that the magnitude of the time rate of change when the capacitor has 50 mC of charge is 0.4 A/s.

The first time the capacitor has 50 mC of charge on it the capacitor energy is dropping and the inductor energy is increasing. That tells you that its current must be increasing (provided that you choose your reference direction for current so it's initially positive).
 
  • #13
Thanks! Problem solved! Answers for three parts are below:

(i.) 0.2205 Joules

(ii.) 0.2970 Amps

(iii.) magnitude of 0.4 Amps/second
 
  • #14
Underhill said:
Thanks! Problem solved! Answers for three parts are below:

(i.) 0.2205 Joules

(ii.) 0.2970 Amps

(iii.) magnitude of 0.4 Amps/second

Looks good.
 

FAQ: Finding the Maximum Current in an LC-only Circuit

What is an LC-only circuit?

An LC-only circuit is a type of electrical circuit that contains only inductors and capacitors. It does not contain any resistors or other components that dissipate energy.

How does an LC-only circuit work?

An LC-only circuit relies on the interaction between the inductor and capacitor to store and release energy. When the circuit is first turned on, the capacitor begins to charge and the inductor begins to build up a magnetic field. As the capacitor reaches its maximum charge, the inductor's magnetic field is at its maximum. The energy is then released back into the circuit, causing the capacitor to discharge and the inductor's magnetic field to collapse. This cycle repeats, creating a sinusoidal current.

What is the maximum current in an LC-only circuit?

The maximum current in an LC-only circuit occurs at the resonant frequency, which is determined by the values of the inductor and capacitor. At this frequency, the reactance of the inductor and capacitor cancel each other out, resulting in a purely resistive circuit. This allows the maximum amount of current to flow through the circuit.

How do you find the maximum current in an LC-only circuit?

The maximum current in an LC-only circuit can be found using the formula I = V/R, where V is the voltage across the circuit and R is the equivalent resistance. The equivalent resistance can be calculated using the formula R = √(L/C), where L is the inductance and C is the capacitance. The resonant frequency can also be calculated using the formula f = 1/(2π√(LC)).

What are some real-world applications of LC-only circuits?

LC-only circuits have many practical applications, such as in radio frequency (RF) circuits, electronic filters, and oscillators. They are also used in devices like metal detectors, electronic musical instruments, and electronic ballasts for fluorescent lights. Additionally, LC-only circuits are used in wireless charging technology and in the tuning circuits of radio and television receivers.

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