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Finding the Maximum Current in an LC-only Circuit

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data

    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.)

    2. Relevant 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)

    3. 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. Mar 25, 2012 #2

    Redbelly98

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  4. Mar 26, 2012 #3
    Sorry, this is the 2011 test, E&M
     
  5. Mar 26, 2012 #4

    gneill

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    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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  6. Mar 29, 2012 #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
     
    Last edited: Mar 29, 2012
  7. Mar 29, 2012 #6

    gneill

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    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+?
    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.
    Fix your energy calculation to reflect the actual starting voltage on the capacitor and you'll be able to calculate the correct maximum current.
    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".
    Nope, as I explained above :smile:
     
  8. Mar 29, 2012 #7
    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?
     
  9. Mar 29, 2012 #8

    gneill

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    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.
     
  10. Mar 29, 2012 #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.
     
    Last edited: Mar 29, 2012
  11. Mar 29, 2012 #10

    gneill

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    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.
     
  12. Mar 29, 2012 #11
    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?
     
  13. Mar 29, 2012 #12

    gneill

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    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).
     
  14. Mar 29, 2012 #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
     
  15. Mar 29, 2012 #14

    gneill

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    Looks good.
     
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