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Parallel RL circuit

  1. May 28, 2013 #1
    1. The problem statement, all variables and given/known data
    There's a DC supplying a current V over resistance R and inductor L in parallel. The circuit looks like this except it's DC http://sub.allaboutcircuits.com/images/02262.png

    I need to find the total current.

    3. The attempt at a solution

    I know that the current through the first resistor is IR = V/R per ohms law, however I'm stuck on the current through the inductor. I first guessed it could be calculated from the impedance IL = V/z = V/iωL, but since ω is the angle frequency and equals zero for DC, I got stuck.

    I think I need to apply faraday's law to find the current through the inductor, IL:

    V = -ε = d(flux)/dt = L*dIL/dt, and calculate IL from the differential equation.

    However, I cant find a physical reason as to why I would want to do that. Is it perhaps because the voltage drives a changing magnetic flux through the inductor?

    thanks
     
  2. jcsd
  3. May 28, 2013 #2
    basically, I need somebody to explain why this equation works. 10269.png .

    Why does the voltage drop over an inductor divided by its self inductance equal the change in current?
     
  4. May 29, 2013 #3
    For that circuit, with a constant voltage supply, the current through the supply will tend to infinity as t→∞, assuming ideal components. The voltage across the inductor will be constant and since L > 0, the instantaneous rate of change of current through it will be some nonzero number of constant sign, i.e. the magnitude of the current through the inductor will increase without bound.

    It's 'instantaneous rate of change'.

    If you know Faraday's law then you know how the voltage across the inductor is related to the flux linking its turns. Do you know an expression that relates the current through the inductor to the flux linkage produced by it?
    http://en.wikipedia.org/wiki/Inductance#Inductance_of_a_solenoid
     
  5. May 29, 2013 #4
    Magnetic flux = L*current...

    I know that a changing current produces a changing magnetic flux. But how do I link this to a voltage drop over the inductor producing a changing magnetic flux?

    I don't understand how using the equation ε = -d(flux)/dt = -LdI/dt links the voltage drop to the changing current, since the epsilon is representing induced voltage, not the voltage drop?
     
    Last edited: May 29, 2013
  6. May 29, 2013 #5
    Ohhh, but the voltage has to fall over the parallel circuit. So ε = -d(flux)/dt = -LdI/dt is the induced counter-voltage as the current changes, and since it's the only thing causing the voltage to drop, v = ε = -d(flux)/dt = -LdI/dt.

    However, won't the current be in the wrong direction if this was true? I hate that minus-sign, it always confuses me.
     
  7. May 29, 2013 #6
  8. May 29, 2013 #7
    With regards to emf vs. voltage, there are times where it's useful to make the distinction, but in terms of work done in moving charge through an emf or voltage, there's no difference.

    You need to be aware of the sign convention you've used in assigning reference polarities. ε is just a number (a scalar), does a positive value of ε mean that the top terminal of your inductor is at a higher potential than its lower terminal? Does it mean the opposite? On your circuit diagram there are reference directions assigned for current (there are no reference polarities shown for the element voltages), so it's clear what a positive value for IL means.

    You use Faraday's law to derive element laws (equations that relate the voltage across an element to the current through it). For an ideal inductor:

    V = L*di/dt is valid for the passive sign convention.
    V = -L*di/dt is valid for the active sign convention.

    http://en.wikipedia.org/wiki/Passive_sign_convention#Sign_conventions

    I posted recently on this here:
    https://www.physicsforums.com/showpost.php?p=4384249&postcount=28

    In that post, if v was assigned a reference polarity according to the passive sign convention, then v = -ε = L*di/dt.
     
    Last edited: May 29, 2013
  9. May 29, 2013 #8
    The changing flux links N turns and you get an induced emf for each turn, which are in series, so their emf's add.

    Edit:
    I reread your question and realized that I didn't really answer what you asked me.

    Say you have a coil of N loops with the same area that's within a uniform magnetic field. Each loop then has magnetic flux [itex]\Phi[/itex] through it and since there are N loops, the total flux through the coil is [itex]N\Phi[/itex].
     
    Last edited: May 29, 2013
  10. May 29, 2013 #9
    Sorry, I'm really mentally tired right now so excuse me if I making an obvious mistake.

    By convention, NΦ = L*I instead of Φ = L* I.

    Isn't it kind of stupid that L*I gives out the flux multiplied by N, instead of the actual flux (just Φ) going through the coil? Is there some physical reason for this?
     
  11. May 29, 2013 #10
    In this expression NΦ = L*I, Φ is the magnetic flux through the surface bounded by one loop of the coil. If you were to define Φ = L*I, then the element law for the ideal inductor (using the passive sign convention) would be:

    v = N*L*di/dt

    Doesn't it seem silly to have two numbers in there when one would do?
     
  12. May 29, 2013 #11
    ok! thx for the help! + rep
     
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