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Effects of various voltage drops in passive circuits

  1. Jul 15, 2008 #1
    Voltage drop due to a resistance: Determined by resistance*current
    Voltage drop due to an inductance: Determined by inductance*change in current/change in time
    Voltage drop due to a capacitance: Determined by charge stored / capacitance

    Assuming applied voltage is constant:

    Question 1: Must the voltage drop due to an inductance always fall by one volt every time the voltage drop due to a resistance goes up by one volt?
    My thoughts: I don't think so, not if there's a capacitance.

    Question 2: Must the development of the magnetic flux of circuit correspond to the increase in the voltage drop due to a resistance, or is it more properly related to the time-integrated voltage drop due to inductance?
    My thoughts: Basic teaching implies the former (i.e. when they say "the current produces the magnetic field"), but perhaps more advanced teachers refer to the latter (e.g. "the increase in current produces the magnetic flux at a rate proportional to the inductance"). When capacitance is ignored it would be safe to assume that decrease in the voltage drop due to inductance implies a 1-to-1 increase in the voltage drop due to resistance.

    Question 3: If one takes the derivative of each of the three types of voltage drops (i.e. due to resistance, inductance, and capacitance) with respect the derivative of length, which of those three quantities is useful, and if so, in what way?
    My thoughts: Maybe each quantity has something to do with how effects of the applied electric field are allocated at each point in the circuit?
     
    Last edited: Jul 15, 2008
  2. jcsd
  3. Jul 16, 2008 #2
    Have you had e-m field theory yet? If not, these issues will get covered when you do take e-m fields.

    Regarding the question, "does the current produce the magnetic field?", all I can say is what Maxwell's equations reveal. Under *time-changing* conditions, the electric and magnetic fields cannot exist independently. Likewise with the circuit quantities current and voltage. Under static conditions, a steady dc current has a static magnetic field associated. If, however, the current is "ac", or time-varying, then there must be a non-zero voltage and non-zero electric field as well as the time-varying magnetic field. As far as the current producing the magnetic field goes, it is very hard to say what causes current, H fields, E fields, voltage, charge, flux, etc. There is no real "pecking order". Maxwell showed us that when the energy is static, we can have a static current and H/B field with zero voltage and E/D field. Or a static voltage and E/D field exists with zero current and H/B field. This is only for static energy conditions.

    Under dynamic energy conditions, all of the above mentioned quantities are non-zero. That's the crux of Maxwell's equations.

    Does this help? BR.
     
  4. Jul 16, 2008 #3
    It's very informative as it helps explain the non-linearity of the situation.

    If I think for a moment...

    The change of energy in the magnetic field depends on the change of current, right?

    The derivative of current squared with respect to time would be 2*current*(change of current/change in time) right?

    Now, the voltage drop due to inductance is equal to inductance * (change of current/change in time), so the derivative of current squared is: 2*current*voltage drop due to inductance/inductance

    Since the energy in a magnetic field is equal to (1/2)*inductance*current^2, the resulting rate change of the energy in the magnetic field seems to be equal to current*voltage drop due to inductance.

    However, the potential energy with respect to an external magnetic field is proportional to current*flux area*b-field (or IAB for short).

    What is the proper way to distinguish between the two apparently different energies (one proportional to current squared, and the other, seemingly proportional to current to the first power?).
     
  5. Jul 16, 2008 #4
    Actually, IAB, which equals I*phi, since phi = A*B, is not the first power of current, but the square. The definition of inductance is N*phi = L*I = N*A*B. Integrating N*phi dI gives 0.5*L*(I^2). So "IAB" is not a 1st power relationship since "AB" which equals "phi" contains another factor of I, making the relationship 2nd power.
     
  6. Jul 16, 2008 #5
    I thought A was the area of the turns and B an external field. Does current somehow affect B? I find it hard to believe that it would affect A. I thought [itex]U=-\vec{\mu}\cdot\mathbf{B}[/itex] was the energy potential of the magnetic field while [itex]\vec{\tau} = \vec{\mu} \times\mathbf{B}[/itex] was the torque (energy and torque seem to be separated by 90 degrees). But if you still meant B as the external field, I am at a loss to explain how phi would be proportional to current.

    So if I increase I, A*B should increase proportionally then (assuming that L/N is constant). When then of the formulas above?
     
  7. Jul 17, 2008 #6
    Increasing I increases B as well, but not A. A is the area which is fixed.

    I'm having trouble following your reasoning. Could you state a specific problem with conditions? Then we can communicate as I don't know what you're asking right now. BR.
     
  8. Jul 17, 2008 #7
    http://physics.bu.edu/~duffy/PY106/MagForce.html

    B is supposed to be the external magnetic field (for example see:http://ocw.mit.edu/NR/rdonlyres/Phy...3-296E-405C-B8AD-CC42AB16AF46/0/solving05.pdf)

    I don't know how the external magnetic field is affected by the current, because if the external magnetic field is REALLY strong I don't see how a current such as 1 milliamp is going to increase it that much, you know?

    If the current doubles, the torque should at least double, but you're saying the external magnetic field would have to double too (since B is the external magnetic field). How does that work?

    And if there was no external magnetic field, how would you define the torque on the magnetic field of the loop anyway?

    By the way, the opening post was asking about whether the magnetic field corresponded to the resistive voltage drop or the time-integrated inductive voltage drop, and I think the answers suggested the latter (1/2)LI^2. Now I trying to figure how exactly does the external magnetic field, which is really [itex]\mathbf{B}_{ext}[/itex], is proportional to the current it couples to.
     
    Last edited: Jul 17, 2008
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