Mutual Inductance of Two RC Circuits?

  1. 1. The problem statement, all variables and given/known data

    In the circuit shown in the following figure(Figure 1) the capacitor has capacitance 20 microF and is initially uncharged. The resistor R_0 has resistance 12 Ohms . An emf of 90.0V is added in series with the capacitor and the resistor. The emf is placed between the capacitor and the switch, with the positive terminal of the emf adjacent to the capacitor. The small circuit is not connected in any way to the large one. The wire of the small circuit has a resistance of 1.1 Ohm/m and contains 29 loops. The large circuit is a rectangle 2.0 m by 4.0 m, while the small one has dimensions a = 14.0cm and b = 25.0cm . The distance c is 3.0cm . (The figure is not drawn to scale.) Both circuits are held stationary. Assume that only the wire nearest the small circuit produces an appreciable magnetic field through it.

    I found a picture of the figure here:

    1) The switch is closed at t = 0. When the current in the large circuit is 5.00A , what is the magnitude of the induced current in the small circuit?

    2) What is the direction of the induced current in the small circuit?

    2. Relevant equations

    Equations with RC Circuits:

    [tex]\tau = RC [/tex]

    Other Equations:

    Magnetic Flux: [itex]\Phi[/itex] = Int(B*dA)

    Faraday's Law: Emf_induced = -N(d([itex]\Phi[/itex])/dt)

    3. The attempt at a solution

    So, the first thing I did was try to find the time at which the current in the large circuit would be 5.00 A. I got the t = 9.73*10^-5 s.

    What I'm stuck at now is how to figure out the change in flux with respect to time, or how to break it up so it is easier to solve. I think you might be able to use Ampere's law to draw an Amperian loop (so that you can figure out B), but I'm not sure what shape to use and where to draw the loop.

    For the direction of the induced current in the small circuit, I know the induced current is supposed to result in a magnetic field that opposes the change in the magnetic flux that induces the current (so, the magnetic flux through the big circuit, right?) I'm thinking that the current is flowing cw in the big circuit, and the current decreases over time, so the magnetic flux would be decreasing into the big circuit, right? If that's the case, then the induced current in the small circuit would oppose that decreasing flux, so the induced current should also be clockwise, right? I guess my biggest uncertainty is that I don't know for sure (it says the positive terminal of the emf source is adjacent to the capacitor, so flowing from + to - should mean cw current, right?) which direction the current is flowing in the big circuit. Thank you!
  2. jcsd
  3. gneill

    Staff: Mentor

    If the circuit is as depicted in your figure, then placing the EMF between the capacitor and switch with its + terminal connected to the capacitor will produce a current that is counter clockwise, not clockwise.


    If you know the direction of the field that the wire produces through secondary area and you know that that field is decreasing in strength over time, then you should be able to see which direction a current in the secondary has to flow in order to reinforce that field.

    Since the problem states that only the rightmost wire of the large circuit induces current in the secondary circuit you only have to deal with the magnetic field produced by that wire in terms of the flux in the secondary.

    Apparently you've written an equation for the current in the primary circuit (since you've solved for the time at which the current is 5.0 A). Now, can you write an equation for the magnetic field produced by the "influencing" wire of the primary with respect to distance r from that wire? if so, you should be able to integrate the field over the area of the secondary in order to determine the flux. Start by simply assuming some current 'I' in the primary since it's an instantaneous value that matters (a snapshot in time).

    Afterwards you can plug in the time-dependent current equation. Differentiate the result with respect to time to find the change in flux with respect to time. Be sure to take into account the number of turns in the secondary.

    Attached Files:

  4. hi gneill,

    I have tried your process, and i am not quite understanding it completely.
    So for the magnetic field of a wire, i got:

    B = (μ0*i)/(2∏d) where d is the distance from the wire to the center of the small loop (3m+.07m).
    I then got flux by:
    ∫B*dA, and since B is constant i pulled it out and it just becomes B*A.

    Next I plugged I=(Emf/R)e^(-t/RC) for I in the above equation and then differentiated with respect to t to get:

    dΦ/dt = (μ0Aεe-t/RC)/(2∏dR2C) * N = εinduced

    N = number of turns of small loop

    do the R's correspond to the resistance of the small loop?
    do i just use the capacitance for the bigger loop for C?
  5. gneill

    Staff: Mentor

    Having reread the problem statement I see that I made an error in my diagram concerning the distance separating the two coils. It should be 3cm and not 3m. Big difference!

    Since the coil is near the current-carrying wire and is dimensionally small with respect to the length of the wire, presumably the wire can be approximated as being of infinite length. So your formula B = (μ0*i)/(2∏d) should hold for points inside the small loop. However, since the B falls off as 1/d, I don't think you should assume that B is constant over the whole area of the small loop; You should integrate the field over the area.

    The resistance of the small loop should be calculated; 29 turns and you have the dimensions of the rectangular turn and the wire resistance per unit length. The resistance determining the time constant and current in the primary loop is different, being set by the 12Ω resistor. Yes, the only capacitance you need to deal with is the one in the large loop.
  6. rude man

    rude man 6,085
    Homework Helper
    Gold Member

    just posting to keep track of this thread ... looks like gneill has things well in hand ...
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