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Homework Help: Mutual Inductance of RC Circuit on a Small Coil

  1. Apr 11, 2014 #1
    1. The problem statement, all variables and given/known data
    My textbook has two problems, based on the same premises, but the solutions manual solves each of them differently, so I was wondering why? Here are the two problems...
    1. In the circuit shown in Fig. P29.49, the capacitor has capacitance C = 20μF and is initially charged to 100V with the polarity shown. The resistor R0 has resistance 10Ω. At time t = 0 the switch is closed. The small circuit is not connected in any way to the large one. The wire of the small circuit has a resistance of 1Ω/m and contains 25 loops. The large circuit is a rectangle 2m by 4m, while the small one has dimensions a = .1m and b = .2m. The distance c is .05m. Both circuits are held stationary. Assume that only the wire nearest the small circuit produces an appreciable magnetic field through it. a)Find the current in the large circuit 200μs after S is closed. b)Find the current in the small circuit 200μs after S is closed.
    2. In the circuit in Fig. P29.49, an emf of 90V is added in series with the capacitor and the resistor, and the capacitor is initially uncharged. The emf is placed between the capacitor and the switch, with the positive terminal of the emf adjacent to the capacitor. Otherwise, the two circuits are the same as in the previous problem. The switch is closed at t = 0. When the current in the large circuit is 5A, what are the magnitude and direction of the induced current in the small circuit?

    2. Relevant equations

    ∫B dl = Magnetic Flux
    i = (V/R)e-t/RC
    di/dt = (V/R2C)e-t/RC
    B (caused by the single wire of the large circuit) = (μ0I)/(2pi(r))
    induced emf = -N d(Flux)/dt
    emf = IR

    3. The attempt at a solution

    Okay, so for the first problem...
    i = (V/R)e-t/RC
    i = (100/10)e-(200E-6)/((10)(20E-6))
    i = 3.7A

    Then part B...
    Flux = ∫B dl
    Flux = ∫(μ0I)/(2pi(r)) dr from c to c+a
    Flux = ((μ0I)/(2pi))ln(1 + (a/c))

    induced emf = -N d(Flux)/dt
    induced emf = -N ((μ0)/(2pi))ln(1 + (a/c)) dI/dt
    induced emf = -((25)(4piE-7)ln(3))/(2pi) ((-3.7)/(200E-6))
    induced emf = .0203

    N(perimeter) = total length
    25(.6) = 15
    R = 1Ω/m
    R = 15Ω

    emf = IR
    .0203 = I(15)
    I =.00135A

    Then problem 2...

    induced emf = -N ((μ0)/(2pi))ln(1 + (a/c)) dI/dt
    i = (V/R)e-t/RC
    di/dt = (V/R2C)e-t/RC
    di/dt = (1/RC)i

    induced emf = -N ((μ0)/(2pi))ln(1 + (a/c)) (1/RC)i
    induced emf = -((25)(4piE-7)ln(3))/(2pi) (1/((10)(20E-6)) (5)
    induced emf = .0275

    emf = IR
    .0275 = I(15)
    I = .00183A

    HOWEVER... If you switch methods and attempt to plug the actually derivative into the emf equation in the first problem, the answer is incorrect because RC ≠ 200E-6. And if you use the formula for i to find the amount of time it took to get the current to 5A in the second problem and plug it into the emf equation you get the wrong answer for the same reason. That time doesn't equal RC. Why is this and how am I supposed to know which equation to use?

    Thank you!

    Attached Files:

  2. jcsd
  3. Apr 12, 2014 #2


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    Homework Helper

    As the RC circuit is the same, RC = 200E-6 in both cases. Also, the current decreases exponentially in both cases: I=(V/R) e-t/(RC), but V is the initial voltage of the capacitor in case 1) V= 100 V, and it is the emf of the battery in case 2), which is 90 V.

    So the currents are

    case 1) I1=10 e-t/(RC)

    case 2) I2=9 e-t/(RC).

    You do not need to calculate the current from time in case 2 as the current is given.

  4. Apr 12, 2014 #3

    rude man

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    Homework Helper
    Gold Member

    Just as a footnote: in actuality there are a couple of approximations made in your presentation. Probably you were meant to make them so don't worry, this is just FYI.

    1: the 4m wire is not infinitely long so Ampere's law is an approximation.

    2: more interestingly, the self-inductance of circuit 2 and coupling from circuit 2 back to circuit 1 are ignored.

    What actually happens is that the current in the secondary circuit generates a counter-emf and also induces current back to the primary circuit. The complete picture involves knowing the self-inductances of both circuits, which is no piece of cake. The mutual inductance is easily calculated since M = N2*ø21/I1 where ø21 is the flux in circuit 2 due to the current I1 in circuit 1 if circuit 2 is open (no current), which in fact you have already calculated, and N2 is the number of turns in the secondary circuit of known dimensions.

    The complete picture looks like this:
    V1 = L1dI1/dt + M dI2/dt,
    V2 = L2 dI2/dt + M dI1/dt

    where L1 and L2 are the self-inductances of circuit 1 & 2 resp. and
    V1 = emf induced in loop1
    V2 = emf induced in loop 2.

    You have in effect computed I1 as if circuit 2 were absent, making I1 an independent variable in the equations, and V2 using the M term only.

    So there are actually two levels of refinement:

    1. Include the self-inductance L2 of circuit 2 but assume I1 independent of I2. This involves knowing L2. There's a website that gives you L for a rectangular circuit.

    2. Correct I1 since I2 is finite. Forget that since L1 would also have to be known.
    Last edited: Apr 12, 2014
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