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Archived Single phase question regarding Inductance

  1. Nov 12, 2012 #1
    Hi There, i have the following question. Can anybody start me off and hopefully i can finish it.

    Question;
    Two similar coils have a coupling coefficient of 0.6. Each coil has a resistance of 8 Ω, and a self-inductance of 2 mH. Calculate the current and the power factor of the circuit when the coils are connected in series :

    (a) cumulatively and

    (b) differentially,

    across a 10 V, 5 kHz supply.


    I have researched and looked at similar problems. I understand what cumultively and differentially means ( direction of the current when the inductors are in series) What i cant understand is how to calculate the mutual inductance

    Is it right to say that the mutual inductance would be

    M = k √ L1 L2

    Also;

    Ltotal = L 1 + L 2 + 2M - Total inductance when the circuit is cumulatively
    and;
    Ltotal = L1 + L2 - 2M - Total inductance when circuit is differentially

    Like i say any help would be appreciated o

    thanks
     
  2. jcsd
  3. Feb 7, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    The OP has correct understanding of the coupling constant and total inductance for the given connection scenarios.

    Since the inductors are identical at ##2~mH##, the mutual inductance is ##M = (0.6)(2~mH) = 1.2~mH##, while the total resistance is twice 8 Ω, or 16 Ω in both cases. The angular frequency equivalent to 5 kHz is ##ω = \pi \times 10^4~rad/sec##.


    Part (a): Cumulative connection
    ##L = 2L + 2M = 8.8~mH~~~;~~~R = 16~Ω##

    The impedance at 5 kHz is then:

    ##Z = R + jωL = 16 + j276.5~Ω##

    The current:

    ##I = \frac{E}{Z} = \frac{10~V}{16 + j276.5~Ω} = 2.09 - j36.1~mA~~##, or in polar form: ##~~36.1~mA~∠~-86.7°##

    The power factor is just the cosine of the current phase angle, so in this case it's ##pf = 0.058##

    Part (b): Differential connection
    ##L = 2L - 2M = 15.2~mH~~~;~~~R = 16~Ω##

    The impedance at 5 kHz is then:

    ##Z = R + jωL = 16 + j477.5~Ω##

    The current:

    ##I = \frac{E}{Z} = \frac{10~V}{16 + j477.5~Ω} = 0.701 - j20.92~mA~~##, or in polar form: ##~~20.9~mA~∠~-88.1°##

    And the power factor is ##pf = 0.033##
     
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