Archived Single phase question regarding Inductance

[ric115]

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 can't 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

 

[gneill]

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 $pf = 0.033$