Single phase question regarding Inductance

In summary, the coupling coefficient between two identical coils is 0.6 and each has a resistance of 8 Ω and a self-inductance of 2 mH. When connected in series, the total inductance and impedance are calculated differently depending on whether it is a cumulative or differential connection. The mutual inductance can be calculated using the formula M = k√L1L2. The current and power factor for a 10 V, 5 kHz supply are also calculated for both connection scenarios.
  • #1
ric115
5
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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
 
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  • #2
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##
 

FAQ: Single phase question regarding Inductance

What is inductance?

Inductance is a property of an electrical circuit that describes the ability of the circuit to store energy in the form of a magnetic field.

How is inductance measured?

Inductance is measured in units called Henrys (H), named after the physicist Joseph Henry. It can be measured using a device called an inductance meter or by using mathematical formulas.

What factors affect inductance?

The inductance of a circuit is affected by the number of turns in the coil, the material and shape of the coil, and the presence of any magnetic materials or conductive materials near the coil.

What is the difference between inductance in AC and DC circuits?

In AC circuits, the inductance can cause a phase shift between the voltage and current, resulting in a lag or lead in the current compared to the voltage. This effect is not present in DC circuits.

How is inductance used in practical applications?

Inductance is used in many practical applications, such as in transformers, motors, and generators. It is also used in electronic components like inductors and chokes to filter out unwanted frequencies in a circuit.

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