The operation of magnetically coupled circuits

In summary: You can't short the secondary without toxic smoke! This is a power xfmr; look at the primary and secondary winding impedances. The components in the model referring to the secondary winding may be totally removed without affecting the four things asked for in the problem if there is no load.
  • #1
139
0

Homework Statement



A transformer has 600 primary turns and 150 secondary turns. The primary and the second resistances are 0.25Ω and 0.01 Ω respectively and the corresponding reactances are 1.0Ω and 0.04Ω respectively.
Detemine the equivalent:
1. resistance
2. the reactance and
3. the equivalent impedance reffered to the primary winding
4. the phase angle of the impedance.

Anyone could help me with this question? Thank you

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
  • #2
I have been trying to answer some of them myself.
For equivalent resistance:
2
Re= R1 + R2= R1 + R2 ( V1 : V2)

Re= 0.41Ω

For equivalent reactance:
2
Xe= X1 + X2' = X1 + X2 ( V1: V2)

Xe= 1.64 Ω

Am i correct?
 
  • #3
Is there a load on the secondary? If not, just ignore the secondary winding altogether and solve the circuit with the primary winding only.
 
  • #4
Thanks. I will solve it OK now
 
  • #5
The problem may be referring to the transformer equivalent model referred to the primary. If so, the results reported by Agata78 in post #2 look correct (although the math as presented is not correct as the turns ratio should be squared in both cases).
 
  • #6
gneill said:
The problem may be referring to the transformer equivalent model referred to the primary. If so, the results reported by Agata78 in post #2 look correct (although the math as presented is not correct as the turns ratio should be squared in both cases).

If there is no load there is no secondary circuit. There is only a single inductor. The model, turns ratio, secondary leakage inductance and resistance do not matter.
 
  • #7
rude man said:
If there is no load there is no secondary circuit. There is only a single inductor. The model, turns ratio, secondary leakage inductance and resistance do not matter.

The problem may be referring to the model alone and not a complete circuit. Or, if you like, consider the secondary to be shorted. The OP would have to put the problem in context to clarify (is the problem associated with a chapter covering the transformer model, for example). Otherwise it's hard to see the point of providing the all the information that's given.
 
  • #8
gneill said:
The problem may be referring to the model alone and not a complete circuit. Or, if you like, consider the secondary to be shorted. The OP would have to put the problem in context to clarify (is the problem associated with a chapter covering the transformer model, for example). Otherwise it's hard to see the point of providing the all the information that's given.

You can't short the secondary without toxic smoke! This is a power xfmr; look at the primary and secondary winding impedances. The components in the model referring to the secondary winding may be totally removed without affecting the four things asked for in the problem if there is no load.

My suspicion is that either the OP left out load characteristics, or it's a trick question. Probably the former.
 

1. What is the principle behind magnetically coupled circuits?

The principle behind magnetically coupled circuits is based on the concept of mutual inductance, where a changing current in one circuit induces a voltage in a nearby circuit through the use of a shared magnetic field. This allows for the transfer of energy and signals between the two circuits without direct electrical connection.

2. How does the coupling coefficient affect the performance of a magnetically coupled circuit?

The coupling coefficient, represented by the symbol k, describes the strength of the magnetic coupling between two circuits. A higher coupling coefficient means a stronger magnetic field, resulting in better energy transfer and a more efficient performance of the circuit. However, too high of a coupling coefficient can also lead to unwanted interference and crosstalk between the circuits.

3. What are the advantages of using magnetically coupled circuits?

There are several advantages to using magnetically coupled circuits. One major advantage is the ability to transfer energy and signals without the need for direct electrical connection, which reduces the risk of electrical interference and increases safety. They also have a simple design and can be easily integrated into electronic systems.

4. How do magnetically coupled circuits impact the overall efficiency of a system?

Magnetically coupled circuits can have a significant impact on the efficiency of a system. By allowing for energy transfer without the need for direct electrical connection, they can reduce power losses and improve overall efficiency. They also provide a means for isolation between circuits, which can prevent interference and improve the performance of the system.

5. What are some common applications of magnetically coupled circuits?

Magnetically coupled circuits have a wide range of applications in various industries. They are commonly used in power supplies, transformers, and communication systems. They are also used in medical devices, such as MRI machines, where they help create a strong magnetic field for imaging purposes. Other applications include wireless charging, energy harvesting, and data transmission in electronic systems.

Suggested for: The operation of magnetically coupled circuits

Replies
4
Views
1K
Replies
6
Views
1K
Replies
7
Views
1K
Replies
12
Views
2K
Replies
13
Views
2K
Replies
3
Views
944
Replies
5
Views
1K
Back
Top