# Very confusing question about transformer

• Clara Chung
In summary, the conversation discusses the power output of a load and the relationship between voltage and current in a circuit. It is determined that the load consumes the rated power when operating at the rated voltage, and that the secondary voltage is equal to the rated voltage due to the turns ratio. It is also mentioned that the approach used in (2) is incorrect, and that the secondary voltage may not be equal to 200V. The conversation concludes with a question about the relationship between voltage and current in the circuit.
Clara Chung
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## The Attempt at a Solution

1.Why does the answer take the rated power as the power output?
2.I tried to find the current like this:
R=200^2 / 100 = 400 ohm
I at Q = 400/[2(400)] = 0.5 A
I at P = 0.5/2 = 0.25 A
3. As the approach as shown as (2) is wrong. Either V is not equal to IR in the circuit Q or the secondary voltage is not 400/2 = 200V.
4. As question 36 used a similar method as (2), V should be equal IR. So the only possibility is Vq not equal to Vp [Nq/Np].
5. However as the change of magnetic flux is the same everywhere in the iron core. By V=-NdΦ/dt.
Vp=- Np dΦ/dt...(a)
Vq=- Nq dΦ/dt...(b)
Vr=- Nr dΦ/dt...(c)
by (b)/(a)
Vq = Vp [Nq/Np]
6. So please tell me what's wrong. Thank you VERY much.

Clara Chung said:
1.Why does the answer take the rated power as the power output?
The power rating of the load is nothing but the power consumed when operating at rated voltage. The rated voltage is 200V and secondary voltage happens to be 200V (because of the turns ratio). Hence, the load consumes the rated power.

Clara Chung
cnh1995 said:
The power rating of the load is nothing but the power consumed when operating at rated voltage. The rated voltage is 200V and secondary voltage happens to be 200V (because of the turns ratio). Hence, the load consumes the rated power.
Oh I see. As you have introduce magnetomotive force before, although Vq = Vp Nq/Np ,however Np Ip = Nq Iq + Nr Ir ,therefore 2Ip = Iq + Ir, so Ip = Iq Am I right?

Clara Chung said:
so Ip = Iq Am I right?
Yes.
Clara Chung said:
As the approach as shown as (2) is wrong. Either V is not equal to IR in the circuit Q or the secondary voltage is not 400/2 = 200V.
Could you elaborate? Why do you think the secondary voltage is not 200V?

## 1. What is a transformer and what does it do?

A transformer is an electrical device that is used to transfer electrical energy from one circuit to another through electromagnetic induction. It increases or decreases the voltage of an alternating current (AC) power supply, depending on the design and purpose of the transformer.

## 2. How does a transformer work?

A transformer works by using two coils of wire, known as the primary and secondary windings, that are wrapped around a core of ferromagnetic material. When an AC current flows through the primary winding, it creates a changing magnetic field which induces a current in the secondary winding, resulting in the transfer of electrical energy.

## 3. What are the different types of transformers?

There are several types of transformers, including step-up and step-down transformers, auto transformers, isolation transformers, and distribution transformers. Each type has a specific purpose and is designed for different voltage levels and applications.

## 4. What are the main components of a transformer?

The main components of a transformer include the core, primary winding, secondary winding, and insulation. The core is typically made of iron or other ferromagnetic materials, while the windings are made of copper or aluminum wire. Insulation is used to prevent electrical leakage and ensure safety.

## 5. What are the common uses of transformers?

Transformers have many uses in everyday life, including power transmission, voltage regulation, and electronic devices. They are also essential in various industries such as manufacturing, transportation, and renewable energy. Additionally, smaller transformers are used in household appliances and electronic devices to convert the voltage for safe use.

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