# Ideal Transformer Homework Solution

• p75213
In summary, the conversation discusses the use of transformers in mesh analysis and the constraints they impose on the total current flowing through the circuit. The suggestion is made to assume a current in the primary winding and use KVL equations to solve for the mesh current in the secondary loop. This approach leads to a solution where I3 is 3A, resulting in a voltage of 24V.
p75213

See attached

## Homework Equations

V2/V1=I1/I2=n n=2

## The Attempt at a Solution

$$\begin{array}{l} {\rm{Mesh 1: }} & 4{I_1} + {V_1} = 60 \\ {\rm{Mesh 2:}} & 10{I_2} - {V_2} - 2{I_3} = 0 \\ {\rm{Mesh 3:}} & - 2{I_2} + 10{I_3} + {V_2} - {V_1} = 0 \\ {I_1} = - 2{I_2} & & {V_1} = \frac{{{V_2}}}{{ - 2}} \\ {\rm{Mesh 1: }} & - 8{I_2} - 0.5{V_2} + 0{I_3} = 60 \\ {\rm{Mesh 2:}} & 10{I_2} - {V_2} - 2{I_3} = 0 \\ {\rm{Mesh 3:}} & - 2{I_2} + 1.5{V_2} + 10{I_3} = 0 \\ \end{array}$$

When I put these equations into a maths program I get I3=7.5A which means that Vo=60V not 24V.

#### Attachments

• ScreenHunter_01 Mar. 21 22.27.gif
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Transformers can cause some problems for standard mesh analysis; transformer coupling relates their primary and secondary loops more intimately than a simple component like a resistor shared by the loop would. The constraint implied by the transformer equation I1 = -2I2 is not just for the mesh currents, but the actual (total) currents that must flow through the transformer windings. That means that the actual current flowing through the 2Ω resistor also has this constraint. Your mesh current I3 cannot be allowed to affect this total current flowing through this resistor!

If I may offer a suggestion: Assume that the current in the primary winding is i1 and the current flowing out of the dot on the secondary is i1/2. That means there is a mesh current of i1/2 flowing upwards through the 8Ω resistor in the second loop. Then write a KVL equation for the path around the outside of the circuit (let the current through the upper 8Ω resistor be i2). You should be able to solve for i2 rather handily...

#### Attachments

• Fig1.gif
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Hi GNeill,
From your hints I went back to the equation V1I1=V2I2:

$$\begin{array}{l} {V_1}\left( {{I_1} - {I_3}} \right) = {V_2}\left( {{I_2} - {I_3}} \right) \\ \frac{{{V_2}}}{{{V_1}}} = \frac{{\left( {{I_1} - {I_3}} \right)}}{{\left( {{I_2} - {I_3}} \right)}} = - n \\ {I_1} = - n\left( {{I_2} - {I_3}} \right) + {I_3} \\ {I_1} = - 2\left( {{I_2} - {I_3}} \right) + {I_3} \\ {I_1} = 3{I_3} - 2{I_2} \\ {\rm{Mesh Equations:}} \\ - 8{I_2} - 0.5{V_2} + 12{I_3} = 60 \\ 10{I_2} - {V_2} - 2{I_3} = 0 \\ - 2{I_2} + 1.5{V_2} + 10{I_3} = 0 \\ \end{array}$$I3 is now 3A which makes the voltage 24V.

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## 1. What is an ideal transformer?

An ideal transformer is a theoretical device that can transfer electrical energy from one circuit to another without any loss or distortion. It consists of two coils, a primary and a secondary, that are magnetically coupled but electrically isolated.

## 2. How does an ideal transformer work?

An ideal transformer works based on the principle of electromagnetic induction. When an alternating current flows through the primary coil, it creates a changing magnetic field, which induces a voltage in the secondary coil. The voltage in the secondary coil is directly proportional to the number of turns in the coil and the rate of change of the magnetic field.

## 3. What is the purpose of an ideal transformer?

An ideal transformer is used to change the voltage and current levels in an electrical circuit. It can step up or step down the input voltage depending on the ratio of the number of turns in the primary and secondary coils. This is essential in power transmission and distribution since it allows for efficient transfer of energy over long distances.

## 4. What are the advantages of using an ideal transformer?

An ideal transformer has several advantages, including high efficiency, low cost, and compact size. It does not have any moving parts, which makes it reliable and requires minimal maintenance. It also provides electrical isolation between the primary and secondary circuits, making it safer to use.

## 5. Can an ideal transformer exist in real life?

No, an ideal transformer is a theoretical concept and cannot exist in real life. In practice, there will always be some energy loss due to factors such as resistive losses in the coils and eddy current losses in the core. However, modern transformers are designed to minimize these losses and come close to the ideal transformer's performance.

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