# Transformer Equivalent Circuit

How to Draw the Transformer Equivalent Circuit? In order to draw the equivalent circuit of the transformer, all real loses that occur in it must be considered. The terms that are used for drawing the equivalent circuit of the transformer are:

- Copper Losses
- Eddy Current Losses
- Hysteresis Losses
- Leakage Flux

## Copper Losses in Transformer

Copper losses are calculated as I^{2}R. They are produced because of the resistive heating losses in primary and secondary windings of the transformers. Copper losses are directly proportional to the square of currents in the windings.

## Eddy Current Losses

These losses are produced because of the resistive heating in the core of transformer. These losses are directly proportional to the square of the voltages that are applied to the transformers.

## Hysteresis Losses

During each half cycle, magnetic domains are rearranged in the core of the transformer which actually cause the hysteresis losses. These are complex and are non-linear function of the applied voltages to the transformers.

## Leakage Flux

Leakage flux is the flux that escape the core and passes only through one of the transformer windings. They are Φ_{LP }and Φ_{LS}. Self induction is caused because of these escaped fluxes in the primary and secondary coils. The effects of this inductances must be accounted for.

### The Exact Equivalent Circuit of a Transformer

By taking into account all the major imperfections it s possible to design an equivalent circuit of a transformer. Every major imperfection is considered in turn and the effect is introduced in the transformer model.

Coper losses are easy to be modeled. Since they are the resistive losses in the primary and secondary coils, so they are modeled by introducing a resistor R_{P }in primary circuit and R_{S} in secondary circuit.

The voltage produced in the primary winding because of leakage flux Φ_{LP }is given as:

e_{LP}(t)=N_{P}(d Φ_{LP}/dt)

and the voltage produced in the secondary windings because of leakage flux is given as:

e_{LS}(t)=N_{S}(d Φ_{LS}/dt)

Because air makes up a significant amount of the leakage flux path and because air has a constant reluctance that is significantly higher than the core reluctance, the flux Φ_{LP }and flux Φ_{LS }are directly proportional to the primary circuit current and secondary current, i_{P} and i_{S} respectively.

$$

\begin{aligned}

&\phi_{\mathrm{LP}}=\left(\mathscr{P} N_{P}\right) i_{P} \\

&\phi_{\mathrm{LS}}=\left(\mathscr{P} N_{S}\right) i_{S}

\end{aligned}

$$

where $\mathscr{P}=$ permeance of flux path

$N_{P}=$ number of turns on primary coil

$N_{S}=$ number of turns on secondary coil

By substitution and simplification, we will get the following equations :

$$

\begin{aligned}

&e_{\mathrm{LP}}(t)=N_{P} \frac{d}{d t}\left(\mathscr{P} N_{P}\right) i_{P}=N_{P}^{2} \mathscr{P P} \frac{d i_{P}}{d t} \\

&e_{\mathrm{LS}}(t)=N_{S} \frac{d}{d t}\left(\mathscr{P} N_{S}\right) i_{S}=N_{S}^{2} \mathscr{P P} \frac{d i_{S}}{d t}

\end{aligned}

$$

The constants in these equations can be lumped together. Then

$$

\begin{aligned}

&e_{\mathrm{LP}}(t)=L_{P} \frac{d i_{P}}{d t} \\

&e_{\mathrm{LS}}(t)=L_{S} \frac{d i_{S}}{d t}

\end{aligned}

$$

where $L_{P}=N_{P}^{2 \mathscr{P}}$ is the self-inductance of the primary coil and $L_{S}=N_{S}^{2 \mathscr{P}}$ is the self-inductance of the secondary coil. Therefore, the leakage flux will be modeled by primary and secondary inductors.

### How to Model the Effects of Core Excitation of Transformer?

It is a little tricky. As we know that the magnetization current i_{m }is directly proportional to the voltage applied to the core and it lags by 90 degree to the applied voltages. By connecting a reactance X_{m }across the primary voltage source this magnetization current can be modeled.

A resistance Rc connected across the primary voltage source can be used to mimic the core-loss current i_{h+e} , which is a current that is proportional to the voltage applied to the core and is in phase with the applied voltage. The inductance X_{m }and the resistance Rc are, at best, approximations of the actual excitation effects (keep in mind that both of these currents are truly nonlinear).

The equivalent circuits that results with these assumptions is shown below:

The components of the excitation branch are positioned inside the primary resistance Rp and the primary inductance Lp, as can be seen.

**Voltage applied to the core=input voltages-internal voltages drops of the windings **

The model displayed above is not used much although it is an accurate model. For drawing an equivalent practical circuit of transformer it is necessary to model it at a single voltage level.

Therefore we can refer the equivalent circuit either on its primary side or on its secondary side as shown below:

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