In a practical transformer -

(a) Some leakage flux is present at both primary and secondary sides. This leakage gives rise to leakage reactances at both sides, which are denoted as X

(b) Both the primary and secondary winding possesses resistance, denoted as R

(c) Permeability of the core can not be infinite, hence some magnetizing current is needed. Mutual flux also causes core loss in iron parts of the transformer.

We need to consider all the above things to derive

###

The no load current I

But, using this equivalent circuit does not simplifies the calculations. To make calculations simpler, it is preferable to transfer current, voltage and impedance either to primary side or to the secondary side. In that case, we would have to work with only one winding which is more convenient.

From the voltage transformation ratio, it is clear that,

E

Now, lets refer the parameters of secondary side to primary.

Z

where, Z

that is, R

equating real and imaginary parts,

R

And V

The following figure shows the

Now, as the values of winding resistance and leakage reactance are so small that, V

Now, let R

Then the

(a) Some leakage flux is present at both primary and secondary sides. This leakage gives rise to leakage reactances at both sides, which are denoted as X

_{1}and X_{2}respectively.(b) Both the primary and secondary winding possesses resistance, denoted as R

_{1}and R_{2}respectively. These resistances causes voltage drop as, I_{1}R_{1}and I_{2}R_{2}and also copper loss I_{1}^{2}R_{1}and I_{2}^{2}R_{2}.(c) Permeability of the core can not be infinite, hence some magnetizing current is needed. Mutual flux also causes core loss in iron parts of the transformer.

We need to consider all the above things to derive

**equivalent circuit of a transformer.**###
**Equivalent circuit of transformer**

Resistances and reactances of transformer, which are described above, can be imagined separately from the windings (as shown in the figure below). Hence, the function of windings, thereafter, will only be the transforming the voltage.

_{0}is divided into, pure inductance X_{0}(taking magnetizing components I_{μ}) and non induction resistance R_{0}(taking working component I_{w}) which are connected into parallel across the primary. The value of E_{1}can be obtained by subtracting I_{1}Z_{1}from V_{1}. The value of R_{0}and X_{0}can be calculated as, R_{0}= E_{1}/ I_{w}and X_{0}= E_{1}/ I_{μ}.But, using this equivalent circuit does not simplifies the calculations. To make calculations simpler, it is preferable to transfer current, voltage and impedance either to primary side or to the secondary side. In that case, we would have to work with only one winding which is more convenient.

From the voltage transformation ratio, it is clear that,

E

_{1}/ E_{2}= N_{1}/ N_{2}= KNow, lets refer the parameters of secondary side to primary.

Z

_{2}can be referred to primary as Z_{2}'where, Z

_{2}' = (N_{1}/N_{2})^{2}Z_{2}= K^{2}Z_{2}. ............where K= N_{1}/N_{2}.that is, R

_{2}'+jX_{2}' = K^{2}(R_{2}+jX_{2})equating real and imaginary parts,

R

_{2}' = K^{2}R_{2}and X_{2}' = K^{2}X_{2}.And V

_{2}' = KV_{2}The following figure shows the

**equivalent circuit of transformer with secondary parameters referred to the primary**.Now, as the values of winding resistance and leakage reactance are so small that, V

_{1}and E_{1}can be assumed to be equal. Therefore, the exciting current drawn by the parallel combination of R_{0}and X_{0}would not affect significantly, if we move it to the input terminals as shown in the figure below.Now, let R

_{1}+ R_{2}' = R'eq and X_{1}+ X_{2}' = X'eqThen the

**equivalent circuit of transformer**becomes as shown in the figure below
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