Equivalent Circuit of a Transformer | Electrical Engineering

The equivalent circuit of any device can be quite helpful in predetermination of the behaviour of the device under various conditions of operation and it can be drawn if the equations describing its behaviour, are known. If any electrical device is to be analysed and investigated further for suitable modifications, its appropriate equivalent circuit is necessary.

The equivalent circuit for electromagnetic devices consists of a combination of resistances, inductances, capacitances, voltage etc. Such an equivalent circuit (or circuit model) can, therefore, be analysed and studied easily by the direct application of electric circuit theory.

As stated above, equivalent circuit is simply a circuit representation of the equations describing the performance of the device. The behaviour of the transformer under load and are helpful in arriving at the transformer equivalent circuit.

Equivalent circuit of a transformer having transformation ratio K = E₂/E₁ is shown in Fig. 10.17.

The induced emf in primary winding E₁ is primary applied voltage V₁ less primary voltage drop. This voltage causes iron loss current I₀ and magnetising current I_m and we can, therefore, represent these two components of no-load current by the current drawn by a non-inductive resistance R₀ and pure reactance X₀ having the voltage E₁ or (V₁—primary voltage drop) applied across them, as shown in Fig. 10.17.

Secondary current, I₁ = I’₁/K = I₁ – I₀ / K

Terminal voltage V₂ across load is induced emf E₂ in secondary winding less voltage drop in secondary winding.

The equivalent circuit can be simplified by transferring the voltage, current and impedance to the primary side. After transferring the secondary voltage, current and impedance to primary side equivalent circuit is reduced to that shown in Fig. 10.18.

The equivalent circuit diagram can further be simplified by transferring the resistance R₀ and reactance X₀ towards left end, as shown in Fig. 10.19. The error introduced by doing so is very small and can be neglected.

No-load current I₀ is hardly 3 to 5 percent of the full-load rated current, the parallel branch consisting of resistance R₀ and reactance X₀ can be omitted without introducing any appreciable error in the behaviour of the transformer under loaded condition. Such a circuit is shown in Fig. 10.16 (a). The equivalent circuit referred to secondary side (neglecting no-load current I₀) is illustrated in Fig. 10.15 (a).

Equivalent Resistance and Reactance:

The two independent circuits of a transformer can be resolved into an equivalent circuit to make the calculations simple.

Let resistances and reactance of primary and secondary windings be R₁ and R₂ and X₁ and X₂ ohms respectively and let transformation ratio be K.

Resistive -drop in secondary winding = I₂ R₂

Reactive drop in secondary winding = I₂ X₂

Resistive drop in primary winding = I₁ R₁

Reactive drop in primary winding = I₁ X₁

Referred To Secondary Side:

Since transformation ratio is K, so primary resistive and reactive drops as referred to secondary will be K times, i.e., K I₁ R₁ and K I₁ X₁ respectively. If I₁ is substituted equal to KI₂, then we have primary resistive and reactive drops referred to secondary equal to K² I₂ R₁ and K² I₂ X₁ respectively.

Total resistive drop in a transformer = K² I₂ R₁ + I₂ R₂ = I₂ (K² R₁ + R₂) = I₂ R₀₂

Total reactive drop in a transformer = K² I₂ X₁ + I₂ X₂ = I₂ (K² X₁ + X₂) = I₂ X₀₂

The term (K² R₁ + R₂) and (K² X₁ + X₂) represent the equivalent resistance and reactance respectively of the transformer referred to secondary and let these be represented by R₀₂ and X₀₂ respectively. Equivalent circuit referred to secondary has been shown in Fig. 10.15 (a).

From phasor diagram [Fig. 10.15 (b)]

Where V₂ is secondary terminal voltage, I₂ is secondary current lagging behind the terminal voltage V₂ by ɸ.

Since term (I₂ X₀₂ cos ɸ – I₂, R₀, sin ɸ) is very small as compared to the term (V₂ + I₂ R₀₂ cos ɸ + I₂ X₀₂ sin ɸ), so neglecting the former we have-

K V₁ = V₂ + I₂ R₀₂ cos ɸ + I₂ X₀₂ sin ɸ

Or V₂ = KV₁ – I₂ R₀₂ cos ɸ – I₂ X₀₂ sin ɸ … (10.23)

Where V₁ is applied voltage to primary winding.

If load is pure resistive, ɸ = 0 and V₂ = KV₁ – I₂ R₀₂ … (10.24)

If load is capacitive then ɸ should be taken as -ve hence we have:

V₂ = KV₁ – I₂ R₀₂ cos ɸ + I₂ X₀₂ sin ɸ … (10.25)

Referred To Primary Side:

Secondary resistive drop referred to primary:

Equivalent circuit referred to primary is shown in Fig. 10.16 (a)

From phasor diagram shown in Fig. 10.16 (b) we have: