# Iron Core Transformer Equivalent Circuit   Whatsapp  For the nonideal or practical iron-core transformer, the equivalent circuit appears as in Fig. 1. As indicated, part of this equivalent circuit includes an ideal transformer. The remaining elements of Fig. 1 are those elements that contribute to the non-ideal characteristics of the device. The resistances $Rp$ and $Rs$ are simply the dc or geometric resistance of the primary and secondary windings, respectively. Fig. 1: Equivalent circuit for the practical iron-core transformer.
For the primary and secondary coils of a transformer, there is a small amount of flux that links each coil but does not pass through the core, as shown in Fig. 2 for the primary winding. This leakage flux, representing a definite loss in the system, is represented by an inductance $Lp$ in the primary circuit and an inductance $Ls$ in the secondary. Fig. 2: Identifying the leakage flux of the primary.
The resistance $Rc$ represents the hysteresis and eddy current losses (core losses) within the core due to an ac flux through the core. The inductance $Lm$ (magnetizing inductance) is the inductance associated with the magnetization of the core, that is, the establishing of the flux $\phi_m$ in the core. The capacitances $Cp$ and $Cs$ are the lumped capacitances of the primary and secondary circuits, respectively, and $Cw$ represents the equivalent lumped capacitances between the windings of the transformer.
Since $i'p$ is normally considerably larger than $i_{\phi_{m}}$ (the magnetizing current), we will ignore $i_{\phi_{m}}$ for the moment (set it equal to zero), resulting in the absence of $Rc$ and $Lm$ in the reduced equivalent circuit of Fig. 3. The capacitances $Cp$, $Cw$, and $Cs$ do not appear in the equivalent circuit of Fig. 3 since their reactance at typical operating frequencies will not appreciably affect the transfer characteristics of the transformer. Fig. 3: Reduced equivalent circuit for the nonideal iron-core transformer.
If we now reflect the secondary circuit through the ideal transformer, as shown in Fig. 4(a), we will have the load and generator voltage in the same continuous circuit. The total resistance and inductive reactance of the primary circuit are determined by
$$\bbox[10px,border:1px solid grey]{Re = R_p + a^2 R_s}$$
and
$$\bbox[10px,border:1px solid grey]{Xe = X_p + a^2 X_s}$$  Fig. 4: Reflecting the secondary circuit into the primary side of the iron-core transformer.
which result in the useful equivalent circuit of Fig. 4(b). The load voltage can be obtained directly from the circuit of Fig. 4(b) through the voltage divider rule:
$$a V_L = { (R_i) V_g \over (Re+Ri) + jX_e}$$
and
$$\bbox[10px,border:1px solid grey]{ V_L = { a^2 R_L V_g \over (Re+Ri) + jX_e}} \tag{1}$$
The network of Fig. 4(b) will also allow us to calculate the generator voltage necessary to establish a particular load voltage. The voltages across the elements of Fig. 4(b) have the phasor relationship indicated in Fig. 5(a). Note that the current is the reference phasor for drawing the phasor diagram. That is, the voltages across the resistive elements are in phase with the current phasor, while the voltage across the equivalent inductance leads the current by $90^circ$ . The primary voltage, by Kirchhoff’s voltage law, is then the phasor sum of these voltages, as indicated in Fig. 5(a). Fig. 5: Phasor diagram for the iron-core transformer with
(a) unity power-factor load (resistive) and
For an inductive load, the phasor diagram appears in Fig. 5(b). Note that $a V_L$ leads I by the power-factor angle of the load. The remainder of the diagram is then similar to that for a resistive load. (The phasor diagram for a capacitive load will be left to the reader as an exercise.) The effect of $Re$ and $Xe$ on the magnitude of $Vg$ for a particular $V_L$ is obvious from Eq. (1) or Fig. 5. For increased values of $Re$ or $Xe$, an increase in $Vg$ is required for the same load voltage. For $Re$ and $Xe = 0$, $V_L$ and $Vg$ are simply related by the turns ratio.
Example 1: For a transformer having the equivalent circuit of Fig. 6:
a. Determine $Re$ and $Xe$.
b. Determine the magnitude of the voltages $VL$ and $Vg$.
c. Determine the magnitude of the voltage Vg to establish the same load voltage in part (b) if $Re$ and $Xe = 0 Ω$. Compare with the result of part (b). Fig. 6: Example 1.
Solution:
a. $R_{e}=R_{p}+a^{2} R_{s}=1 \Omega+(2)^{2}(1 \Omega =\mathbf{5} \Omega$
$X_{e}=X_{p}+a^{2} X_{s}=2 \Omega+(2)^{2}(2 \Omega =10 \Omega$
b. The transformed equivalent circuit appears in Fig. 7. $$V_{L}=\left(I_{p}\right)\left(a^{2} R_{L}\right)=2400 \mathrm{~V}$$ Thus, $$V_{L}=\frac{2400 \mathrm{~V}}{a}=\frac{2400 \mathrm{~V}}{2}=1200 \mathrm{~V}$$ and \begin{aligned} V_{g} &=\mathbf{I}_{p}\left(R_{e}+a^{2} R_{L}+j X_{e}\right) \\ &=10 \mathrm{~A}(5 \Omega+240 \Omega+j 10 \Omega)\\ &=10 \mathrm{~A}(245 \Omega+j 10 \Omega) \\ V_{g} &=2450 \mathrm{~V}+j 100 \mathrm{~V}\\ &=2452.04 \mathrm{~V} \angle 2.34^{\circ}=2452.04 \mathrm{V} \angle 2.34^{\circ}\\ \end{aligned} Fig. 7:
c. For $R_{e}$ and $X_{e}=0$, $$V_{g}=a V_{L}=(2)(1200 \mathrm{~V})=2400 \mathrm{~V}$$ Therefore, it is necessary to increase the generator voltage by $52.04 \mathrm{~V}$ (due to $R_{e}$ and $X_{e}$ ) to obtain the same load voltage.

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