A transformer is a static electrical device that transfers electrical energy from one circuit to another through electromagnetic induction, without any physical connection. It works only on AC supply and can step up or step down voltage levels without changing its frequency. The transformer consists of primary and secondary windings wound around a common magnetic core, which helps in efficient energy transfer while maintaining electrical isolation between the circuits.
A basic transformer has two coils or windings, wound around a common iron core. However, more than two-winding transformers exist, but in the basic design, there are mainly two windings. One winding is connected to the input alternating voltage source, while the other is connected to the output load.
Now, to understand the working principle of a transformer, we shall first consider an ideal transformer. The ideal transformer is an imaginary transformer that has no core losses, no copper losses, and perfect magnetic coupling.
- No core losses mean hysteresis loss plus eddy current loss is zero.
- No copper losses imply that the windings have zero resistance.
- Perfect magnetic coupling means there is no leakage flux, i.e. all flux links both windings.
Suppose there is an ideal transformer with an open secondary winding. Now, an alternating voltage is applied to the primary winding. Say, the voltage is \( v_{1} \).
\[ v_1 = v_{1m} \sin2\pi f t \;\;Where,\;v_{1m}\;is\;the\;peak\;voltage. \]
Under this condition, the primary winding draws a small current (magnetizing current) to establish an equal and opposite EMF across the winding. This is self-induced EMF in the primary. That means, self-induced EMF \( e_{1} \) is, at any instant, equal to and in opposition to \( v_{1} \). The primary draws only magnetizing current to establish the counter EMF. Since the winding is purely inductive, the magnetizing current \( I_{\mu} \) lags behind the applied voltage \( v_{1} \) by 90°. This alternating current \( I_{\mu} \) produces an alternating flux \( \phi \). This flux is proportional to the current and hence it is in phase with the current.
\[ \phi = \phi_{m} \sin2\pi f t \;\;Where,\;\phi_{m}\;is\;the\;peak\;flux. \]
This changing flux is linked with both windings. Hence, in the secondary winding, an induced EMF \( e_{2} \) is produced, which is known as mutually induced EMF. The EMF is in phase opposition with \( v_{1} \), and its magnitude is proportional to the rate of change of flux and the number of secondary turns.
\[ e_2 = -N_2 \frac{d\phi}{dt} \;\;volt \]
The self induced primary emf can be similarly written as\[ e_1 = -N_1 \frac{d\phi}{dt} \;\;volt \]
Hence, the self induced primary emf and mutually induced secondary emf are\[ e_1 = -N_1 \frac{d\phi}{dt} = -N_1 \frac{d(\phi_{m} \sin2\pi f t)}{dt} = -N_1 \phi_m \cos 2\pi f t \times 2\pi f \]\[ e_2 = -N_2 \frac{d\phi}{dt} = -N_2 \frac{d(\phi_{m} \sin2\pi f t)}{dt} = -N_2 \phi_m \cos 2\pi f t \times 2\pi f \]
These equations are expressed as vectors i.e. instantaneous values. If \( E_{1} \) and \( E_{2} \) are the RMS values of \( e_{1} \) and \( e_{2} \). To obtain the RMS value of EMF \( e_1 \) and \( e_2 \), divide their maximum value by \( \sqrt{2} \) \[ E_1 = \frac{2\pi}{\sqrt{2}} f N_1 \phi_m = 4.44 f N_1 \phi_m \;\;\&\;\; E_2 = \frac{2\pi}{\sqrt{2}} f N_2 \phi_m = 4.44 f N_2 \phi_m \] The cosine term has no significance except to derive the instantaneous values.
In an ideal transformer, \[ V_1 = E_1 \]\[ V_2 = E_2 \] Where, \( V_1 \) is the primary source terminal voltage and \( V_2 \) is the secondary terminal voltage.
\[ \frac{E_2}{E_1} = \frac{N_2}{N_1} = K \] This constant is known as the voltage transformation ratio.
Considering Core Losses
Now, we shall consider the core losses of a transformer. Suppose this transformer is in a no-load condition. The primary input current under no-load condition has to supply the iron loss in the core. Iron loss is the sum of hysteresis loss and eddy current loss. A very small amount of copper loss also occurs in the primary winding. This happens because the no-load current has to flow through the primary winding. However, we normally neglect this ohmic loss. Because of core losses, the no-load primary input current is not exactly 90° behind \( v_1 \) but lags it by an angle less than 90°.
No-load primary input power will be \[ W_o = V_1 I_o \cos \theta_o \] No load primary current \( I_o \) has two components. One in phase with \( v_1 \). This is known as the active component or working component or iron loss component \( I_w \). It supplies the iron-loss and that neglected small quantity of primary copper loss. \[ I_w = I_o \cos \theta_o \] Other one is in quadrature with \( v_1 \). This is known as the magnetizing component \( I_{mu} \). This component of no load current sustains the alternating flux in the core and is wattless. \[ I_{\mu} = I_o \sin \theta_o \] Thus, \( I_o \) is the vector sum of \( I_o \) and \( I_{mu} \) \[ I_o = \sqrt{I_{\mu}^2 + I_w^2} \]The no-load current is quite small compared to the full-load current of a transformer.
Now the secondary is loaded. The, secondary current \( I_2 \) starts flowing. The secondary current contributes its own mmf (\(I_2N_2\)). Because of, this mmf the secondary winding creates its own flux \(\phi_2\). This flux is in opposition to the main primary \(\phi\). The opposite secondary flux \(\phi_2\) weakens the primary flux momentarily. Hence, the primary back emf \( E_1 \) tends to reduce. For a moment, \( V_1 \) gains the upper hand over \( E_1 \). This causes more current \( I_1′ \) to flow in primary. This current \( I_1′ \) is known as the load component of primary current. This current is in phase opposition to current \( I_2 \). Therefore, the primary mmf (\(I_1’N_1\)) sets up a flux \(\phi_1\) which opposes \(\phi_2\). Obviously, \(\phi_1\) is in the same direction as \(\phi \). It is needless to say, \(\phi_1\) = \(\phi_2\). Thus, the magnetic effect of secondary current gets neutralized immediately by additional primary current \( I_1′ \). Hence, whatever may be the load conditions, the net flux remains constant.
Considering Winding Resistance
An ideal transformer has no resistance. But in reality, a transformer has resistance in its windings. The primary and secondary windings have some resistance. Because of this resistance, there is a voltage drop in the windings.
The secondary terminal voltage \( V_2 \) is equal to the vector difference of the secondary induced emf \( E_2 \) and the voltage drop across the secondary winding resistance \( I_2 R_2\). The direction of this voltage drop will be along the direction of secondary current \( I_2 \). \[ V_2 = E_2 – I_2 R_2 \]
The primary induced emf \( E_1 \) is given by the vector difference of the source voltage \( V_1 \) and the voltage drop across the primary winding resistance \( I_1 R_1\). The direction of this voltage drop will be along the direction of total primary current \( I_1 \). Here, \( I_1 \) is the vector sum of transforming current \( I_1′ \) and no load current \( I_o \). The transforming current \( I_1′ \) refers to that current which is actually transformed to secondary. . \[ E_1 = V_1 – I_1 R_1 \]
Considering Winding Reactance
In an ideal transformer, all the flux links the both primary and secondary winding. But in real transformers, this is not fully possible. Most of the flux links both windings, but some flux does not. This is called leakage flux. Leakage flux does not help in transferring energy between the windings.
On account of the leakage flux, both the primary and secondary windings have leakage reactance. Both windings will have self-induced emf. The magnitude of the emf is a small fraction of rated voltage. The primary terminal voltage \( V_1 \) therefore, has a component \( I_1X_1 \). As this is a pure reactance, the primary current \( I_1 \) lags the leakage reactance emf by 90o. Similarly, an emf of self-induction \( I_2X_2 \) is developed in the secondary winding. Similarly, the secondary current \( I_2 \) lags the secondary leakage reactance emf by 90o.
Considering Both Winding Resistance and Reactance
The primary impedance considering winding resistance and leakage reactance is given by\[ Z_1 = \sqrt{R_1^2 + X_1^2} \]Similarly, the secondary impedance considering winding resistance and leakage reactance is given by\[ Z_2 = \sqrt{R_2^2 + X_2^2} \]Therefore,\[ V_1 = E_1 + I_1 (R_1 + jX_1) = E_1 + I_1 Z_1 \] \[ E_2 = V_2 + I_2 (R_2 + jX_2) = V_2 + I_2 Z_2 \]