# The Iron Core Transformer

An iron-core transformer under loaded conditions is shown in Fig. 1. The iron core will serve to increase the coefficient of coupling between the coils by increasing the mutual flux fm. Recall from Chapter 10 that magnetic flux lines will always take the path of least reluctance, which in this case is the iron core.
Fig. 1: Iron-core transformer
In the analyses to follow in this chapter, we will assume that all of the flux linking coil 1 will link coil 2. In other words, the coefficient of coupling is its maximum value, 1, and $\phi_m = \phi_p = \phi_s$.
In addition, we will first analyze the transformer from an ideal viewpoint; that is, we will neglect losses such as the geometric or dc resistance of the coils, the leakage reactance due to the flux linking either coil that forms no part of fm, and the hysteresis and eddy current losses. This is not to convey the impression, however, that we will be far from the actual operation of a transformer. Most transformers manufactured today can be considered almost ideal.
When the current ip through the primary circuit of the iron-core transformer is a maximum, the flux $\phi_m$ linking both coils is also a maximum. In fact, the magnitude of the flux is directly proportional to the current through the primary windings. Therefore, the two are in phase, and for sinusoidal inputs, the magnitude of the flux will vary as a sinusoid also. That is, if $$i_p = V_m \sin wt = \sqrt{2} I_p \sin wt$$ then $$\phi_m = \Phi_m \sin wt$$ The induced voltage across the primary due to a sinusoidal input can be determined by Faraday's law: $$e_p = N_p { d \phi_p \over dt} = N_p { d \phi_m \over dt}$$ Substituting for $\phi_m$ gives us $$e_p = N_p { d \over dt} (\Phi_m \sin wt)$$ and differentiating, we obtain $$e_p = w N_p \Phi_m \cos wt)$$ or $$e_p = w N_p \Phi_m \sin (wt+90^\circ)$$ indicating that the induced voltage $e_p$ leads the current through the primary coil by $90^\circ$. The effective value of $e_p$ is $$E_p = { w N_p \Phi_m \over \sqrt{2}}= { 2 \pi f N_p \Phi_m \over \sqrt{2}}$$ and $$\bbox[10px,border:1px solid grey]{E_p = 4.44 f N_p \Phi_m} \tag{1}$$ which is an equation for the rms value of the voltage across the primary coil in terms of the frequency of the input current or voltage, the number of turns of the primary, and the maximum value of the magnetic flux linking the primary.
For the case under discussion, where the flux linking the secondary equals that of the primary, if we repeat the procedure just described for the induced voltage across the secondary, we get $$\bbox[10px,border:1px solid grey]{E_s = 4.44 f N_s \Phi_m} \tag{2}$$ Dividing Eq. (1) by Eq. (2), as follows: $${ E_p \over E_s} = { 4.44 f N_p \Phi_m \over 4.44 f N_s \Phi_m}$$ we obtain $$\bbox[10px,border:1px solid grey]{{ E_p \over E_s} = { N_p \over N_s}} \tag{3}$$ revealing an important relationship for transformers:
The ratio of the magnitudes of the induced voltages is the same as the ratio of the corresponding turns.
The ratio $N_p /N_s$, usually represented by the lowercase letter $a$, is referred to as the transformation ratio: $$\bbox[10px,border:1px solid grey]{a = { N_p \over N_s}}$$ If $a < 1$, the transformer is called a step-up transformer since the voltage $Es > Ep$;
and, if $a > 1$, the transformer is called a step-down transformer since $Ep > Es$; that is, $$E_p = a E_s$$
Example 1: For the iron-core transformer of Fig. 2:
a. Find the maximum flux $\phi_m$.
b. Find the secondary turns $N_s$.
Fig. 2: Example 1.
Solution:
a. $E_p = 4.44N_p f \phi_m$ Therefore $$\phi_m = { E_p \over 4.44N_p f} = { 200V \over (4.44)(50t)(60Hz)}$$ and $$\phi_m = 15.02mWb$$ b. ${E_p \over E_s} = {N_p \over N_s}$ Therefore, $$N_s = {N_p E_s \over E_p} = { (50t)(2400V) \over 200V}$$ $$N_s = 600 turns$$
The induced voltage across the secondary of the transformer of Fig. 1 will establish a current is through the load $Z_L$ and the secondary windings. This current and the turns $N_s$ will develop an $mmf = N_s i_s$ that would not be present under no-load conditions since $i_s = 0$ and $N_s i_s = 0$.
Under loaded or unloaded conditions, however, the net ampere-turns on the core produced by both the primary and the secondary must remain unchanged for the same flux fm to be established in the core.
The flux $\phi_m$ must remain the same to have the same induced voltage across the primary and to balance the voltage impressed across the primary. In order to counteract the mmf of the secondary, which is tending to change $\phi_m$, an additional current must flow in the primary. This current is called the load component of the primary current and is represented by the notation $i_p^'$.
For the balanced or equilibrium condition, $$N_p i_p^' = N_si_s$$ The total current in the primary under loaded conditions is $$i_p = i_p^' + i_{\phi_m}$$ where $i_{\phi_m}$ is the current in the primary necessary to establish the flux $\phi_m$. For most practical applications, $$i_p^' > i_{\phi_m}$$ For our analysis, we will assume $$i_p \cong i_p^' ,$$ so that $$N_p i_p = N_si_s$$ Since the instantaneous values of $i_p$ and $i_s$ are related by the turns ratio, the phasor quantities $I_p$ and $I_s$ are also related by the same ratio: $$N_p I_p = N_sI_s$$ or $$\bbox[10px,border:1px solid grey]{{I_p \over I_s}= {N_s \over N_p }} \tag{1}$$
The primary and secondary currents of a transformer are therefore related by the inverse ratios of the turns.
Keep in mind that Equation (1) holds true only if we neglect the effects of ifm. Otherwise, the magnitudes of $I_p$ and $I_s$ are not related by the turns ratio, and $I_p$ and $I_s$ are not in phase.
For the step-up transformer, $a < 1$, and the current in the secondary, $$I_s = a I_p,$$ is less in magnitude than that in the primary. For a step-down transformer, the reverse is true.