Ideal Transformer

An ideal transformer is one which has no losses, i.e. its windings have no ohmic resistance, there is no magnetic leakage, and hence has no \(I^2R\) and core losses

In other words, an ideal transformer consists of two purely inductive coils wound on a loss-free core

It is impossible to realize such a transformer in practice

Consider an ideal transformer whose secondary is open and primary is connected to sinusoidal alternating voltage \(V_1\)

Potential difference cause an alternating current to flow in the primary

Since, the primary coil is purely inductive and the secondary is open (no output) the primary draws a magnetizing current \(I_\mu\) only

\(I_\mu\) is very small in magnitude, magnetize the core, and lags \(V_1\) by \(90^0\)

The alternating current \(I_\mu\) produces an alternating flux \(\Phi\), which at all times is proportional to the current and is in phase with it.

This changing flux is linked with both \(N_1\) and \(N_2\)

Therefore, it produced self induced emf in the primary (\(E_1\)) which at every instant oppose \(V_1\) known as counter emf or back emf of the primary

Similarly, there is secondary mutually induced emf (\(E_2\)), antiphase with \(V_1\)

The magnitude of \(E_2\) is proportional to the rate of change of flux and the number of secondary turns

EMF Equation of a transformer

Let,

\(N_{1}=\) | No. of turns in primary |

\(N_{2}=\) | No. of turns in secondary |

\(\Phi_{m}=B_{m}\times A\) | Maximum flux in core in webers |

\(f=\) | Frequency of a.c. input in Hz |

Flux increases from zero to maximum value \(\Phi_{m}\) in one quarter of the cycle i.e. \(1/4~f\) s

\(\therefore\) Average rate of change of flux \(=\dfrac{\Phi_{m}}{1/4f}\) \(=4f\Phi_{m}\) volt Rate of change of flux per turn means induced emf in volts

\(\therefore\) Average emf/turn \(=4f\Phi_{m}\) volt If flux varies sinusoidally, then rms value of induced emf is obtained by multiplying the average value with form factor

Form factor = \(\dfrac{\mbox{r.m.s value}}{\mbox{average value}}\)= 1.11 \(\therefore\) r.m.s value of e.m.f/turn = \(1.11 \times 4f\Phi_{m} = 4.44f\Phi_{m} volt\) Now, r.m.s value of the induced e.m.f in the whole primary winding

equation* E_1 = 4.44fN_1_m = 4.44 fN_1B_mA

Similarly, for secondary winding

equation* E_2 = 4.44fN_2_m = 4.44 fN_2B_mA

From above two equations

equation* E_1/N_1 = E_2/N_2 = 4.44f_m

emf/turn is same for both the primary and secondary winding

Voltage Transformation Ratio

\[\dfrac{E_{2}}{E_{1}}=\dfrac{N_{2}}{N_{1}}=K\] Step-Up Transformer: \[N_{2}>N_{1}\Rightarrow K>1\]

Step-Down Transformer: \[N_{2}<N_{1}\Rightarrow K<1\]

For an ideal transformer, input VA = output VA \[\begin{aligned} V_{1}I_{1} & =V_{2}I_{2}\\ \Rightarrow\dfrac{I_{2}}{I_{1}} & =\dfrac{V_{1}}{V_{2}}=\dfrac{1}{K} \end{aligned}\] Currents is in inverse proportional of the voltage transformation ratio