Lecture 04Electro Mechanical System1 Ideal Transformer  An ideal transformer  Transformer has no losses and core is infinitely permeable  All fluxes link all coils and here are no leakage fluxes  Voltage relationship  Consider a transformer with two coils of N1 and N2 turns  A magnetizing current I m creates a flux  m  The flux varies sinusoidally and has a peak value of  max  The induced voltages are  From these equations, it can be deduced that :  The ratio of the primary and secondary voltages is equal to the ratio of the number of turns  E1 and E2 are in-phase  polarity marks show the terminal on each coil that have a peak positive voltage simultaneously

Lecture 04Electro Mechanical System2 Ideal Transformer  Current relationship  Let a load be connected across the secondary of an ideal transformer  Current I 2 will immediately flow I 2 = E 2 /Z  Coil voltages E1 and E2 cannot change when connected to a fixed voltage source and hence flux  m cannot change  Current I 2 produces an mmf mmf 2 =N 2 I 2  If mmf2 acts alone, it would profoundly change  m   m can only remain fixed if the primary circuit develops a mmf which exactly counter balances mmf2  Current I 1 must flow such that:  I 1 and I 2 must be in-phase when I 1 flows into the positive polarity marking of the primary, I 2 flows out of the positive polarity marking of the secondary

Lecture 04Electro Mechanical System3 Ideal Transformer  An ideal transformer is a lossless device with an input and an output winding. The relationships between input voltage and output voltage, and between input current and output current, are given by:

Lecture 04Electro Mechanical System4 Impedance Ratio

Lecture 04Electro Mechanical System5 Shifting Impedances  Impedances located on the secondary side of a transformer can be relocated to the primary side.  The circuit configuration remains the same (series or shunt connected) but the shifted impedance values are multiplied by the turns ratio squared.  Impedance on the primary side can be moved to the secondary side in reverse manner.  The impedance values are divided by the turns ratio squared.  In general, as an impedance is shifted across the transformer  The real voltage across the impedance increases by the turns ratio  The actual current through the impedance decreases by the turns ratio  The required equivalent impedance increases by the square of the turns ratio

Lecture 04Electro Mechanical System6 Shifting Impedances

Lecture 04Electro Mechanical System7 Example Calculate the voltage E and current I in the circuit, knowing that the turns ratio is 1:100 Shift all the impedances to the primary side, since primary side has 100 fewer turns, so impedences values are devided by and voltage E becomes E/100