2.0 Transformers
2.3 The ideal transformer (Pg 13)
2.1 Why transformers are important to modern life (Pg 4)
2.2 Type and construction of transformers (Pg 7)
2.4 Theory of operation of real single phase transformers (Pg 30)
2.6 The equivalent circuit of a transformers (Pg 41)
2.5 Transformer losses (Pg 40)
2.8 Transformer voltage regulation and efficiency (Pg 59)
2.7 The per-unit system (Pg 55)
2.9 Three phase transformer (Pg 65)
After transformer
Transmission losses
Previous problem
transmission losses in the lines of a power system are proportional to the square of the current in the lines
Eliminated these restriction on the range and power level of power systems
Power system generated and transmitted power at such low voltages that very large currents were necessary to supply significant amounts of power
High currents caused huge voltage drops and power losses in transmission line
it changes one AC voltage level to another level voltage without affecting the actual power supplied
P = IV
If the transformer steps up the voltage, it must decrease the current to keep the power into the device equal to the power out of it.
P = IV
P = I^2 R
Type of construction
Type of special-purpose transformer
Core form
Shell form
Potential transformer
Current transformer
Consists a simple rectangular laminated piece of steel
Consists three-legged laminated core
with transformer windings wrapped around two sides of the rectangle
with the winding wrapped around the center leg
Sampling a high voltage and produce a low secondary voltage proportionally
Provide much smaller secondary current than but directly proportional to its primary current
To handle very small current
Assumption
Formula
变形金刚很重（Important）是因为它的构造（construction）太过完美（ideal）
不过身边有一只真的喜欢运动的单身狗（operation real single...）流了很多汗（losses）导致电路损坏（circuit）
所以就分件拿去卖（per-unit system）有调节电压和效率高的比较好卖（voltage regulation and efficiency）卖了三个部分（three phase）
Power, P in transformer
Impedance transformation through a transformer
Relationship between vp&vs and ip&is
No magnetic flux leaks out between the two windings
The winding resistance is zero
It is infinitely easy to set up a magnetic flux in the core
There is no hysteresis or eddy-current power loss in the core
the reluctance of the core is zero
the magnetizing current required is almost zero
core permeability is infinite
Φ = Ni / R
R is reluctance of circuit = lc / μA
vp(t) / vs(t) = Np / Ns = a
ip(t) / is(t) = Ns / Np = 1/a
apparent power, S = VI
power factor, pf = cosθ
reactive power, Q = VI cosθ
P = VI cosθ
Q = S cosθ
Apparent impedance of the primary circuit if trasformer
Load from secondary part
Thevenin equivalent of transformer circuit
ZL = Vs / Is
Z'L = Vp / Ip = a^2 ZL
Vp = a Vs
Ip = Is / a
Primary part 
Secondary part
Equivalent circuit when secondary impedance is transferred to primary side
Equivalent circuit when primary source is transferred to secondary side
Characteristic
Formula
The magnetizing current is not zero
There is power loss in the iron core
Flux leakage between the windings
Capacitive effects are present in high voltage transformers
Resistance of the windings
leading to inductive reactance effects
but it may be as little as 3% of the load currents
due to hysteresis and eddy current effects
affecting their ability to withstand strikes of lightning
Voltage supply (Faraday's Law)
Excitation current, iex
Induced emf (Faraday's Law)
Average flux in the core
e(ind) = N dΦ/dt
Primary
i(ex) = im + i(h+e)
arrange and integration of vp(t) = Np dΦ/dt
Secondary
vp(t) = e(P) + e(LP)
vs(t) = e(S) + e(LS)
vp(t) = Np dΦM/dt + Np dΦ(LP)/dt
Φ = ΦM + ΦL
im = magnetization current to produce flux in transformer core
i(h+e) = Core-loss current to make up for hysteresis and eddy current losses
Φ = (Vm/ωNp) sin ωt
Vp(t) = Vm cos ωt
Hysteresis losses
Eddy current losses
Leakage flux
Copper (I^2R) losses
resistive heating losses in the primary and secondary windings
resistive heating losses in the core
associated with the rearrangement of the magnetic domains in core
fluxes ΦLP and ΦLS that escape the core
Primary part
Secondary part
experimentally determine the values of the reactance and resistances in the transformer model
Open circuit test
Short circuit test
resistance, Req = Rp/a^2 + Rs
resistance, Req = Rp + a^2*Rs
reactance, Xeq = Xp + a^2*Xs
reactance, Xeq = Xp/a^2 + Xs
rectangular form, Y(E) = 1/R(C) + j 1/X(M)
polar form, Y(E) = [ I(OC)/V(OC) ] < - cos-1 PF
rectangular form, Z(SE) = Req + jXeq
polar form, Z(SE) = [ I(SC)/V(SC) ] < cos-1 PF
PF = P(OC)/[V(OC)*I(OC)]
Negative θ
PF = P(SC)/[V(SC)*I(SC)]
Positive θ
Per unit basis equation
In single-phase
Reactive Power, Q(base)
Apparent Power, S(base)
Power, P(base)
Impedance, Z(base) = V(base) / I(base) = [V(base)]^2 / S(base)
Admittance, Y(base) = I(base) / V(base)
S(base) = V(base)*I(base)
quantity per unit = actual value / base value of quantity
In secondary circuit
Transformer efficiency, η
Voltage Regulation, VR
VR = {[V(s, nl) - V(s, fl)] / V(s, fl)} x 100%
Vp/a = Vs + Req Is + j Xeq Is
η = {(Vs Is cosθ) / [Vs Is cosθ + I^2 R(eq,s) + (Vs^2 / Rc)]} x 100%