Transformer

An ideal transformer has no winding resistance, no leakage flux, no core loss and a core of infinite permeability (so it needs negligible magnetising current). Its primary and secondary are linked by the same mutual flux φ.

No-load operation

          phi (mutual flux)
     +---------------------+
  I0 |                     |
 o-->|]  N1           N2  [|---o
 V1  |]  E1           E2  [|   open
 o---|]               [   |---o
     +---------------------+

• The primary is connected to $V_1$; the secondary is open, so $I_2 = 0$.
• Since the winding has no resistance, the applied voltage is balanced only by the induced emf: $V_1 = E_1 = 4.44 f N_1\phi_m$. This fixes the flux.
• A very small current $I_0$ (the magnetising current $I_m$) flows. It is in phase with φ and lags $V_1$ by 90°, so the input power is zero.
• The same flux induces $E_2 = 4.44 f N_2\phi_m$ in the secondary. $E_1$ and $E_2$ are in phase, both lagging φ by 90°, and

$$\frac{E_1}{E_2} = \frac{N_1}{N_2}$$

Phasor diagram (no load): $\phi$ along the reference, $I_0$ in phase with φ, $V_1$ leading φ by 90°, $E_1$ and $E_2$ opposite to $V_1$.

Loaded operation

1. A load is connected; current $I_2$ flows in the secondary. It produces mmf $N_2 I_2$ which, by Lenz's law, opposes the mutual flux.
2. If φ fell, $E_1$ would fall below $V_1$. The primary therefore draws an extra load component of current $I_2'$ whose mmf exactly cancels the secondary mmf:

$$N_1 I_2' = N_2 I_2 \Rightarrow I_2' = \frac{N_2}{N_1}I_2$$

3. Total primary current $I_1 = I_0 + I_2'$. Power drawn from the supply rises to match the load: $V_1 I_1 \cos\phi_1 = V_2 I_2\cos\phi_2$.

Proof that the core flux remains constant at any load

Net mmf acting on the core:

$$\begin{aligned} F_{net} &= N_1 I_1 - N_2 I_2 \\ &= N_1 (I_0 + I_2') - N_2 I_2 \\ &= N_1 I_0 + (N_1 I_2' - N_2 I_2) \\ &= N_1 I_0 \end{aligned}$$

So the net mmf, and hence the flux $\phi = N_1 I_0 / S$, is the same as on no load. The same result follows from the voltage equation: since $V_1 = E_1 = 4.44 f N_1\phi_m$ at every load and $V_1$, $f$ and $N_1$ are fixed,

$$\phi_m = \frac{V_1}{4.44 f N_1} = \text{constant}$$

Hence the mutual flux of a (ideal) transformer is independent of load from no load to full load. In a practical transformer it falls only very slightly because of the small leakage impedance drop.

An ideal transformer is one with zero winding resistance, no leakage flux (all flux links both windings), no hysteresis or eddy-current loss, and a core of infinite permeability, so it needs no magnetising current. Its efficiency is 100% and its voltage regulation is zero.

Operating principle

A transformer works on mutual induction (Faraday's law).

           phi (alternating)
     +---------------------+
 o---|]                   [|---o
 V1  |]  N1     core  N2  [|   V2 -> load
 o---|]                   [|---o
     +---------------------+
   primary             secondary

1. An ac voltage $V_1$ is applied to the primary winding of $N_1$ turns.
2. The current sets up an alternating flux φ in the laminated iron core.
3. The flux links the secondary winding of $N_2$ turns and induces in it an emf $E_2$ of the same frequency. The same flux induces a back emf $E_1$ in the primary that balances $V_1$.
4. When a load is connected, $I_2$ flows; the primary automatically draws a matching current so that $N_1I_1 = N_2I_2$ and $V_1I_1 = V_2I_2$.
5. Since emf per turn is the same in both windings, $V_1/V_2 = N_1/N_2$: the voltage is stepped up or down without changing frequency.

EMF equation

Let the flux be sinusoidal: $\phi = \phi_m\sin\omega t$, $\omega = 2\pi f$.

$$\begin{aligned} e_1 &= -N_1\frac{d\phi}{dt} = -N_1\omega\phi_m\cos\omega t \\ &= N_1\omega\phi_m\sin(\omega t - 90^\circ) \end{aligned}$$

So $e_1$ lags φ by 90°. Its maximum value is

$$E_{1m} = N_1\omega\phi_m = 2\pi f N_1\phi_m$$

RMS value:

$$\begin{aligned} E_1 &= \frac{E_{1m}}{\sqrt2} = \frac{2\pi}{\sqrt2} f N_1 \phi_m \\ E_1 &= 4.44\, f N_1 \phi_m = 4.44\, f N_1 B_m A \end{aligned}$$

Similarly for the secondary:

$$E_2 = 4.44\, f N_2 \phi_m$$

Transformation ratio

$$\frac{E_2}{E_1} = \frac{N_2}{N_1} = K, \qquad \frac{E_1}{N_1} = \frac{E_2}{N_2} = 4.44 f\phi_m \ (\text{emf per turn})$$

For an ideal transformer $V_1 = E_1$, $V_2 = E_2$ and $I_1/I_2 = N_2/N_1 = K$.

Example: a 50 Hz transformer with $N_1 = 500$ and $\phi_m = 9$ mWb gives $E_1 = 4.44\times50\times500\times0.009 = 999$ V.

A transformer has no moving parts, so it has no friction or windage loss. Its losses are core (iron) losses and copper losses.

Losses in a transformer

1. Core or iron loss ($P_i$) – occurs in the core due to the alternating flux. It depends on $B_m$ and $f$, which are almost constant, so it is a constant loss (same at all loads). It is measured by the open-circuit test.
   • Hysteresis loss: $P_h = \eta B_m^{1.6} f V$; reduced by using silicon steel / CRGO.
   • Eddy current loss: $P_e = K_e B_m^2 f^2 t^2 V$; reduced by thin insulated laminations.
2. Copper loss ($P_{cu}$) – $I_1^2R_1 + I_2^2R_2 = I_2^2 R_{02}$ in the windings. It varies with the square of the load current, so it is a variable loss. Full-load value is measured by the short-circuit test. At a fraction x of full load, $P_{cu} = x^2 P_{cu,fl}$.
3. Stray loss – from leakage flux causing eddy currents in the tank, clamps and conductors. Usually small and included in copper loss.
4. Dielectric loss – in the insulation (oil, paper); important only at very high voltage.

Efficiency

$$\eta = \frac{\text{output}}{\text{output} + \text{losses}} = \frac{V_2 I_2\cos\phi_2}{V_2 I_2\cos\phi_2 + P_i + I_2^2 R_{02}}$$

Condition for maximum efficiency

For constant $V_2$ and $\cos\phi_2$, η is maximum when the denominator of the following is minimum. Divide numerator and denominator by $I_2$:

$$\eta = \frac{V_2\cos\phi_2}{V_2\cos\phi_2 + \dfrac{P_i}{I_2} + I_2R_{02}}$$

η is maximum when $D = \dfrac{P_i}{I_2} + I_2R_{02}$ is minimum:

$$\begin{aligned} \frac{dD}{dI_2} &= -\frac{P_i}{I_2^2} + R_{02} = 0 \\ I_2^2 R_{02} &= P_i \end{aligned}$$

($d^2D/dI_2^2 = 2P_i/I_2^3 > 0$, so it is a minimum of D.)

Efficiency is maximum when variable copper loss equals constant iron loss.

Load at maximum efficiency

$$x = \frac{I_2}{I_{2,fl}} = \sqrt{\frac{P_i}{P_{cu,fl}}}, \qquad \text{kVA at } \eta_{max} = \text{rated kVA}\times\sqrt{\frac{P_i}{P_{cu,fl}}}$$ $$\eta_{max} = \frac{x S\cos\phi}{x S\cos\phi + 2P_i}$$

Also, for a given kVA, η is highest at unity power factor.

Example: a transformer with $P_i = 1$ kW and $P_{cu,fl} = 4$ kW has maximum efficiency at $\sqrt{1/4} = 0.5$, i.e. half load. Distribution transformers are designed with low $P_i$ so that η is maximum at about 50–70% load, as they are lightly loaded much of the day.

A transformer is a static device that transfers electrical energy from one ac circuit to another at the same frequency, usually changing the voltage level, by mutual induction between two windings on a common magnetic core.

Construction (in brief)

Two windings — primary ($N_1$ turns) connected to the supply and secondary ($N_2$ turns) connected to the load — are placed on a laminated silicon-steel core. They are electrically isolated but magnetically coupled.

           phi ---->
     +---------------------+
 o---|]                   [|---o
 V1  |]  N1           N2  [|   V2 --> load
 o---|]  I1           I2  [|---o
     +---------------------+

Operating principle

1. An alternating voltage $V_1$ applied to the primary drives an alternating current, which sets up an alternating flux φ in the core.
2. Almost all of this flux is confined to the core and links the secondary winding.
3. By Faraday's law the changing flux induces an emf $E_2$ in the secondary (mutually induced emf) and an emf $E_1$ in the primary (self-induced back emf, which nearly balances $V_1$).
4. If a load is connected, current $I_2$ flows and power is delivered. The secondary mmf tends to reduce φ, so the primary draws extra current to restore it; in this way power passes from primary to secondary through the magnetic field.
5. Emf per turn is the same in both windings, so the voltages are in the ratio of turns: step-up if $N_2 > N_1$, step-down if $N_2 < N_1$. A dc supply gives no changing flux, so a transformer does not work on dc.

Derivation of emf equation

Let the core flux be $\phi = \phi_m \sin\omega t$, where $\omega = 2\pi f$.

Instantaneous emf in the primary:

$$\begin{aligned} e_1 &= -N_1\frac{d\phi}{dt} = -N_1 \phi_m \omega \cos\omega t \\ &= 2\pi f N_1\phi_m \sin(\omega t - 90^\circ) \end{aligned}$$

Maximum value: $E_{1,max} = 2\pi f N_1 \phi_m$.

RMS value:

$$\begin{aligned} E_1 &= \frac{2\pi f N_1\phi_m}{\sqrt2} = 4.44\, f N_1 \phi_m \ \text{V} \end{aligned}$$

Similarly $E_2 = 4.44\, f N_2 \phi_m$ V. Since $\phi_m = B_m A$, $E = 4.44\, f N B_m A$.

Voltage transformation ratio

$$K = \frac{E_2}{E_1} = \frac{N_2}{N_1}$$

The emf lags the flux by 90°.

Example: a 230/115 V, 50 Hz transformer with 400 primary turns has $\phi_m = 230/(4.44\times50\times400) = 2.59$ mWb, and needs $400\times115/230 = 200$ secondary turns.

Shunt branch parameters come from the OC test (done on LV side, so they are directly LV values); series parameters come from the SC test (done on HV side, so they are HV values). Turns ratio $a = 2500/250 = 10$.

Rated currents: LV = 20000/250 = 80 A, HV = 20000/2500 = 8 A, so the SC test was at full load.

From the OC test (LV side)

$$\begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0 I_0} = \frac{105}{250\times1.4} = 0.30 \\ I_c &= I_0\cos\phi_0 = 1.4\times0.30 = 0.42\ \text{A} \\ I_m &= I_0\sin\phi_0 = 1.4\times0.954 = 1.336\ \text{A} \\ R_0 &= \frac{V_0}{I_c} = \frac{250}{0.42} = 595.2\ \Omega \\ X_0 &= \frac{V_0}{I_m} = \frac{250}{1.336} = 187.2\ \Omega \end{aligned}$$

From the SC test (HV side)

$$\begin{aligned} R_{eq,HV} &= \frac{P_{sc}}{I_{sc}^2} = \frac{320}{8^2} = 5\ \Omega \\ Z_{eq,HV} &= \frac{V_{sc}}{I_{sc}} = \frac{120}{8} = 15\ \Omega \\ X_{eq,HV} &= \sqrt{15^2 - 5^2} = 14.14\ \Omega \end{aligned}$$

Referred to LV side (divide HV impedances by $a^2 = 100$)

Approximate equivalent circuit referred to HV side

        Req = 5 ohm    Xeq = 14.14 ohm
  o---+----/\/\/----mmmmmm-----o
      |                         +
  +---+---+
  |       |                    V2'
 [R0]    [X0]
 59.52k  18.72k
  |       |                     -
  +---+---+
      |
  o---+-------------------------o
  V1 = 2500 V

The shunt branch ($R_0 \parallel X_0$) is placed across the supply terminals, and the total series impedance $5 + j14.14\ \Omega$ is in the line.

Answer: referred to LV: $R_0 = 595.2\ \Omega$, $X_0 = 187.2\ \Omega$, $R_{eq} = 0.05\ \Omega$, $X_{eq} = 0.1414\ \Omega$; referred to HV: $R_0 = 59.52$ kΩ, $X_0 = 18.72$ kΩ, $R_{eq} = 5\ \Omega$, $X_{eq} = 14.14\ \Omega$.

The OC test (LV side) gives the shunt branch; the SC test (HV side) gives the series impedance referred to HV. Rated HV current $= 40000/2200 = 18.18$ A, so the SC test (18.2 A) is practically at full load.

(i) $R_0$ and $X_0$ (LV side)

$$\begin{aligned} \cos\phi_0 &= \frac{320}{110\times8.4} = 0.3463 \\ I_c &= 8.4\times0.3463 = 2.909\ \text{A}, \quad I_m = 8.4\times\sin(69.74^\circ) = 7.880\ \text{A} \\ R_0 &= \frac{110}{2.909} = 37.81\ \Omega, \quad X_0 = \frac{110}{7.880} = 13.96\ \Omega \end{aligned}$$

SC test values (HV side)

$$\begin{aligned} R_{eq} &= \frac{650}{18.2^2} = 1.962\ \Omega, \quad Z_{eq} = \frac{85}{18.2} = 4.670\ \Omega \\ X_{eq} &= \sqrt{4.670^2 - 1.962^2} = 4.238\ \Omega \end{aligned}$$

(ii) Full-load efficiency at 0.85 pf lag

Full-load copper loss (corrected to rated current 18.18 A):

$$\begin{aligned} P_{cu,fl} &= 650\left(\frac{18.18}{18.2}\right)^2 = 648.7\ \text{W} \\ P_{out} &= 40000\times0.85 = 34000\ \text{W} \\ \eta &= \frac{34000}{34000 + 320 + 648.7} = \frac{34000}{34968.7} = 97.23\% \end{aligned}$$

(iii) Approximate voltage regulation (0.85 pf lag, $\sin\phi = 0.527$)

$$\begin{aligned} \%R &= \frac{I(R_{eq}\cos\phi + X_{eq}\sin\phi)}{V}\times100 \\ &= \frac{18.18(1.962\times0.85 + 4.238\times0.527)}{2200}\times100 \\ &= \frac{30.33 + 40.59}{2200}\times100 = 3.22\% \end{aligned}$$

Answer: $R_0 = 37.81\ \Omega$, $X_0 = 13.96\ \Omega$ (LV side); full-load efficiency = 97.23%; regulation ≈ 3.22%.

Need for parallel operation

Two or more transformers are connected in parallel (primaries to a common supply bus, secondaries to a common load bus) because:

• Load growth: a second unit can be added when the load exceeds the rating of the existing one.
• Reliability: if one transformer fails or is taken out for maintenance, the others still supply (part of) the load.
• Efficiency: at light load some units can be switched off so the rest run near maximum efficiency.
• Transport and spares: one very large unit is hard to transport, and a small spare unit is cheaper to keep.

  HV bus  ==========+==============+=====
                    |              |
                  [T1]           [T2]
                    |              |
  LV bus  ==========+==============+===== --> load

Conditions for parallel operation

1. Same voltage ratio (same rated primary and secondary voltages). If the ratios differ, the no-load secondary emfs are unequal and a circulating current $I_c = (E_A - E_B)/(Z_A + Z_B)$ flows round the loop even with no load. Since the impedances are small, even 1% difference gives a large current, which causes extra copper loss and heating and reduces the useful capacity.

2. Same polarity (essential). The terminals must be connected with correct polarity. With wrong polarity the two secondary emfs add in the closed loop, which acts like a dead short circuit, and very large currents flow.

3. Same per-unit (percentage) impedance. Load is shared in the inverse ratio of impedances: $S_A/S_B = Z_B/Z_A$. If the per-unit impedances are equal, each transformer takes load in proportion to its kVA rating, so none is overloaded while the others are under-loaded.

4. Same X/R ratio. If X/R differ, the currents in the two units are not in phase; the total current is less than the arithmetic sum, and the units work at different power factors, so the combined capacity is not fully used.

5. For three-phase transformers, additionally:

   • Same phase sequence (essential) – otherwise pairs of phases are short-circuited.
   • Same vector group / phase displacement (essential) – e.g. a Dyn11 unit cannot be paralleled with a Dyn1 unit, as the 60° phase difference between secondary voltages would cause huge circulating currents. Units of the same group (0°, 180°, −30°, +30°) can be paralleled.

Conditions 2 and 5 (polarity, phase sequence, phase displacement) must be met exactly; conditions 1, 3 and 4 should be met as closely as possible.

Sumpner's test (back-to-back test) is a method of loading two identical transformers at full load to measure their temperature rise and efficiency, while drawing from the supply only the power needed for their losses.

Principle and connection

 V1 supply        T1             T2
 o--[W1]--[A]--+--]  [--+-----+--]  [--+
               |  ]  [  |     |  ]  [  |
 o-------------+--]  [--+     |  ]  [  |
  primaries in          |  [W2][A2]    |
  parallel              |     |        |
                        +----(Vr)------+
         secondaries in series opposition,
         fed by low-voltage regulating supply

1. Primaries in parallel on rated voltage and frequency. Wattmeter $W_1$ reads the input on this side.
2. Secondaries in series opposition. With correct opposition, their emfs cancel round the secondary loop, so the voltmeter across the open point reads zero (no current flows). This is the "open-circuit" condition, and $W_1$ reads the core loss of both transformers ($2P_i$).
3. A small auxiliary (regulating) voltage $V_r$ is injected into the secondary loop and adjusted until rated current flows in the secondaries. Because the loop impedance is only the two leakage impedances, a small voltage (about 2× the SC voltage) is enough. Wattmeter $W_2$ reads the full-load copper loss of both transformers ($2P_{cu}$). The primaries carry the matching load current, which circulates between the two units.
4. Both units are now at rated voltage and rated current simultaneously, i.e. working at full load, but the supply provides only the losses.

$$\eta_{each} = \frac{\text{output}}{\text{output} + \dfrac{W_1 + W_2}{2}}$$

Running for several hours gives the temperature rise (heat run).

Why two identical transformers are needed

• The test relies on the two secondary emfs being equal and opposite, so that no current flows in the loop without the injected voltage. Unequal ratios would cause an uncontrolled circulating current.
• Losses must divide equally between the two units, so that each gets half of $W_1$ and $W_2$; with different units the share could not be found.
• Each unit must be loaded to its own rated current at the same time; with the same current in series secondaries this is possible only if the ratings are equal.

Advantage: a large transformer can be heat-run at full load without a large load bank and with very small energy use.

Referred quantities are values of voltage, current and impedance of one winding expressed as equivalent values on the other side, chosen so that power, losses and per-unit drops are unchanged. Referring lets the two windings be joined into a single equivalent circuit without the ideal transformer.

Let $K = N_2/N_1$ (transformation ratio).

Referred voltage

Emf per turn is the same in both windings: $E_1/N_1 = E_2/N_2$. So a secondary voltage seen on the primary side is

$$V_2' = \frac{N_1}{N_2}V_2 = \frac{V_2}{K}$$

and a primary voltage referred to the secondary is $V_1' = K V_1$.

Referred current

The secondary load current is balanced by the primary load component: $N_1 I_2' = N_2 I_2$ (mmf balance). Hence

$$I_2' = \frac{N_2}{N_1}I_2 = K I_2$$

and $I_1' = I_1 / K$ when the primary current is referred to the secondary. Volt-amperes are unchanged: $V_2' I_2' = (V_2/K)(K I_2) = V_2 I_2$.

Referred resistance

The copper loss must be the same before and after referring:

$$\begin{aligned} I_2'^2 R_2' &= I_2^2 R_2 \\ R_2' &= R_2\left(\frac{I_2}{I_2'}\right)^2 = \frac{R_2}{K^2} \end{aligned}$$

Referred reactance and impedance

The reactive power (or the voltage drop as a fraction of voltage) must be the same:

$$\begin{aligned} I_2'^2 X_2' &= I_2^2 X_2 \Rightarrow X_2' = \frac{X_2}{K^2} \\ Z_2' &= \frac{V_2'}{I_2'} = \frac{V_2/K}{K I_2} = \frac{Z_2}{K^2} \end{aligned}$$

Likewise, primary impedance referred to the secondary: $Z_1' = K^2 Z_1$.

Summary

Rule of thumb: to move an impedance from LV to HV multiply by (turns ratio)², i.e. HV side always has the larger impedance.

Example: for 2000/200 V ($K = 0.1$), $R_2 = 0.02\ \Omega$ referred to HV is $0.02/0.01 = 2\ \Omega$, so $R_{01} = R_1 + 2\ \Omega$.

All-day efficiency = (energy output in 24 h) / (energy output + energy losses in 24 h). Iron loss lasts all 24 hours; copper loss varies as the square of kVA load.

Data: 150 kVA, $P_i = 0.4$ kW, $P_{cu,fl} = 1.7$ kW.

Energy table

Copper loss at each load $= 1.7\times(\text{kVA}/150)^2$.

Losses and efficiency

$$\begin{aligned} \text{Iron energy} &= 0.4\times24 = 9.6\ \text{kWh} \\ \text{Total losses} &= 9.6 + 5.957 = 15.557\ \text{kWh} \\ \eta_{all\text{-}day} &= \frac{990}{990 + 15.557}\times100 = 98.45\% \end{aligned}$$

Answer: all-day efficiency ≈ 98.45%.

Per-unit resistance and impedance drops are given: $\varepsilon_r = 5\%$, $\varepsilon_z = 10\%$, so the reactance drop is

$$\varepsilon_x = \sqrt{10^2 - 5^2} = 8.66\%$$

a) Regulation at full load, 0.8 pf lagging

$$\begin{aligned} \%\text{Reg} &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi \\ &= 5\times0.8 + 8.66\times0.6 \\ &= 4.0 + 5.196 = 9.20\% \end{aligned}$$

b) Power factor for zero regulation

Regulation is zero for a leading pf, when

$$\begin{aligned} \varepsilon_r\cos\phi - \varepsilon_x\sin\phi &= 0 \\ \tan\phi &= \frac{\varepsilon_r}{\varepsilon_x} = \frac{5}{8.66} = 0.577 \Rightarrow \phi = 30^\circ \\ \cos\phi &= \frac{\varepsilon_x}{\varepsilon_z} = \frac{8.66}{10} = 0.866 \end{aligned}$$

Answer: (a) regulation = 9.20%; (b) zero regulation at pf = 0.866 leading.

Characteristics of an ideal transformer

1. Winding resistances are zero, so there is no copper loss.
2. There is no leakage flux; all the flux links both windings (coupling coefficient k = 1).
3. The core has infinite permeability, so no magnetising current is needed.
4. There is no core loss (no hysteresis or eddy current loss).
5. Hence efficiency is 100%, voltage regulation is zero, and

$$\frac{V_1}{V_2} = \frac{N_1}{N_2} = \frac{I_2}{I_1}, \qquad Z_1 = \left(\frac{N_1}{N_2}\right)^2 Z_2$$

Coupled circuit model of a transformer

A real (linear) transformer is treated as two magnetically coupled coils with self inductances $L_1$, $L_2$, mutual inductance $M$, and winding resistances $R_1$, $R_2$.

   i1    R1            M             R2    i2
  o-->--/\/\--+    .        .    +--/\/\--<--o
              |   (L1)    (L2)   |
  v1          )    )      (      (         v2
              |                  |
  o-----------+                  +-----------o

The flux linking each coil has a self and a mutual part:

$$\lambda_1 = L_1 i_1 + M i_2, \qquad \lambda_2 = M i_1 + L_2 i_2$$

Voltage equations (with both currents entering the dotted terminals):

$$\begin{aligned} v_1 &= R_1 i_1 + L_1\frac{di_1}{dt} + M\frac{di_2}{dt} \\ v_2 &= R_2 i_2 + L_2\frac{di_2}{dt} + M\frac{di_1}{dt} \end{aligned}$$

In sinusoidal steady state: $V_1 = (R_1 + j\omega L_1)I_1 + j\omega M I_2$ and $V_2 = j\omega M I_1 + (R_2 + j\omega L_2) I_2$.

Link with leakage and magnetising inductances

With $a = N_1/N_2$, the self inductances are split into leakage and magnetising parts:

$$\begin{aligned} L_1 &= L_{l1} + aM, \qquad L_2 = L_{l2} + \frac{M}{a}, \qquad L_m = aM \\ M &= k\sqrt{L_1 L_2}, \quad k \le 1 \end{aligned}$$

Substituting gives the familiar T-equivalent circuit: $R_1$ and $X_{l1} = \omega L_{l1}$ in series, the magnetising reactance $X_m = \omega aM$ in shunt, and $R_2' = a^2R_2$, $X_{l2}' = a^2 X_{l2}$ followed by an ideal transformer of ratio $a$.

  R1   Xl1           a^2 R2  a^2 Xl2      ideal a:1
 o-/\/-mmm--+--------/\/\----mmm-----+--||--o
            |                        |  ||
           Xm=w aM                   |  ||  V2
            |                        |  ||
 o----------+------------------------+--||--o

When $k \to 1$, $R \to 0$ and $L_1, L_2 \to \infty$ with $L_1/L_2 = a^2$, the coupled-circuit model reduces to the ideal transformer. Core loss is added later as a shunt resistance $R_c$ across $X_m$.

Maximum efficiency occurs when copper loss equals iron loss. Use this with the full-load efficiency to find both losses.

Losses at full load

$$\begin{aligned} P_{out,fl} &= 200\times0.8 = 160\ \text{kW} \\ \text{Total losses} &= \frac{160}{0.98} - 160 = 3.2653\ \text{kW} \\ P_i + P_{cu,fl} &= 3.2653 \end{aligned}$$

Condition for maximum efficiency at 3/4 load

$$\begin{aligned} P_i &= \left(\tfrac34\right)^2 P_{cu,fl} = 0.5625\,P_{cu,fl} \\ 1.5625\,P_{cu,fl} &= 3.2653 \Rightarrow P_{cu,fl} = 2.0898\ \text{kW} \\ P_i &= 3.2653 - 2.0898 = 1.1755\ \text{kW} \end{aligned}$$

Efficiency at half load

$$\begin{aligned} P_{out} &= 0.5\times160 = 80\ \text{kW} \\ P_{cu} &= (0.5)^2\times2.0898 = 0.5224\ \text{kW} \\ \eta &= \frac{80}{80 + 1.1755 + 0.5224} = \frac{80}{81.698} = 97.92\% \end{aligned}$$

Answer: efficiency at half load ≈ 97.92%.

In a Δ–Y bank each HV winding gets the full 2400 V line voltage, and each LV winding gives 120 V phase, so LV line voltage is √3 × phase voltage. The SC test was done on the HV side (LV shorted), so it gives impedance referred to HV. Turns ratio per transformer $a = 2400/120 = 20$.

Equivalent impedance (HV side, per transformer)

$$\begin{aligned} Z_{eq} &= \frac{53.4}{41.7} = 1.2806\ \Omega \\ R_{eq} &= \frac{850}{41.7^2} = 0.4888\ \Omega \\ X_{eq} &= \sqrt{1.2806^2 - 0.4888^2} = 1.1836\ \Omega \end{aligned}$$

Rated HV winding current $I = 100000/2400 = 41.67$ A.

LV voltage at rated current, 0.8 pf lag

Using $V_1 = V_2' + I(R_{eq}\cos\phi + X_{eq}\sin\phi)$ in phasor form with $V_1 = 2400$ V:

$$\begin{aligned} I(R\cos\phi + X\sin\phi) &= 41.67(0.4888\times0.8 + 1.1836\times0.6) = 45.88\ \text{V} \\ I(X\cos\phi - R\sin\phi) &= 41.67(1.1836\times0.8 - 0.4888\times0.6) = 27.23\ \text{V} \\ V_2' &= \sqrt{2400^2 - 27.23^2} - 45.88 = 2353.96\ \text{V} \end{aligned}$$

Referred to LV and changed to line value:

$$\begin{aligned} V_{2,ph} &= \frac{2353.96}{20} = 117.70\ \text{V} \\ V_{2,LL} &= \sqrt3\times117.70 = 203.86\ \text{V} \end{aligned}$$

(Regulation = (2400 − 2353.96)/2400 = 1.92%.)

Currents at rated load

Answer: LV line-to-line voltage ≈ 203.9 V; HV winding current 41.67 A, LV winding current 833.3 A.

No-load operation of a single-phase transformer

With the secondary open ($I_2 = 0$), the primary draws a small no-load current $I_0$ (2–5% of rated) from the supply. It does two jobs, so it has two components:

• Magnetising component $I_m = I_0\sin\phi_0$: in phase with the flux, lags $V_1$ by 90°; it sets up the core flux.
• Core-loss (working) component $I_c = I_0\cos\phi_0$: in phase with $V_1$; it supplies hysteresis and eddy losses (and a very small $I_0^2 R_1$).

$$I_0 = \sqrt{I_m^2 + I_c^2}, \qquad P_0 = V_1 I_0\cos\phi_0 \approx \text{core loss}$$

        V1
        ^
        |    I0
        |   /
     Ic |  /  phi0 (about 75-85 deg)
        | /
        +--------> phi, Im
        |
        v E1, E2

The no-load power factor is low (0.1–0.3 lag). The shunt branch of the equivalent circuit is $R_0 = V_1/I_c$ and $X_0 = V_1/I_m$.

Properties of an ideal transformer

1. Zero winding resistance (no $I^2R$ loss).
2. No leakage flux; all flux links both windings.
3. Core of infinite permeability, so negligible magnetising current.
4. No hysteresis or eddy current loss.
5. 100% efficiency, zero regulation, and $V_1/V_2 = N_1/N_2 = I_2/I_1$.

The Scott (T) connection uses two single-phase transformers to convert a balanced three-phase supply into a balanced two-phase supply (two voltages equal in magnitude and 90° apart), or the reverse. It is used for two-phase loads such as electric furnaces and for traction supply.

Connections

• Main transformer: primary of $N_1$ turns connected across two lines B and C, with a centre tap D.
• Teaser transformer: primary of $\frac{\sqrt3}{2}N_1$ turns (86.6%), connected between line A and the centre tap D.
• Both secondaries have the same number of turns $N_2$ and feed the two phases separately.

            A
            |
            |  Teaser primary
            |  (0.866 N1 turns)
            |
   B -------D------- C
     N1/2       N1/2
   Main primary (N1 turns,
   centre tapped at D)

  Secondaries (N2 turns each):
   Teaser sec -> V_T (phase 1)
   Main sec   -> V_M (phase 2)

Voltage relations (phasor diagram)

Let the line voltage be $V_L$.

              A
             /|\
            / | \
           /  |  \   VAD = 0.866 VL
          /   |   \  (perpendicular
         /    |    \   to BC)
        B-----D-----C
          VL/2  VL/2

• Voltage across the main primary: $V_{BC} = V_L$.
• D is the midpoint of BC, so $V_{AD}$ is the height of the equilateral triangle:

$$V_{AD} = \sqrt{V_L^2 - \left(\frac{V_L}{2}\right)^2} = \frac{\sqrt3}{2}V_L = 0.866\,V_L$$

and $V_{AD}$ is at 90° to $V_{BC}$.

• Volts per turn on the teaser: $\frac{0.866V_L}{0.866N_1} = \frac{V_L}{N_1}$, the same as on the main. Therefore the secondary voltages are

$$V_{M} = \frac{N_2}{N_1}V_L, \qquad V_{T} = \frac{N_2}{0.866N_1}(0.866V_L) = \frac{N_2}{N_1}V_L$$

They are equal in magnitude and 90° apart — a balanced two-phase supply.

Current relations (balanced two-phase load)

• Teaser primary current $I_A = \frac{2}{\sqrt3}\frac{N_2}{N_1}I_2 = 1.155\,\frac{N_2}{N_1} I_2$, flowing in line A.
• It divides equally at D into the two halves of the main primary, where it combines with the main load current; the resulting line currents $I_B$ and $I_C$ are equal in size to $I_A$ and 120° apart, so the three-phase side sees a balanced load.

Points to note

• If the neutral of the three-phase side is wanted, it is at the point on the teaser one-third of its length from D (i.e. at $\frac{N_1}{2\sqrt3}$ turns from D).
• The two transformers can be identical units if each has 50% and 86.6% tappings, so they are interchangeable.

OC test on LV gives the shunt branch directly on LV. SC test on HV gives series values on HV, which are referred to LV by $a^2 = (220/400)^2 = 0.3025$. Rated HV current $= 5000/400 = 12.5$ A, so the SC reading at 11.4 A must be scaled to full load.

i) Equivalent circuit parameters referred to LV

OC test:

$$\begin{aligned} \cos\phi_0 &= \frac{100}{220\times2} = 0.2273 \\ I_c &= 2\times0.2273 = 0.4545\ \text{A}, \quad I_m = 2\times0.9738 = 1.9477\ \text{A} \\ R_0 &= \frac{220}{0.4545} = 484.0\ \Omega, \quad X_0 = \frac{220}{1.9477} = 112.96\ \Omega \end{aligned}$$

SC test (HV side):

$$\begin{aligned} R_{eq,HV} &= \frac{200}{11.4^2} = 1.539\ \Omega, \quad Z_{eq,HV} = \frac{40}{11.4} = 3.509\ \Omega \\ X_{eq,HV} &= \sqrt{3.509^2 - 1.539^2} = 3.153\ \Omega \end{aligned}$$

Referred to LV:

$$R_{eq,LV} = 1.539\times0.3025 = 0.4655\ \Omega, \quad X_{eq,LV} = 3.153\times0.3025 = 0.9539\ \Omega$$

ii) Efficiency at full load, 0.9 pf lag

$$\begin{aligned} P_{cu,fl} &= 200\left(\frac{12.5}{11.4}\right)^2 = 240.46\ \text{W} \\ P_{out} &= 5000\times0.9 = 4500\ \text{W} \\ \eta &= \frac{4500}{4500 + 100 + 240.46} = 92.97\% \end{aligned}$$

Approximate regulation (on HV side, $\sin\phi = 0.4359$)

$$\begin{aligned} \%\text{Reg} &= \frac{I(R\cos\phi + X\sin\phi)}{V}\times100 \\ &= \frac{12.5(1.539\times0.9 + 3.153\times0.4359)}{400}\times100 \\ &= \frac{17.31 + 17.18}{400}\times100 = 8.62\% \end{aligned}$$

Answer: $R_0 = 484\ \Omega$, $X_0 = 112.96\ \Omega$, $R_{eq} = 0.466\ \Omega$, $X_{eq} = 0.954\ \Omega$ (all LV side); efficiency = 92.97%; regulation ≈ 8.62%.

Reading of the data: the load is 5 MW in total (balanced, three-phase) at 0.8 pf lag, supplied at 11 kV line voltage from the delta secondary; the 132 kV primary is star-connected. Winding R and X are per phase. Copper loss is found from winding currents; the voltage drop is neglected when finding currents.

Secondary currents (delta, 11 kV)

$$\begin{aligned} I_{L2} &= \frac{P}{\sqrt3 V_L\cos\phi} = \frac{5\times10^6}{\sqrt3\times11000\times0.8} = 328.04\ \text{A} \\ I_{ph2} &= \frac{I_{L2}}{\sqrt3} = \frac{328.04}{\sqrt3} = 189.39\ \text{A} \end{aligned}$$

Primary current (star, 132 kV)

Phase voltages: primary $132/\sqrt3 = 76.21$ kV, secondary 11 kV, so the turns ratio per phase is $76.21/11 = 6.928$.

$$I_{L1} = I_{ph1} = \frac{189.39}{6.928} = 27.34\ \text{A}$$

Check: $5\times10^6/(\sqrt3\times132000\times0.8) = 27.34$ A.

Losses

$$\begin{aligned} P_{cu1} &= 3 I_{ph1}^2 R_1 = 3\times27.34^2\times10 = 22419\ \text{W} \\ P_{cu2} &= 3 I_{ph2}^2 R_2 = 3\times189.39^2\times0.02 = 2152\ \text{W} \\ P_{cu} &= 22419 + 2152 = 24571\ \text{W} \\ P_{loss} &= 24571 + 20000 = 44571\ \text{W} \end{aligned}$$

(The leakage reactances cause voltage drop but no loss, so they do not enter the efficiency.)

Efficiency

$$\eta = \frac{5000}{5000 + 44.571}\times100 = 99.12\%$$

Answer: secondary phase current = 189.4 A, secondary line current = 328.0 A, primary line current = 27.34 A, efficiency = 99.12%.

Method. The no-load (OC) test gives the shunt branch: $\cos\phi_0 = \frac{P_0}{V_0 I_0}$, $I_w = I_0\cos\phi_0$, $I_\mu = I_0\sin\phi_0$, $R_0 = \frac{V_0}{I_w}$, $X_0 = \frac{V_0}{I_\mu}$. The SC test gives the series branch: $Z = \frac{V_{sc}}{I_{sc}}$, $R = \frac{P_{sc}}{I_{sc}^2}$, $X = \sqrt{Z^2 - R^2}$. Values found on one side are moved to the other side by multiplying by the square of the voltage ratio.

Ratio: HV = 400 V, LV = 120 V, so to refer an LV value to HV multiply by $\left(\frac{400}{120}\right)^2 = 11.11$.

From OC test (on LV side): 120 V, 4 A, 75 W

$$\begin{aligned} \cos\phi_0 &= \frac{75}{120 \times 4} = 0.1563, \quad \sin\phi_0 = 0.9877 \\ I_w &= 4 \times 0.1563 = 0.625\ \text{A}, \quad I_\mu = 4 \times 0.9877 = 3.951\ \text{A} \\ R_0 (\text{LV}) &= \frac{120}{0.625} = 192\ \Omega, \quad X_0 (\text{LV}) = \frac{120}{3.951} = 30.37\ \Omega \end{aligned}$$

Referred to HV:

$$\begin{aligned} R_0 (\text{HV}) &= 192 \times 11.11 = 2133.3\ \Omega \\ X_0 (\text{HV}) &= 30.37 \times 11.11 = 337.5\ \Omega \end{aligned}$$

From SC test (on HV side): 9.5 V, 20 A, 110 W

The test is on the HV side, so the results are already referred to HV. (Rated HV current = 8000/400 = 20 A, so this is a full-load test.)

$$\begin{aligned} Z_{01} &= \frac{9.5}{20} = 0.475\ \Omega \\ R_{01} &= \frac{110}{20^2} = 0.275\ \Omega \\ X_{01} &= \sqrt{0.475^2 - 0.275^2} = 0.3873\ \Omega \end{aligned}$$

Equivalent circuit referred to HV side

 I1 ->        I0                 I2'
 o-----+--------+---[ R01 ]---[ jX01 ]---o
 +     |        |                        +
     [ R0 ]  [ jX0 ]                   load
 V1    |        |                       V2'
 -     |        |                        -
 o-----+--------+------------------------o
 All values referred to HV side:
 R0 = 2133.3 ohm, X0 = 337.5 ohm
 R01 = 0.275 ohm, X01 = 0.387 ohm

Answer: $R_0 = 2133.3\ \Omega$, $X_0 = 337.5\ \Omega$, $R_{01} = 0.275\ \Omega$, $X_{01} = 0.387\ \Omega$, $Z_{01} = 0.475\ \Omega$ (all referred to HV).

Regulation is worked out per phase. The primary is delta (phase voltage = line voltage) and the secondary is star (phase voltage = line voltage/$\sqrt{3}$).

Step 1: Per-phase voltages and turns ratio

$$\begin{aligned} V_{1ph} &= 33000\ \text{V}, \quad V_{2ph} = \frac{6600}{\sqrt{3}} = 3810.5\ \text{V} \\ K &= \frac{V_{2ph}}{V_{1ph}} = \frac{3810.5}{33000} = 0.11547 \end{aligned}$$

Step 2: Equivalent resistance referred to primary (per phase)

$$\begin{aligned} R_2' &= \frac{R_2}{K^2} = \frac{0.08}{0.11547^2} = 6.0\ \Omega \\ R_{01} &= R_1 + R_2' = 8 + 6 = 14\ \Omega \end{aligned}$$

Step 3: Percentage resistance and reactance

Full-load primary phase current:

$$I_{1ph} = \frac{2 \times 10^6}{3 \times 33000} = 20.2\ \text{A}$$ $$\begin{aligned} \%R &= \frac{I_{1ph} R_{01}}{V_{1ph}} \times 100 = \frac{20.2 \times 14}{33000} \times 100 = 0.857\% \\ \%Z &= 7\% \ (\text{given}) \\ \%X &= \sqrt{7^2 - 0.857^2} = 6.947\% \end{aligned}$$

Step 4: Regulation at 0.75 pf lagging

$\cos\phi = 0.75$, $\sin\phi = \sqrt{1 - 0.75^2} = 0.6614$.

$$\begin{aligned} \%\text{Reg} &\approx \%R\cos\phi + \%X\sin\phi \\ &= 0.857 \times 0.75 + 6.947 \times 0.6614 \\ &= 0.643 + 4.595 = 5.24\% \end{aligned}$$

If the more exact formula is used, adding the second-order term $\frac{(\%X\cos\phi - \%R\sin\phi)^2}{200}$ gives about 5.35%.

Answer: Full-load voltage regulation at 0.75 pf lagging is about 5.24% (approximate formula).