Transformer

Losses in a transformer

A transformer has no moving parts, so there are only two kinds of loss.

1. Core (iron) losses $P_i$: present whenever the primary is energised at rated voltage and frequency. They are practically constant from no load to full load.
   • Hysteresis loss: $P_h = \eta B_{max}^{1.6} f V$, from the repeated reversal of magnetisation. It is reduced by using silicon steel.
   • Eddy current loss: $P_e \propto B_{max}^2 f^2 t^2$, from currents induced in the core. It is reduced by thin insulated laminations.
2. Copper losses $P_{cu}$: $I_1^2R_1 + I_2^2R_2 = I_2^2R_{02}$ in the windings. They vary as the square of the load current.

Stray (leakage flux) and dielectric losses are small and usually ignored.

Condition for maximum efficiency

Let the load current be $I_2$, secondary voltage $V_2$ (assumed constant), power factor $\cos\phi_2$ and $R_{02}$ the total resistance referred to the secondary.

$$\eta = \frac{V_2I_2\cos\phi_2}{V_2I_2\cos\phi_2 + P_i + I_2^2R_{02}}$$

Divide numerator and denominator by $I_2$:

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

For constant $V_2$ and $\cos\phi_2$, $\eta$ is maximum when the denominator is minimum:

$$\frac{d}{dI_2}\left(\frac{P_i}{I_2} + I_2R_{02}\right) = 0 \Rightarrow -\frac{P_i}{I_2^2} + R_{02} = 0$$ $$P_i = I_2^2R_{02}$$

Condition: efficiency is maximum when the variable (copper) loss equals the constant (iron) loss.

Load at maximum efficiency, with $P_{cu,fl}$ the full-load copper loss:

$$x = \frac{I_2}{I_{2,fl}} = \sqrt{\frac{P_i}{P_{cu,fl}}},\qquad \text{kVA at max } \eta = \text{rated kVA}\times\sqrt{\frac{P_i}{P_{cu,fl}}}$$

The second derivative is positive, so this is a minimum of the denominator and a maximum of $\eta$. The condition does not depend on the power factor, but the value of the maximum efficiency is higher at a higher power factor.

An ideal transformer has no winding resistance, no leakage flux, no core loss and infinite core permeability, so it needs no magnetising current.

No-load operation

The secondary is open. The primary is connected to a sinusoidal supply $V_1$.

  I1=0      +------+      I2=0
 o--->---+  |||||||  +--- o
  V1     ) core     (  E2   open
 o-------+  |||||||  +--- o
         N1         N2

• The primary current is only the tiny magnetising current $I_\mu$ which sets up the core flux $\phi$ (taken as zero for an ideal one).
• The flux induces $E_1 = -N_1\,d\phi/dt$ in the primary. For an ideal transformer $E_1 = V_1$ (equal and opposite).
• The same flux links the secondary, so $E_2 = V_2 = -N_2\,d\phi/dt$ and $\dfrac{V_1}{V_2} = \dfrac{N_1}{N_2}$.

Loaded operation

A load is connected on the secondary and a current $I_2$ flows.

• $I_2$ sets up a secondary mmf $N_2I_2$ which tends to oppose (by Lenz's law) the core flux.
• The primary draws an extra current $I_1'$ so that its mmf $N_1I_1'$ cancels the secondary mmf:

$$N_1I_1' = N_2I_2 \Rightarrow \frac{I_1}{I_2} = \frac{N_2}{N_1}$$

• The input power equals the output power, $V_1I_1 = V_2I_2$.

Proof that the net flux remains constant

The primary equation gives $V_1 \approx E_1 = 4.44fN_1\phi_m$. Since $V_1$ and $f$ are fixed by the supply, $\phi_m = V_1/(4.44fN_1)$ is fixed, whatever the load.

Also, from the mmf balance, the net mmf acting on the core on load is:

$$\begin{aligned} F_{net} &= N_1I_1 - N_2I_2 \\ &= N_1(I_\mu + I_1') - N_2I_2 \\ &= N_1I_\mu + (N_1I_1' - N_2I_2) = N_1I_\mu \end{aligned}$$

The net mmf is $N_1I_\mu$, the same as at no load. The core flux is $\phi = N_1I_\mu/S$, so the flux is constant. Any load only adds the balancing component $I_1'$ in the primary, which cancels the secondary mmf and leaves the flux unchanged.

Efficiency is the ratio of output power to input power for a transformer:

$$\eta = \frac{\text{output}}{\text{input}} = \frac{\text{output}}{\text{output} + P_i + P_{cu}}$$

where $P_i$ is the core loss (constant) and $P_{cu} = I_2^2R_{02}$ is the copper loss (varies as the square of the load).

For load current $I_2$, secondary voltage $V_2$ and power factor $\cos\phi_2$:

$$\eta = \frac{V_2I_2\cos\phi_2}{V_2I_2\cos\phi_2 + P_i + I_2^2R_{02}}$$

Condition for maximum efficiency

Dividing by $I_2$ gives

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

With $V_2$ and $\cos\phi_2$ constant, $\eta$ is maximum when the denominator is minimum:

$$\frac{d}{dI_2}\left(\frac{P_i}{I_2} + I_2R_{02}\right) = -\frac{P_i}{I_2^2} + R_{02} = 0$$ $$P_i = I_2^2R_{02}\quad\Rightarrow\quad \text{iron loss} = \text{copper loss}$$

Current at maximum efficiency

$$I_{2,\eta max} = \sqrt{\frac{P_i}{R_{02}}}$$

In terms of the full-load current $I_{2,fl}$ and full-load copper loss $P_{cu,fl} = I_{2,fl}^2R_{02}$:

$$I_{2,\eta max} = I_{2,fl}\sqrt{\frac{P_i}{P_{cu,fl}}}$$

The corresponding kVA load is $\text{kVA}_{\eta max} = \text{kVA}_{rated}\sqrt{P_i/P_{cu,fl}}$ and the maximum efficiency is

$$\eta_{max} = \frac{V_2I_2\cos\phi_2}{V_2I_2\cos\phi_2 + 2P_i}$$

Distribution transformers (on load all day) are designed with a small $P_i$ so that maximum efficiency occurs near the usual load, which is below full load.

An ideal transformer has no resistance, no leakage flux, no core loss, and a core of infinite permeability.

(i) No-load operation

   I1 ~ 0     N1 : N2     I2 = 0
 o---->--+   ||     ||   +---o
         )   ||     ||   (        
  V1     )   ||core ||   (  V2 (open)
 o-------+   ||     ||   +---o

The secondary is open, so $I_2 = 0$. A sinusoidal voltage $V_1$ on the primary draws a very small magnetising current $I_\mu$, which creates the alternating flux $\phi = \phi_m\sin\omega t$ in the core. This flux links both windings and induces:

$$e_1 = -N_1\frac{d\phi}{dt},\qquad e_2 = -N_2\frac{d\phi}{dt}$$

In an ideal transformer $V_1 = E_1$ and $V_2 = E_2$. Dividing,

$$\frac{V_1}{V_2} = \frac{E_1}{E_2} = \frac{N_1}{N_2} = a$$

where $a$ is the turns ratio.

(ii) Loaded operation

When a load is connected, the secondary current $I_2$ flows and its mmf $N_2I_2$ opposes the flux. The flux would fall, so the primary draws an extra current $I_1$ to restore it, until the two mmfs balance (neglecting $I_\mu$):

$$N_1I_1 = N_2I_2 \Rightarrow \frac{I_1}{I_2} = \frac{N_2}{N_1} = \frac{1}{a}$$

Since there are no losses, the power in equals the power out:

$$V_1I_1 = V_2I_2 \Rightarrow \frac{V_1}{V_2} = \frac{I_2}{I_1}$$

Results

$$\boxed{\frac{V_1}{V_2} = \frac{E_1}{E_2} = \frac{N_1}{N_2} = \frac{I_2}{I_1}}$$

The voltage ratio equals the turns ratio, and the current ratio is its inverse. In phasor terms, $V_1$ leads the flux by 90° (in the sign convention $V_1 = -E_1$), and on load $I_1$ and $V_1$ have the same phase angle $\phi$ as $I_2$ and $V_2$.

Given: 20 kVA, 250 V/2500 V. Primary = LV (250 V), secondary = HV (2500 V), $K = 10$, $K^2 = 100$.

Open-circuit test (on the LV side)

The no-load power is almost entirely core loss, so the shunt branch is found from it.

$$\begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{105}{250\times1.4} = 0.3 \\ I_w &= I_0\cos\phi_0 = 0.42\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 1.336\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 595.2\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 187.2\ \Omega\quad(\text{on the LV side}) \end{aligned}$$

Short-circuit test (on the HV side)

At the reduced test voltage the core loss is negligible, so the wattmeter reading is the copper loss.

$$\begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{120}{8} = 15\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{320}{8^2} = 5\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 14.14\ \Omega\quad(\text{on the HV side}) \end{aligned}$$

Referring to the LV side

The turns ratio is $K = 2500/250 = 10$, so $K^2 = 100$. Impedances are multiplied by $K^2$ when moved from LV to HV, and divided by $K^2$ when moved from HV to LV.

$$\begin{aligned} R_{eq} &= 5 \div 100 = 0.05\ \Omega \\ X_{eq} &= 14.14 \div 100 = 0.1414\ \Omega \end{aligned}$$

Equivalent circuit referred to the LV side

   Req=0.05 ohm   Xeq=0.141 ohm
 o--+-----/\/\/\-----UUUU------o
    |                          
   [R0=595.2] [X0=187.2]   Load
    |      (in parallel)
 o--+---------------------------o

$R_{eq} = 0.05\ \Omega$, $X_{eq} = 0.1414\ \Omega$ ($Z_{eq} = 0.15\ \Omega$), shunt branch $R_0 = 595.2\ \Omega$ in parallel with $X_0 = 187.2\ \Omega$.

Answer (referred to primary): $R_0 = 595.2\ \Omega$, $X_0 = 187.2\ \Omega$, $R_{01} = 0.05\ \Omega$, $X_{01} = 0.1414\ \Omega$.

Given: 25 kVA, 250 V/2500 V. Primary = LV (250 V), secondary = HV (2500 V), $K = 10$, $K^2 = 100$.

Open-circuit test (on the LV side)

The no-load power is almost entirely core loss, so the shunt branch is found from it.

$$\begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{105}{250\times1.4} = 0.3 \\ I_w &= I_0\cos\phi_0 = 0.42\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 1.336\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 595.2\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 187.2\ \Omega\quad(\text{on the LV side}) \end{aligned}$$

Short-circuit test (on the HV side)

At the reduced test voltage the core loss is negligible, so the wattmeter reading is the copper loss.

$$\begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{120}{8} = 15\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{320}{8^2} = 5\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 14.14\ \Omega\quad(\text{on the HV side}) \end{aligned}$$

Referring to the LV side

The turns ratio is $K = 2500/250 = 10$, so $K^2 = 100$. Impedances are multiplied by $K^2$ when moved from LV to HV, and divided by $K^2$ when moved from HV to LV.

$$\begin{aligned} R_{eq} &= 5 \div 100 = 0.05\ \Omega \\ X_{eq} &= 14.14 \div 100 = 0.1414\ \Omega \end{aligned}$$

Equivalent circuit referred to the LV side

   Req=0.05 ohm   Xeq=0.141 ohm
 o--+-----/\/\/\-----UUUU------o
    |                          
   [R0=595.2] [X0=187.2]   Load
    |      (in parallel)
 o--+---------------------------o

$R_{eq} = 0.05\ \Omega$, $X_{eq} = 0.1414\ \Omega$ ($Z_{eq} = 0.15\ \Omega$), shunt branch $R_0 = 595.2\ \Omega$ in parallel with $X_0 = 187.2\ \Omega$.

Answer (referred to primary): $R_0 = 595.2\ \Omega$, $X_0 = 187.2\ \Omega$, $R_{01} = 0.05\ \Omega$, $X_{01} = 0.1414\ \Omega$.

Note: the 25 kVA rating is not needed for the parameters. The SC test current of 8 A is the rated HV current only for a 20 kVA unit (a 25 kVA unit has 10 A on the HV side), so the test is below full load. The equivalent circuit parameters are the same, because they do not depend on the test current if the transformer is linear. The copper loss at the true full load would be $320\times(10/8)^2 = 500$ W.

Losses in a transformer

The first two together are the iron (core) loss $P_i$. Stray load loss and dielectric loss are small and neglected. Because there is no rotating part, there is no friction or windage loss.

Expression for efficiency

$$\eta = \frac{\text{output power}}{\text{input power}} = \frac{\text{output}}{\text{output} + \text{losses}}$$

For a load that takes a fraction $x$ of the full-load current, at power factor $\cos\phi$ and rated secondary voltage $V_2$:

$$\begin{aligned} \text{Output} &= x\,S\cos\phi\quad (S = V_2I_{2,fl}\ \text{rated VA}) \\ P_{cu} &= x^2P_{cu,fl} \\ \text{Losses} &= P_i + x^2P_{cu,fl} \end{aligned}$$ $$\boxed{\eta = \frac{xS\cos\phi}{xS\cos\phi + P_i + x^2P_{cu,fl}}\times100\%}$$

Remarks:

• Efficiency is zero at no load, rises with load, reaches a maximum when $P_i = x^2P_{cu,fl}$, and then falls.
• For a good power transformer $\eta$ is 95 to 99%.
• The efficiency is higher at a higher power factor, since the output power is larger for the same losses.

The core flux of a transformer is fixed by the primary applied voltage, not by the load. Therefore it stays practically constant from no load to full load.

Reason 1: voltage balance in the primary

$$V_1 = E_1 + I_1Z_1 \approx E_1 = 4.44fN_1\phi_m$$

The primary impedance drop $I_1Z_1$ is only 1 to 3% of $V_1$ at full load. Since the supply $V_1$ and $f$ are constant, $\phi_m = \dfrac{V_1}{4.44fN_1}$ stays practically constant.

Reason 2: mmf balance

When a load current $I_2$ flows, its mmf $N_2I_2$ tends to reduce the flux. The primary current immediately increases by $I_1'$, so that

$$N_1I_1' = N_2I_2$$

The net mmf in the core is then again $N_1I_\mu$, the no-load value, so the flux does not change.

Conclusion: the only change in flux comes from the small primary leakage drop, which is about 1 to 3% at full load. Because the flux is constant, the core (iron) loss is constant at all loads, and the no-load core loss can be taken as the core loss at full load.

Operating principle

A transformer works on mutual induction. Two windings (primary of $N_1$ turns, secondary of $N_2$ turns) are wound on a common laminated core. An alternating voltage on the primary sends a current that sets up an alternating flux $\phi$ in the core. This flux links the secondary and, by Faraday's law, induces an emf in it. Energy is transferred from primary to secondary at the same frequency, with voltage and current changed in the ratio of turns.

An ideal transformer has no winding resistance, no leakage flux, no core loss, and a core of infinite permeability.

        core (laminated)
     +--------------------+
  ~  |  N1            N2  |  load
 V1  |  primary   secondary|  Z
     +--------------------+
        <----- flux ----->

emf equation

Let the flux be $\phi = \phi_m\sin\omega t$, where $\phi_m$ is the maximum flux (Wb).

$$\begin{aligned} e_1 &= -N_1\frac{d\phi}{dt} = -N_1\omega\phi_m\cos\omega t \\ E_{m1} &= N_1\omega\phi_m = 2\pi fN_1\phi_m \end{aligned}$$

The rms value is

$$E_1 = \frac{E_{m1}}{\sqrt2} = \frac{2\pi}{\sqrt2}fN_1\phi_m = 4.44\,fN_1\phi_m$$

Similarly,

$$E_2 = 4.44\,fN_2\phi_m$$ $$\boxed{E_1 = 4.44fN_1\phi_m = 4.44fN_1B_mA\ \text{V},\qquad \frac{E_1}{E_2} = \frac{N_1}{N_2}}$$

where $B_m$ is the maximum flux density and $A$ the core cross-section. The induced emf lags the flux by 90°. For the ideal case $V_1 = E_1$ and $V_2 = E_2$, hence $V_1/V_2 = N_1/N_2 = a$.

A practical transformer has winding resistances $R_1, R_2$, leakage reactances $X_1, X_2$ and a core that draws an exciting current $I_0$.

No-load operation

The secondary is open ($I_2 = 0$). The primary draws only $I_0$, which has two components:

$$I_0 = I_c + I_\mu,\qquad I_c = I_0\cos\phi_0\ (\text{core loss}),\qquad I_\mu = I_0\sin\phi_0\ (\text{magnetising})$$

              -E1 (= V1 approx)
                 ^
                 |       phi0 is about 75-80 deg
                 |
        Ic <-----+       Phasors with V1 as reference:
         \       |       Im lags V1 by 90 deg,
       I0 \      |       Ic is in phase with V1,
           \     |       flux phi is in phase with Im
            v    v
                Im

• The flux $\phi$ is in phase with $I_\mu$ and lags $V_1$ by about 90°.
• $E_1$ and $E_2$ lag $\phi$ by 90°.
• $I_0$ is small (2 to 5% of full-load current) and lags $V_1$ by $\phi_0$, so the no-load power factor is low (0.1 to 0.2).

On load

The secondary current $I_2$ flows, and the primary takes the additional load component $I_1' = I_2/a$, so $I_1 = I_0 + I_1'$ (phasor sum). The voltage equations are

$$V_1 = -E_1 + I_1(R_1 + jX_1),\qquad V_2 = E_2 - I_2(R_2 + jX_2)$$

Resistive load. $I_2$ is in phase with $V_2$. Draw $E_2$ as the sum of $V_2$, $I_2R_2$ (in phase with $I_2$) and $I_2X_2$ (90° ahead of $I_2$). So $E_2 > V_2$ and the voltage falls on load. $I_1'$ is in phase with $I_2$ reversed, and $I_1$ is close to $I_1'$. The primary power factor is nearly unity.

  Resistive load (phasors referred to secondary)

          E2
         /|
        / | I2X2
       /  |
      /___|______
      V2  I2R2    I2 along V2

Capacitive load. $I_2$ leads $V_2$ by $\phi_2$. The drops $I_2R_2$ and $I_2X_2$ are drawn along and 90° ahead of $I_2$. As $I_2$ is now ahead of $V_2$, the drop $I_2X_2$ points partly opposite to $V_2$ and $E_2$ becomes smaller than $V_2$. So the secondary terminal voltage rises on load (negative regulation) when the leading power factor is low enough.

  Capacitive load

       I2 (leads V2)
        \
         \  I2X2 almost against V2
   E2 ----V2 (longer than E2)

In each case, $I_1$ is obtained as $I_0 + I_1'$, and $V_1 = -E_1 + I_1(R_1 + jX_1)$ completes the primary diagram.

The transformer is treated as a practical one with winding resistances $R_1,R_2$ and leakage reactances $X_1,X_2$. For any load, with the turns ratio $a = N_1/N_2$:

$$V_2 = E_2 - I_2(R_2 + jX_2),\qquad I_1 = I_0 + \frac{I_2}{a},\qquad V_1 = -E_1 + I_1(R_1 + jX_1)$$

Circuit diagram

 I1   R1   X1        R2   X2   I2
 o--/\/\--UUU--+  +--/\/\--UUU--o
               |  |             |
   V1     E1 ( )  ( ) E2        Z (load)
               |  |             |
 o-------------+--+-------------o
        ideal transformer N1:N2

Phasor diagrams (drawn with $V_2$ as the reference)

For each load, add to $V_2$ the drop $I_2R_2$ (parallel to $I_2$) and then $I_2X_2$ (90° ahead of $I_2$) to obtain $E_2$.

 Inductive          Resistive         Capacitive
 (I2 lags V2)       (I2 = V2)         (I2 leads V2)

   E2 (big)           E2                I2 /
   /|                 /|                  /
  / | I2X2           / | I2X2        E2 /__ V2 (> E2)
 /__|                /__|
 V2 I2R2             V2 I2R2
 (I2 below V2)      (I2 along V2)

Primary side: in each case, the load component $I_1' = I_2/a$ has the same phase angle $\phi_2$ relative to $-E_1$ as $I_2$ has relative to $E_2$. It is added to the no-load current $I_0$ (which lags $-E_1$ by about 78°) to give $I_1$. Then $V_1$ is the phasor sum of $-E_1$, $I_1R_1$ and $jI_1X_1$.

Result: the inductive load gives the poorest regulation, the resistive load an intermediate value, and a capacitive load can give a voltage rise. The primary power factor follows the load: lagging for an inductive load, nearly unity for a resistive load, and leading for a strongly capacitive load.

Practical vs ideal transformer

Operation of a practical transformer on load

When load current $I_2$ flows, the secondary terminal voltage and primary current are given by

$$V_2 = E_2 - I_2(R_2 + jX_2),\qquad I_1 = I_0 + I_1',\ \ I_1' = \frac{N_2}{N_1}I_2,\qquad V_1 = -E_1 + I_1(R_1 + jX_1)$$

Phasor diagram (inductive load, lagging $\phi_2$):

          V1 (applied)
         /|
   I1X1/ |             Steps:
      /  | I1R1        1. Take flux phi as reference.
 -E1 /___|              2. E1, E2 lag phi by 90 deg.
     |                  3. I2 lags V2 by phi2.
     |  E2              4. E2 = V2 + I2R2 + jI2X2.
     |   \ I2X2         5. I1' = I2/a opposes the I2 mmf.
 phi |    \ I2R2        6. I1 = I0 + I1'.
     v     \V2          7. V1 = -E1 + I1R1 + jI1X1.

Steps in words:

1. Draw the flux $\phi$ as reference. The emfs $E_1$ and $E_2$ lag it by 90°, and $-E_1$ is drawn opposite to $E_1$.
2. Draw $V_2$, then $I_2$ lagging it by $\phi_2$. Add $I_2R_2$ (parallel to $I_2$) and $I_2X_2$ (90° ahead) to reach $E_2$.
3. The load component $I_1'$ is drawn opposite to $I_2$ (scaled by $N_2/N_1$). The no-load current $I_0$ is drawn leading the flux by the hysteresis angle, with $I_\mu$ along $\phi$ and $I_c$ along $-E_1$.
4. $I_1 = I_0 + I_1'$. Add $I_1R_1$ and $I_1X_1$ to $-E_1$ to get $V_1$.

The angle between $V_1$ and $I_1$ is the primary power factor angle $\phi_1$, slightly different from $\phi_2$ because of $I_0$ and the drops.

Voltage regulation is the change in secondary terminal voltage from no load to full load (at the same primary voltage and a given power factor), expressed as a fraction of the no-load voltage:

$$\text{VR} = \frac{E_2 - V_2}{E_2}\times100\%$$

where $E_2$ is the no-load secondary voltage and $V_2$ the full-load terminal voltage.

Derivation (lagging power factor)

Refer all quantities to the secondary side: $R_{02} = R_2 + R_1/a^2$ and $X_{02} = X_2 + X_1/a^2$. With the exciting branch ignored, the circuit is a source $E_2$ in series with $R_{02} + jX_{02}$, feeding the load current $I_2$ at angle $\phi$ lagging $V_2$.

 Phasor diagram (inductive load, V2 as reference)

                  E2
                 /|
         I2X02  / |
               /  |
          V2  /___|
   O--------->  I2R02
    \ phi
     \ I2 (lags V2)

Using $V_2$ as reference: $V_2 = V_2\angle0°$, $I_2 = I_2\angle-\phi$.

$$\begin{aligned} \bar E_2 &= \bar V_2 + \bar I_2(R_{02} + jX_{02}) \\ &= (V_2 + I_2R_{02}\cos\phi + I_2X_{02}\sin\phi) + j(I_2X_{02}\cos\phi - I_2R_{02}\sin\phi) \end{aligned}$$

The quadrature term is small, so

$$E_2 \approx V_2 + I_2R_{02}\cos\phi + I_2X_{02}\sin\phi$$ $$E_2 - V_2 = I_2(R_{02}\cos\phi + X_{02}\sin\phi)$$

Hence

$$\boxed{\text{VR} = \frac{I_2R_{02}\cos\phi + I_2X_{02}\sin\phi}{E_2}\times100\%}$$

For the exact value: $E_2 = \sqrt{(V_2 + I_2R_{02}\cos\phi + I_2X_{02}\sin\phi)^2 + (I_2X_{02}\cos\phi - I_2R_{02}\sin\phi)^2}$.

For a leading load, the sign of the $X_{02}\sin\phi$ term changes (VR $= I_2(R_{02}\cos\phi - X_{02}\sin\phi)/E_2$), and the regulation can become negative. The maximum regulation occurs when $\cos\phi = R_{02}/Z_{02}$.

Inrush current is the large transient magnetising current drawn by a transformer when it is first connected to the supply. It can reach 5 to 10 times the full-load current (and up to 20 times for large units) and dies away in a few cycles to a few seconds.

Doubling effect

If the supply voltage is $v = V_m\sin(\omega t + \theta)$, the flux needed in steady state is $\phi = -\phi_m\cos(\omega t + \theta)$, because flux lags voltage by 90°. At the instant of switching the flux cannot change suddenly, so it starts from the residual flux $\phi_r$ and a decaying transient flux is added to the steady-state value.

Case: switching on at the instant when the supply voltage is zero ($\omega t = 0$).

• In steady state, the flux at this instant would be at its negative maximum ($-\phi_m$).
• But the core flux just before switching is the residual flux $\phi_r$ (nearly zero, and it cannot change suddenly).
• So the flux starts from $\phi_r \approx 0$ and follows

$$\phi = \phi_r + \phi_m(1 - \cos\omega t)$$

• Half a cycle later ($\omega t = \pi$), the flux reaches

$$\phi_{max} = 2\phi_m + \phi_r$$

The peak flux is therefore about twice the normal peak, and with residual flux it can be more. This is the doubling effect.

 flux
  2phi_m |        .-.
         |      /     \
   phi_m |    /   steady-state  
         |  /  .-"-.
      0 -+-/-"------"-------> t
         voltage zero when switched

Consequence

• A normal transformer is designed to work at a flux density near the knee of the $B$–$H$ curve (about 1.5 to 1.7 T). Twice the flux drives the core deep into saturation, where permeability falls sharply.
• A very large magnetising current is needed to produce this flux, and this is the inrush current. It has a high harmonic content (mainly second harmonic) and decays with the winding resistance.

Case: switching on at voltage maximum gives no transient: the flux starts at zero and is already at its steady-state value, so the inrush is minimal.

Effects and remedies: inrush can trip overcurrent relays and stress the windings. It is reduced by switching with a series resistor, point-on-wave closing, and harmonic restraint in differential relays.

A three-phase transformer bank can be formed by three single-phase transformers or one three-phase unit. The primary and secondary windings can be connected in star (Y) or delta (Δ), giving four basic connections.

Let $V_L$ be the primary line voltage, and $a = N_1/N_2$ the turns ratio per phase.

1. Star–Star (Y–Y)

  A o--+--(N1)--+       a o--(N2)--+
  B o--+--(N1)--+ N     b o--(N2)--+ n
  C o--+--(N1)--+       c o--(N2)--+

• Line voltage ratio $V_{L1}/V_{L2} = a$. No phase shift between primary and secondary (0°).
• Needs a neutral connection, otherwise unbalanced loads and third harmonic voltages cause trouble.
• Applications: high-voltage, small-current transmission, where insulation per phase is lower; small distribution loads with neutral (4-wire).

2. Delta–Delta (Δ–Δ)

  Each primary phase winding between line terminals;
  same for the secondary (closed triangles).

• $V_{L1}/V_{L2} = a$. No phase shift (0°). Line current = $\sqrt3\times$ phase current.
• Third harmonics circulate in the delta and are suppressed. If one transformer fails, the other two can supply 58% of the load in open delta (V–V).
• Applications: large low and medium-voltage industrial loads where continuity and balanced loads are needed.

3. Star–Delta (Y–Δ)

• $V_{L1}/V_{L2} = \sqrt3\,a$, with a 30° phase shift (secondary lags primary by 30°).
• Stepping down: the star primary has a lower phase voltage so less insulation is required.
• Applications: step-down transformers at the receiving end of a transmission line, and generator transformers' LV side.

4. Delta–Star (Δ–Y)

• $V_{L2}/V_{L1} = \sqrt3/a$ (step-up), with a 30° shift. The star secondary gives a neutral.
• Applications: step-up transformers at the generating station and distribution transformers (11 kV/415 V, Dyn11), giving a 4-wire supply. The delta suppresses third harmonic currents.

Comparison

Other connections: open-delta (V–V) for 58% capacity, Scott connection for 3-phase to 2-phase conversion, and zig-zag for earthing.

An equivalent circuit replaces the magnetically coupled windings by an electric network that has the same terminal behaviour. Winding resistance and leakage reactance are put in series and the core is represented by a shunt branch.

Equivalent circuit (ideal transformer $N_1:N_2$ in the centre)

   R1     X1                R2      X2
 o-/\/\--UUU--+---+----+--/\/\--UUU--o
              |   |    |  ideal      
  V1        Rc   Xm    ) N1:N2 (       V2   Load
              |   |    |             
 o------------+---+----+-------------o

• $R_1, X_1$: primary resistance and leakage reactance. $R_2, X_2$: secondary values.
• $R_c$: core-loss resistance. $X_m$: magnetising reactance, with $I_c = V_1/R_c$ and $I_\mu = V_1/X_m$ (both form $I_0$).
• $a = N_1/N_2$.

Referring all parameters to the primary side

Energy in each element must remain the same after referring. Secondary quantities are referred to the primary as follows:

$$V_2' = aV_2,\qquad I_2' = \frac{I_2}{a}$$

For the secondary impedance, equating the $I^2Z$ drop (power and VAr) before and after:

$$I_2'^{\,2}R_2' = I_2^2R_2 \Rightarrow R_2' = R_2\left(\frac{I_2}{I_2'}\right)^2 = a^2R_2$$ $$X_2' = a^2X_2,\qquad Z_L' = a^2Z_L$$

The referred circuit has all the elements on the primary side:

   R1    X1     R2'     X2'
 o-/\/\--UUU--+--/\/\--UUU--+---o
              |            |
  V1        Rc   Xm       ZL' (= a^2 ZL)
              |            |
 o------------+------------+---o

Then the total referred values are

$$R_{01} = R_1 + a^2R_2,\qquad X_{01} = X_1 + a^2X_2,\qquad Z_{01} = \sqrt{R_{01}^2 + X_{01}^2}$$

The shunt branch ($R_c, X_m$) is often moved to the input terminals (approximate circuit) because $I_0$ is small. With this approximation the circuit is: $V_1$ across $R_c \parallel X_m$, followed by $R_{01} + jX_{01}$ in series with $Z_L'$.

Referring to the secondary side

Primary quantities are referred to the secondary by dividing by $a$ (voltage) and $a^2$ (impedance):

$$R_1' = \frac{R_1}{a^2},\quad X_1' = \frac{X_1}{a^2},\quad R_{02} = R_2 + \frac{R_1}{a^2},\quad X_{02} = X_2 + \frac{X_1}{a^2}$$

Similarly $R_c' = R_c/a^2$, $X_m' = X_m/a^2$, and the primary current becomes $aI_1$.

Rule: an impedance is multiplied by $a^2$ when moved from the secondary to the primary, and divided by $a^2$ when moved from the primary to the secondary, where $a = N_1/N_2$.

When the load is more than one transformer can carry, or for reliability and maintenance, two or more transformers are connected in parallel (primaries to the same supply, secondaries to the same bus). The following conditions must hold.

Essential conditions

1. Same voltage ratio (turns ratio). Reason: if the ratios differ, the secondary no-load emfs differ. A circulating current $I_c = \Delta E/(Z_{A} + Z_{B})$ flows between the secondaries even at no load, and heats the windings and wastes power. Since the impedance is small, even a 1% difference gives a large current.

2. Same polarity (terminals connected correctly). Reason: with wrong polarity the two secondary emfs add in the loop and act as a short circuit, causing a very large current that can burn the windings.

3. Same phase sequence and zero relative phase displacement (three-phase units). Reason: the secondary line voltages must coincide at every instant. Otherwise there is a voltage difference between the two bus-bar connections, giving large circulating currents. Transformers with the same vector group (for example Dyn11 with Dyn11) can be paralleled; a group with a 30° shift cannot be paralleled with 0°.

4. Equal per-unit (percentage) impedance on their own kVA bases. Reason: load is shared in inverse proportion to the per-unit impedance. If the per-unit impedances differ, the one with the lower value is overloaded while the other is lightly loaded, so the combined rating is not fully used.

5. Same ratio of resistance to reactance ($R/X$). Reason: if the $X/R$ ratios differ, the two load currents are out of phase with each other. Each transformer then works at a different power factor, and the total current is less than the arithmetic sum, so the combined capacity is reduced.

Desirable but not essential

• Same frequency (always satisfied if on the same supply).
• Kva ratings that are not too different (ratio less than 3:1), since the impedance requirement is harder to meet.

In short: equal ratio, correct polarity, same phase relation, and equal per-unit impedance with the same $X/R$ ratio.

Parallel operation requires that the secondary line voltages of the two banks be exactly equal in magnitude and in phase. A star–delta transformer does not meet this condition with a star–star or a delta–delta bank.

Reason: phase displacement

• In a star–star or delta–delta transformer, the secondary line voltage is in phase with the primary line voltage (0° shift).
• In a star–delta (or delta–star) transformer, the secondary line voltage is displaced by 30° from the primary (lagging for Yd1, leading for Dy11).

When the primaries are on the same supply, the secondary line voltages of the two banks differ by 30°.

  V (Y-Y secondary)
   ^
   |    30 deg
   |  /  V (Y-D secondary)
   | /
   +----------------->

The voltage difference across the paralleled terminals is

$$\Delta V = 2V\sin\frac{30^\circ}{2} = 0.518\,V$$

This is about 52% of the rated secondary voltage. It appears across the very low internal impedances of the two transformers, so a huge circulating current flows, even at no load. The windings overheat and may be damaged, and the load cannot be shared properly.

Also: the voltage ratios differ in form. Star–delta has $V_{L2}/V_{L1} = 1/(\sqrt3\,a)$ compared with $1/a$ for star–star, so the turns would have to be redesigned. Even if the magnitudes matched, the 30° displacement cannot be removed by any change of turns ratio.

Hence star–delta can be paralleled only with another star–delta, or with a transformer that has the same phase displacement (the same vector group), such as delta–star of the right group.

Two transformers with equal turns ratio have the same secondary voltage, and they share the load current in inverse proportion to their impedances.

Given: $Z_A = 0.5 + j3\ \Omega$, $Z_B = 0.6 + j10\ \Omega$ (referred to secondary), total load $P = 100$ kW at 0.8 pf lagging.

Total load

$$S = \frac{100}{0.8} = 125\ \text{kVA},\qquad \phi = \cos^{-1}0.8 = 36.87^\circ\ \text{lagging}$$

Current division. Let the total current be $I = I\angle-36.87^\circ$ with $V_2$ as reference. For parallel branches:

$$I_A = I\frac{Z_B}{Z_A + Z_B},\qquad I_B = I\frac{Z_A}{Z_A + Z_B}$$ $$\begin{aligned} Z_A + Z_B &= 1.1 + j13\ \Omega \\ \frac{Z_B}{Z_A + Z_B} &= 0.7676 + j0.0188 = 0.7679\angle1.40^\circ \\ \frac{Z_A}{Z_A + Z_B} &= 0.2324 - j0.0188 = 0.2331\angle-4.63^\circ \end{aligned}$$

Load shared (kVA = fraction of 125 kVA, since $V_2$ is common):

$$\begin{aligned} S_A &= 0.7679\times125 = 95.98\ \text{kVA at angle } 36.87^\circ - 1.40^\circ = 35.47^\circ \\ P_A &= 95.98\cos35.47^\circ = 78.17\ \text{kW},\quad \text{pf}_A = 0.814\ \text{lag} \\ S_B &= 0.2331\times125 = 29.14\ \text{kVA at angle } 36.87^\circ + 4.63^\circ = 41.50^\circ \\ P_B &= 29.14\cos41.50^\circ = 21.83\ \text{kW},\quad \text{pf}_B = 0.749\ \text{lag} \end{aligned}$$

Check: $P_A + P_B = 78.17 + 21.83 = 100$ kW.

Answer: A supplies about 78.2 kW (96.0 kVA) and B about 21.8 kW (29.1 kVA). The transformer of lower impedance (A) carries the greater share.

An instrument transformer is a special transformer used to step down high voltages or currents to safe, measurable values for meters and relays. It isolates the measuring circuit from the high-voltage line. There are two types: the current transformer (CT) and the potential transformer (PT).

Types and working

Current transformer. The primary has few turns (often a single bar) and is connected in series with the line. The secondary has many turns and feeds an ammeter or relay coil (low impedance burden). Its secondary is nearly short-circuited. Secondary current is 5 A or 1 A.

$$I_1N_1 = I_2N_2 \Rightarrow \frac{I_1}{I_2} = \frac{N_2}{N_1} = \text{CT ratio}$$

Potential transformer. The primary is connected across the line, and the secondary feeds a voltmeter or pressure coil (high impedance burden). It works almost at no load, with secondary voltage 110 V.

$$\frac{V_1}{V_2} = \frac{N_1}{N_2}$$

  Line ----+----- Load       Line ------+------
           |                            |
          [CT]--(A)                    [PT]--(V)

Difference from the power transformer

Errors

For a CT, the exciting current $I_0$ is drawn from the primary but does not appear in the secondary, so the actual ratio $I_1/I_2$ differs from the turns ratio $N_2/N_1$ (ratio error), and $I_2$ is not exactly 180° from $I_1$ (phase-angle error). Both errors are kept small by a low-loss, high-permeability core operated at low flux density and by a low burden.