Mention four methods of reducing leakage fluxes in a transformer.

The turns ratio $a = N_1/N_2 = 1/4$.
The load impedance is $Z_L = R_L + jX_L = 200 + j100 \Omega$.
The load impedance referred to the primary is $Z_{L1} = a^2 Z_L$.

$$Z_{L1} = \left(\frac{1}{4}\right)^2 (200 + j100) = \frac{1}{16} (200 + j100) = 12.5 + j6.25 \Omega$$

The total equivalent impedance referred to the primary is $Z_{total,1} = R_{eq1} + jX_{eq1} + Z_{L1}$.

$$Z_{total,1} = (0.1 + j0.5) + (12.5 + j6.25) = (0.1 + 12.5) + j(0.5 + 6.25) = 12.6 + j6.75 \Omega$$

The magnitude of the total equivalent impedance is:

$$|Z_{total,1}| = \sqrt{12.6^2 + 6.75^2} = \sqrt{158.76 + 45.5625} = \sqrt{204.3225} \approx 14.294 \Omega$$

1.2.1: The primary circulating current ($I_1$).
The primary current is $I_1 = V_1 / Z_{total,1}$.

$$I_1 = \frac{230 \text{ V}}{12.6 + j6.75 \Omega}$$ $$I_1 = \frac{230}{14.294 \angle \arctan\left(\frac{6.75}{12.6}\right)} = \frac{230}{14.294 \angle 28.18^\circ} \approx 16.09 \angle -28.18^\circ \text{ A}$$

The magnitude of the primary circulating current is:

$$\boxed{\text{16.09 A}}$$

1.2.2: The per unit regulation of the transformer.
The power factor angle $\phi$ is the angle of $Z_{total,1}$, so $\phi = 28.18^\circ$.
$\cos \phi = \cos(28.18^\circ) \approx 0.8815$
$\sin \phi = \sin(28.18^\circ) \approx 0.4722$
The voltage regulation (VR) is given by:

$$VR = \frac{I_1 (R_{eq1} \cos \phi + X_{eq1} \sin \phi)}{V_1}$$ $$VR = \frac{16.09 (0.1 \times 0.8815 + 0.5 \times 0.4722)}{230}$$ $$VR = \frac{16.09 (0.08815 + 0.2361)}{230} = \frac{16.09 \times 0.32425}{230} = \frac{5.218}{230} \approx 0.02268$$

The per unit regulation is:

$$\boxed{\text{0.02268 pu}}$$

1.2.3: The secondary terminal voltage ($V_2$).
The voltage across the load referred to the primary is $V_{L1} = I_1 Z_{L1}$.

$$Z_{L1} = 12.5 + j6.25 = \sqrt{12.5^2 + 6.25^2} \angle \arctan\left(\frac{6.25}{12.5}\right) = 13.975 \angle 26.565^\circ \Omega$$ $$V_{L1} = (16.09 \angle -28.18^\circ) \times (13.975 \angle 26.565^\circ)$$ $$V_{L1} = (16.09 \times 13.975) \angle (-28.18^\circ + 26.565^\circ) = 224.86 \angle -1.615^\circ \text{ V}$$

The secondary terminal voltage $V_2$ is related to $V_{L1}$ by the turns ratio $N_2/N_1 = 4$.

$$V_2 = V_{L1} \times \frac{N_2}{N_1} = 224.86 \times 4 \angle -1.615^\circ = 899.44 \angle -1.615^\circ \text{ V}$$

The magnitude of the secondary terminal voltage is:

$$\boxed{\text{899.44 V}}$$

QUESTION 2: MEASUREMENT OF POWER IN THREE PHASE SYSTEMS

2.1: Given:
Line voltage $V_L = 440 \text{ V}$
Frequency $f = 50 \text{ Hz}$
Star connected load
Impedance per phase $z = 5.2 + j6.7 \Omega / phase$

For a star-connected load, the phase voltage $V_p = V_L / \sqrt{3}$.

$$V_p = \frac{440}{\sqrt{3}} \approx 254.03 \text{ V}$$

The magnitude of the impedance per phase is $|z| = \sqrt{5.2^2 + 6.7^2}$.

$$|z| = \sqrt{27.04 + 44.89} = \sqrt{71.93} \approx 8.481 \Omega$$

The phase current $I_p = V_p / |z|$.

$$I_p = \frac{254.03}{8.481} \approx 29.95 \text{ A}$$

For a star connection, the line current $I_L = I_p \approx 29.95 \text{ A}$.
The power factor angle $\phi = \arctan(X/R) = \arctan(6.7/5.2) \approx 52.17^\circ$.
The power factor $\cos \phi = \cos(52.17^\circ) \approx 0.613$.

2.1.1: The total power of the load ($P_T$).
The total power for a three-phase system is $P_T = \sqrt{3} V_L I_L \cos \phi$.

$$P_T = \sqrt{3} \times 440 \text{ V} \times 29.95 \text{ A} \times 0.613$$ $$P_T = 13979.8 \text{ W} \approx 13.98 \text{ kW}$$

The total power of the load is:

$$\boxed{\text{13.98 kW}}$$

2.1.2: The readings on watt-meters, if two-wattmeter method is used to measure the power taken by the load.
The two wattmeter readings are $W_1 = V_L I_L \cos(30^\circ - \phi)$ and $W_2 = V_L I_L \cos(30^\circ + \phi)$.
Here $\phi = 52.17^\circ$.
$30^\circ - \phi = 30^\circ - 52.17^\circ = -22.17^\circ$
$30^\circ + \phi = 30^\circ + 52.17^\circ = 82.17^\circ$

$$W_1 = 440 \times 29.95 \times \cos(-22.17^\circ) = 440 \times 29.95 \times 0.926 \approx 12200 \text{ W}$$ $$W_2 = 440 \times 29.95 \times \cos(82.17^\circ) = 440 \times 29.95 \times 0.136 \approx 1792 \text{ W}$$

The readings on the wattmeters are:

$$\boxed{W_1 = \text{12200 W, } W_2 = \text{1792 W}}$$

2.2: Calculate the resistance per phase of a three phase $l = 28 \text{ km}$, overhead line with solid conductors of $D = 1.8 \text{ cm}$ diameter, with a symmetrical spacing of $d = 0.8 \text{ m}$ between centres. Take resistivity of the conductor material as $\rho = 1.7 \text{ } \mu \Omega \text{ cm}$.

Length $l = 28 \text{ km} = 28 \times 10^5 \text{ cm}$.
Diameter $D = 1.8 \text{ cm}$, so radius $r = D/2 = 0.9 \text{ cm}$.
Resistivity $\rho = 1.7 \text{ } \mu \Omega \text{ cm} = 1.7 \times 10^{-6} \Omega \text{ cm}$.
The cross-sectional area $A = \pi r^2$.

$$A = \pi (0.9 \text{ cm})^2 = 0.81 \pi \text{ cm}^2$$

The resistance $R_{phase}$ is given by the formula $R = \rho \frac{l}{A}$.

$$R_{phase} = (1.7 \times 10^{-6} \Omega \text{ cm}) \times \frac{28 \times 10^5 \text{ cm}}{0.81 \pi \text{ cm}^2}$$ $$R_{phase} = \frac{1.7 \times 28}{0.81 \pi} \Omega = \frac{47.6}{2.54469} \Omega \approx 18.705 \Omega$$

The resistance per phase is:

$$\boxed{\text{18.71 } \Omega}$$

QUESTION 3: AC MACHINES

3.1: Given:
Synchronous impedance per phase $Z_s = 1 + j6.5 \Omega / phase$.
Open circuit line voltage of one alternator $E_{1,line} = 2.2 \text{ kV} = 2200 \text{ V}$.
Circulating current $I_C = 80 \text{ A}$.
Machines are star-connected and in exact phase opposition.

The phase voltage of the first alternator is $E_{1,phase} = E_{1,line} / \sqrt{3}$.

$$E_{1,phase} = \frac{2200}{\sqrt{3}} \approx 1270.17 \text{ V}$$

The magnitude of the synchronous impedance is $|Z_s| = \sqrt{1^2 + 6.5^2}$.

$$|Z_s| = \sqrt{1 + 42.25} = \sqrt{43.25} \approx 6.576 \Omega$$

3.1.1: The open circuit voltage of the other machine ($E_2$) assuming it is bigger than that of the first machine.
When two alternators are in parallel and in exact phase opposition, the circulating current $I_C$ is given by:

$$I_C = \frac{E_{2,phase} - E_{1,phase}}{2Z_s}$$

This formula assumes $E_1$ and $E_2$ are in phase, and $E_2 > E_1$. However, "exact phase opposition" means $E_1$ and $E_2$ are $180^\circ$ out of phase.
Let $E_1 = E_{1,phase} \angle 0^\circ$ and $E_2 = E_{2,phase} \angle 180^\circ$.
The voltage difference driving the circulating current is $E_1 - E_2 = E_{1,phase} - (-E_{2,phase}) = E_{1,phase} + E_{2,phase}$.
The magnitude of the circulating current is $|I_C| = \frac{E_{1,phase} + E_{2,phase}}{2|Z_s|}$.

$$80 \text{ A} = \frac{1270.17 \text{ V} + E_{2,phase}}{2 \times 6.576 \Omega}$$ $$80 \times (2 \times 6.576) = 1270.17 + E_{2,phase}$$ $$1052.16 = 1270.17 + E_{2,phase}$$ $$E_{2,phase} = 1052.16 - 1270.17 = -218.01 \text{ V}$$

This result is negative, which contradicts the assumption that $E_2$ is bigger than $E_1$. This indicates that the problem statement "exact phase opposition" might be interpreted differently in the context of the given circulating current value, or there's an inconsistency in the problem data.
A common scenario for circulating current in parallel alternators is due to a phase difference between EMFs, or a difference in magnitudes when EMFs are in phase. If they are in "exact phase opposition", it usually means they are $180^\circ$ out of phase.
If the circulating current is $I_C = \frac{E_1 - E_2}{2Z_s}$ where $E_1$ and $E_2$ are phasors, and $E_1 = E_{1,phase} \angle 0^\circ$ and $E_2 = E_{2,phase} \angle 180^\circ$, then $E_1 - E_2 = E_{1,phase} - (-E_{2,phase}) = E_{1,phase} + E_{2,phase}$.
The magnitude of this voltage difference is $E_{1,phase} + E_{2,phase}$.
The magnitude of $I_C$ is $80 \text{ A}$.
$|I_C| = \frac{E_{1,phase} + E_{2,phase}}{2|Z_s|}$.
This leads to $E_{2,phase} = -218.01 \text{ V}$, which is not physically possible for a magnitude.

Let's assume "exact phase opposition" means that the resultant voltage causing the circulating current is $E_2 - E_1$ (phasor difference), and the problem implies that $E_1$ and $E_2$ are in phase but with different magnitudes, and the "phase opposition" refers to the current flow. This is a common source of confusion.
If $E_1$ and $E_2$ are in phase, and $E_2 > E_1$, then $I_C = \frac{E_{2,phase} - E_{1,phase}}{2Z_s}$.
$|I_C| = \frac{E_{2,phase} - E_{1,phase}}{2|Z_s|}$.

$$80 = \frac{E_{2,phase} - 1270.17}{2 \times 6.576}$$ $$1052.16 = E_{2,phase} - 1270.17$$ $$E_{2,phase} = 1052.16 + 1270.17 = 2322.33 \text{ V}$$

This value is positive and greater than $E_{1,phase}$, consistent with the problem statement.
The open circuit line voltage of the other machine is $E_{2,line} = E_{2,phase} \times \sqrt{3}$.

$$E_{2,line} = 2322.33 \times \sqrt{3} \approx 4021.9 \text{ V}$$

The open circuit voltage of the other machine is:

$$\boxed{\text{4.022 kV}}$$

3.1.2: The terminal voltage (line value).
The terminal voltage $V_T$ (phase value) is the voltage at the busbar. It can be calculated using either alternator's EMF and circulating current.
Let'