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1.

(a) To supply a balanced 3-phase load, the transformers are connected in a T configuration, also known as a "zigzag" connection. This connection allows for three-wire output from two transformer windings.

The voltage rating of each coil can be found using the turns ratio equation:

V1/V2 = N1/N2

where V1 is the primary voltage (3300 V), N1 is the number of turns on the primary coil, V2 is the secondary voltage (440 V), and N2 is the number of turns on the secondary coil.

Solving for N1 and N2:

3300/440 = N1/N2

N1 = 20N2

We can assume that the coils have the same number of turns, so N1 = N2.

3300/440 = N1/N1

N1 = N2 = 75

Therefore, each coil has 75 turns.

The current rating of each coil can be found using the kVA equation:

kVA = V x I x sqrt(3)/1000

where V is the voltage (either primary or secondary), I is the current, and sqrt(3) is the square root of 3 (1.732).

Solving for I:

33,000/440/1.732 = 43.4 A

Therefore, each coil has a current rating of 43.4 A.

(b) The main transformer supplies the majority of the power to the load, while the teaser transformer supplies a smaller portion. The kVA rating of each transformer can be found using the following equations:

kVA = V x I x sqrt(3)/1000

where V is the voltage (either primary or secondary), I is the current, and sqrt(3) is the square root of 3 (1.732).

For the main transformer:

kVA = 3300 x 43.4 x sqrt(3)/1000 = 202.5 kVA

For the teaser transformer:

kVA = (3300 - 440) x 43.4 x sqrt(3)/1000 = 160.8 kVA

Therefore, the main transformer has a kVA rating of 202.5 kVA, and the teaser transformer has a kVA rating of 160.8 kVA.

2.

In a delta connection, the line voltage is equal to the phase voltage. Therefore, the line voltage on the transformer is:

V = 160 kW/(100 kVA x sqrt(3)) = 960 V

Since one transformer is now removed, the two remaining transformers must supply the full load, resulting in an overload. To prevent this overload, we can add capacitors in parallel with the load to increase the power factor.

The new load current can be found using the power equation:

P = sqrt(3) x V x I x cos(theta)

where P is the power (160 kW), V is the voltage (960 V), I is the current, and cos(theta) is the power factor (0.8 lagging).

Solving for I:

160,000 = sqrt(3) x 960 x I x 0.8

I = 219.4 A

To prevent an overload, the current through each transformer must be reduced to 100/3 kA, or 33.3 kA.

The new power factor can be found using the following equation:

cos(theta) = P/(sqrt(3) x V x I)

cos(theta) = 160,000/(sqrt(3) x 960 x 33,300) = 0.634

To increase the power factor from 0.8 to 0.634, we need to add capacitive reactance. The capacitive reactance can be found using the following equation:

Xc = 1/(2 x pi x f x C)

where Xc is the capacitive reactance, pi is the mathematical constant pi (approximately 3.14159), f is the frequency (60 Hz), and C is the capacitance.

Solving for C:

Xc = 1/(2 x pi x 60 x C)

C = 1/(2 x pi x 60 x Xc)

We want to add enough capacitance to increase the power factor to 0.634. Using the following equation:

cos(theta) = sqrt(1/(1 + (Xc/R)^2))

where R is the resistance of the load (160 kW/0.8 = 200 kohms).

Solving for Xc:

Xc = R x (1/cos(theta) - 1)

Xc = 200,000 x (1/0.634 - 1) = 124,354 ohms

Solving for C:

C = 1/(2 x pi x 60 x Xc) = 212 uF

Therefore, a capacitor bank with a total capacitance of 212 uF should be added across the load to prevent the transformers from being overloaded.

3.

To find the turns ratio, we can use the following equation:

V1/V2 = N1/N2

where V1 is the primary voltage (6000 V), N1 is the number of turns on the primary coil, V2 is the secondary voltage (415 V line-to-line), and N2 is the number of turns on the secondary coil.

We first need to convert the line-to-line voltage to phase voltage:

Vph = Vll/sqrt(3) = 415/sqrt(3) = 239.6 V

Now we can solve for N1/N2:

6000/239.6 = N1/N2

N1/N2 = 25

The equivalent resistance drop and reactance drop are given as 1% and 5%, respectively. We can use these values to find the total impedance of the transformer:

Z = Vph^2/PF x S

where Z is the impedance, Vph is the phase voltage (239.6 V), PF is the power factor (0.8 lagging), and S is the apparent power (33 kVA).

Solving for Z:

Z = Vph^2/PF x S = 239.6^2/0.8 x 33,000 = 2,757 ohms

The equivalent resistance and reactance can be found using the following equations:

R = Z x 1% = 27.57 ohms

X = Z x 5% = 137.85 ohms

The total impedance can be represented as a series circuit of R and X.

We can use the impedance triangle to find the phase angle:

tan(theta) = X/R = 137.85/27.57 = 5

theta = atan(5) = 78.69 degrees

Therefore, the power factor angle is 78.69 degrees lagging.

To find the turns ratio:

cos(theta) = PF

cos(78.69) = N2/N1

N2/N1 = 0.195

N1/N2 = 1/0.195 = 5.13

Therefore, the turns ratio is approximately 5.13:1.