5.1 Explain the principle of mutual induction with reference to transformers - NSC Electrical Technology Power Systems - Question 5 - 2022 - Paper 1

1. Transformer ratio
2. Voltage rating
3. Current rating

The Star (Y) connection creates a neutral point on the secondary side of a three-phase transformer.

Copper losses are primarily due to the internal resistance of the copper conductors in the coils, which results in heat loss when current flows through them.

Iron losses occur due to eddy currents and the changing magnetic field inside the iron core. These losses are attributed to hysteresis and the heat produced by the alternating magnetic flux in the core.

Insulation failure in dry-type transformers is managed by employing tubular radiators through which air circulates, effectively cooling the windings and preventing overheating.

In a shell-type transformer, the coils are wound around the central section of the core, which has three limbs. Conversely, a core-type transformer has limbs, and the coils are arranged around all three limbs, providing different magnetic pathways.

A balanced earth-fault relay monitors the currents in all three phases. Under normal conditions, the three-phase currents sum to zero. If an earth fault occurs, the difference in currents will detect the fault, triggering the relay to isolate the transformer from the supply.

The secondary line current, $I_L$, can be calculated using the formula:
$I_L = \frac{P}{\sqrt{3} \cdot V_L \cdot \cos \theta}$
Where $P = 200,000 W$, $V_L = 400 V$, and $\, \cos \theta = 0.8$. Thus:
$I_L = \frac{200,000}{\sqrt{3} \cdot 400 \cdot 0.8} = 360.84 A$

In a star connection, the secondary phase current, $I_{PH}$, is equal to the line current, $I_L$. Hence,
$I_{PH} = I_L = 360.84 A$

Apparent power, $S$, can be calculated using the formula:
$S = \frac{P}{\cos \theta} = \frac{200,000}{0.8} = 250,000 VA$

The primary line current, $I_{L1}$, can be calculated using the following formula:
$I_{L1} = \frac{P}{\sqrt{3} \cdot V_{L1} \cdot \cos \theta}$
Substituting the known values:
$I_{L1} = \frac{200,000}{\sqrt{3} \cdot 6000 \cdot 0.8} = 24.06 A$