Question

A single-phase 60−Hz transformer has a nameplate voltage rating of 7.97 kV:120 V based on its known winding turns ratio. The manufacturer calculates that the primary (7.97-kV) leakage inductance is 193mH and the primary magnetizing inductance is 167H. For an applied primary voltage of 7970 V at 60 Hz, calculate the resultant open-circuit secondary voltage.

Answer 1

The question involves calculating the number of turns in the secondary coil of a multipurpose transformer for different output voltages and determining the maximum output currents. Using the transformer equation, we can find the required number of turns in the secondary coil to produce the given output voltages. By applying the transformer equation again, we can calculate the maximum output currents using the input current and the number of turns in the secondary coil.

Step-by-step explanation:

The question asks about the number of turns in the secondary coil of a multipurpose transformer. To calculate this, we need to use the transformer equation. It states that the ratio of the secondary voltage to the primary voltage equals the ratio of the number of turns in the secondary coil to the number of turns in the primary coil. In this case, the input voltage is 240 V and the primary coil has 280 turns. Let's calculate the number of turns in the parts of the secondary used to produce the given output voltages:

1. For an output voltage of 5.60 V, the secondary coil would have [(5.60 V / 240 V) * 280 turns = 6.53 turns]. However, since turns cannot be fractions, we round it to 7 turns.
2. For an output voltage of 12.0 V, the secondary coil would have [(12.0 V / 240 V) * 280 turns = 14 turns].
3. For an output voltage of 480 V, the secondary coil would have [(480 V / 240 V) * 280 turns = 560 turns].

For part (b) of the question, we can use the transformer equation to find the maximum output currents. The ratio of the primary current to the secondary current equals the ratio of the number of turns in the secondary coil to the number of turns in the primary coil. Since the input current is 5.00 A, we can calculate the maximum output currents:

1. For the secondary coil with 7 turns: [(5.00 A / 280 turns) * 7 turns = 0.125 A]
2. For the secondary coil with 14 turns: [(5.00 A / 280 turns) * 14 turns = 0.250 A]
3. For the secondary coil with 560 turns: [(5.00 A / 280 turns) * 560 turns = 10.000 A]