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The transfoner has the following iforation on the nameplate: Primary roltage 13,800 V Secondary voltage 240V Impedance5%% 150 EVA The secondary fuse has a blow rating of 800 A and an interupt rating of 10,000 A. Is this interrupt rating sufficient for this installation? Show your work.

User Sensanaty
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The numbers of turns for the secondary coil to get the desired output voltages of 5.60V, 12.0V, and 480V are approximately 6.53, 14.00, and 560.00, respectively. Applying the power conservation principle and given a maximum primary current of 5.00A, the maximum output currents for each voltage output are roughly 214.29A, 100.00A, and 2.50A, respectively.

The fundamental principle of a transformer is based on Faraday's law of electromagnetic induction, relating the input and output voltages to the number of turns on the primary and secondary coils. This relationship is given by the formula:

Vp/Vs = Np/Ns

where Vp is the primary voltage, Vs is the secondary voltage, Np is the number of turns in the primary coil, and Ns is the number of turns in the secondary coil.

The transformer in question has a primary voltage (Vp) of 240 volts and a primary coil with 280 turns (Np). We can use the equation above to calculate the number of turns needed in the secondary coil (Ns) to produce the desired secondary voltages (5.60, 12.0, and 480V):

For 5.60V: Ns = Np * (Vs/Vp) = 280 * (5.60/240) = 6.53

For 12.0V: Ns = Np * (Vs/Vp) = 280 * (12.0/240) = 14.00

For 480V: Ns = Np * (Vs/Vp) = 280 * (480/240) = 560.00

For the calculation of maximum output currents, we use the principle of conservation of power: Pp = Ps, where Pp is the power on the primary side and Ps is the power on the secondary side. Since P = VI:

Ips = Ipp * (Vpp/Vps).

Considering a maximum primary current (Ipp) of 5.00A:

For 5.60V: Ips = 5.00 * (240/5.60) = 214.29A

For 12.0V: Ips = 5.00 * (240/12.0) = 100.00A

For 480V: Ips = 5.00 * (240/480) = 2.50A

User Yazid
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