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The half-wave rectifier circuit of Fig. 1 ha a tranformer inerted between the
ource and the remainder of the circuit. The ource i 240 V rm at 60 Hz, and the
load reitor i 20 Ω. (a) Determine the required turn ratio of the tranformer uch
that the average load current i 12 A. (b) Determine the average current in the
primary winding of the tranformer

Sagot :

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(a) ratio = 0.32 One the average current with in primary coil of the tranformer.

(b) The load current is 753.6 volts, and the transformer's required turn ratio is 12. The mean load current is 12.

What is current?

Charge moves across a point on such a circuit at a rate known as current. When numerous coulombs of charge pass over a wire's cross section in a circuit, a large current is the result. It is not necessary to have a fast speeds to have a large current if the positive ions are tightly packed inside the wire.

Briefing:

Source = 240Vrms/ 60Hz

Resistor = 20 ohm

Avg. load current = voltage/ resistance * pi

= 12A times (20*3.14)

load current = 753.6 volts

ratio = Vp/Vs

ratio = 240/754

ratio = 0.32 A

To know more about current visit:

https://brainly.com/question/28217993

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