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The secondary coil of a neon sign transformer provides 7500 V at 0.01 A. The primary coil operates on 120 V. What is the input current in the primary coil of this transformer assuming that no energy is lost in the transformer?

Respuesta :

Answer:

0.625 A

Explanation:

Vs = 7500 V, Is = 0.01 A

Vp = 120 V

Let the primary current be Ip.

As the transformer is ideal, so input power is equal to the output power

Vp x Ip = Vs x Is

120 x Ip = 7500 x 0.01

Ip = 0.625 A

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