Question

The secondary coil of an ideal neon sign transformer provides 7500 V at 10.0 mA. The primary coil operates on 120 V. what current does the primary draw?

Answer 1

given quantities :- input voltage = 120 v

output voltage = 7500 v

output current = 10 mA = 10 * 10^-3 A

solution :-

for an ideal transformer the product of input voltage and current and output voltage and current must equal each other.

so,

input power = output power

input current * input voltage = output current * output voltage

Ic * 120 volts = 10 * 10^-3 A * 7500 volts

where Ic is the current in the primary coil,

solving for Ic,

Ic = 75/120

Ic = 0.625 Ampere

so, the current passing through the primary coil is 0.625 Amperes.

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