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?
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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