Question

An ideal transformer with 5 turns on the primary and 10 turns on the secondary is shown in the figure. A sinusoidal voltage source having V(t) = 10 sin(100t) volts is connected across the primary and a 1 Ω resistor is connected across the secondary.

What is the maximum value of the current in the primary circuit (not the secondary)?

(answer is 40 amps)

Answer 1

here,

number of turns in primary , Np = 5 turns

number of turns in seccondary , Ns = 10 turns

the maximum voltage of primary , Vp = 10 V

for ideal transformer ,

the maximum voltage in seccondary , Vs = Vp * ( Np/Na)

Vs = 10 * ( 10/5) = 20V

the current in seccondary , Is = Vs /1 = 20 A

as the power dissipated remains constant

so, the maximum current in primary , Ip = Is * ( Vs /Vp)

Ip = 20 * ( 20 /10) = 40 A