A constant voltage of 10.00 V has been observed over a certain time interval acr

Question

A constant voltage of 10.00 V has been observed over a certain time interval across a 2.40 H inductor. The current through the inductor, measured as 3.00 A at the beginning of the time interval, was observed to increase at a constant rate to a value of 9.00 A at the end of the time interval. How long was this time intervalAn ideal step-down transformer has a primary coil of 390 turns and a secondary coil of 17 turns. Its primary coil is plugged into an outlet with 220 V(AC), from which it draws an rms current of 0.35 A.
What is the voltage in the secondary coil?Calculate the rms current in the secondary coil.Assuming that the transformer secondary is driving a resistive load, calculate the average power dissipated in the resistor.

Explanation / Answer

V = L*I/t

so t = L*I/V = 2.40*(9.0 - 3.0)/10.0 = 1.44s

2. For a step down transformer the current relationship is

Ns/Np =Vs/Vp

Vs=(Ns/Np)*Vp=9.59v

Ip/Is = Ns/Np

so Is = Ip/(Ns/Np) = 0.35/(17/390) = .015A

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