a circuit with an ideal transformer with a source voltage, Vs=10cos(ωt)V, N1:N2=1:4, and RL=50Ω.

What is the magnitude of the current on the primary side of the transformer?

Rp = (N1/N2)^2 * RL = (1/4)^2 * 50 = 3.125 Ohms = Primary resistance.

Ip = Vs/Rp

To find the magnitude of the current on the primary side of the transformer, we need to follow these steps:

Step 1: Determine the primary voltage (V1):
Since the source voltage is given as Vs = 10cos(ωt), and the transformer turns ratio is N1:N2 = 1:4, we can calculate the primary voltage as follows:
V1 = Vs * N2/N1
= 10cos(ωt) * (4/1)
= 40cos(ωt)

Step 2: Calculate the primary current (I1):
Using Ohm's Law, I = V/R, where V is the voltage and R is the resistance, we can calculate the primary current:
I1 = V1 / RL
= (40cos(ωt)) / 50
= (4/5)cos(ωt)

So, the magnitude of the current on the primary side of the transformer is (4/5)A.