A 1,200/240 V rms transformer is connected to a [tex]0.4000\angle10^\circ\ \Omega[/tex] load on the low voltage side.

1. Calculate the turns ratio.

2. Determine the primary and secondary current when the transformer primary winding is connected to [tex]1200 \text{ V rms}[/tex] through a [tex]60\angle-30^\circ \ \Omega[/tex] impedance.

Please report your answer so the magnitude is positive, and all angles are in the range of -180 degrees to +180 degrees.

- Primary current: [tex]I_1 = 19.77 \angle(27.759) \text{ A}[/tex]
- Secondary current: [tex]I_2 = 98.876 \angle(27.759) \text{ A}[/tex]
- Turns ratio: [tex]n = 0.2[/tex]

The turns ratio of the transformer is 0.2. When the primary winding is connected to a 1200V rms source through a 60∠−30∘ Ω impedance, the primary current is I1 = 19.77 ∠(27.759)A and the secondary current is I2 = 98.876 ∠(27.759)A.

Explanation:

In this scenario, we are dealing with a transformer which has a voltage ratio of 1200/240 or 5:1. This means that the turns ratio, which is the ratio of the number of turns in the primary coil to the number of turns in the secondary coil, is also 5:1, or simply 0.2 as given in the question.

So, turns ratio n = 0.2.

Next, we want to calculate the primary and secondary current. If the primary winding is connected to a 1200V rms source through an impedance of 60∠−30∘ Ω, we can calculate the current using Ohm's law (I = V/R).

So, the primary current I1 = 19.77 ∠(27.759)A, and the secondary current I2 = 98.876 ∠(27.759)A. Here, the angle represents the phase difference between the current and voltage.

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