Solution: Find Full-Load Primary Current in Polar Form

[yoamocuy]

Homework Statement

A single phase 500V/250V transformer that supplies a rated load current of 10 A with a power factor of 0.8 lagging. At no load, the supplied current is 0.25 A with a power factor of 0.10 lagging. I am supposed to find the full-load primary current and express it in polar form. Winding impedance voltage drop can be neglected.

Homework Equations

I=I_o + I₂'
I₂'=(N₂/N₁)*I₂
V₂/V₁=N₂/N₁

The Attempt at a Solution

Full Load

500V/250V=N₁/N₂
I think that rated load current is = to I₂
therefore, I₂'=(250/500)*10 A
I₂'=5 A

No Load

I_o=0.25 A

using the equation I=I_o + I₂'
I= 5 A + 0.25 A
I=5.25 A

Does everything look ok so far?
At this point I'm not sure how to find the angle of I.

 

[jbsteele]

I'm assuming that the angle of I at no load is the same as the power factor of 0.10 lagging? Can someone help me with the next step?

 

[tomcue]

Yes, your calculations for the full load and no load currents appear to be correct. To find the angle of I, you can use the power factor formula: cosφ = P/S, where P is the real power and S is the apparent power. In this case, the power factor is given as 0.8 lagging, so cosφ = 0.8. The real power can be calculated using the formula P = V1*I1*cosφ, where V1 is the primary voltage and I1 is the primary current. Plugging in the given values, we get P = (500V)*(I1)*(0.8) = 400*I1. The apparent power can be calculated using the formula S = V1*I1, so S = (500V)*(I1). Now we can solve for I1 by dividing both sides of the power factor formula by S and taking the inverse cosine: I1 = cos^-1 (P/S) = cos^-1 (400*I1 / (500*I1)) = cos^-1 (0.8) = 36.87 degrees. Therefore, the full-load primary current in polar form is I1 = 5.25 A ∠36.87°.