Recent content by Mbert

As the name implies, the no-load current is the current through the primary winding when the secondary winding is not loaded. Look at it this way. Draw your model with a 2300V source and a variable resistor load. The parameters of your transformer model are fixed. If the load is R=infinity...

That seems correct (did not calculate, but the steps are good).

Almost there! You forgot that the magnetizing current is not the same when loaded, so you can't use: Ip' = Imag = 0.15<-72.66° A Instead, when the transformer is not loaded, you simply have a potential divider between Z1=R1+jX1 and Z2=Rmag//jXmag (shunt impedances). The voltage V1' will...

You have to be careful about the image at R=0. As R goes to zero, the image goes to infinity. In the case where the charge is outside, you need 2 images, one at the origin of the sphere and the other at some ratio that depends on the radius of the sphere. If the charge is at infinity, there's no...

I agree, but if the goal is comprehension of the concepts, then why give the formula (that only works out for a given voltage waveform)? Understanding where that formula comes from is the key in understanding, rather than "plug-and-play". If the question is about transformer engineering...

This is false. 1) The flux linkage is the integral of voltage, not volts multiplied by time. If you do the time integral of the previous voltage waveform, the magnitude of this waveform is the answer to the poster's question. Flux is simply the flux linkage divided by the number of turns. 2)...

Not exactly. You forgot to remove Imag before here: Vp' = (2.7+j9.1)*(6.52<36.9°) + 2278.8<1.39° = 2256.6<2.87° V The current in R1+jX1 is not (6.52<36.9°). Apply V1 to Rmag and Xmag. Also, since the load is capacitive, I think it is normal that you get an higher voltage when loaded.

Yes, many thanks.

I'm guessing I have to use the Y-D transformation, so I get 2 triangles in parallel?

Homework Statement Find the Thevenin equivalent network of the lattice network (see attached), viewed from points As and Br (terminals Bs and Ar are not connected to anything). Homework Equations The Attempt at a Solution Without the resistance f, the problem is quite simple. I...

Yes, you have done it correctly. You can bring back impedances on the 2300V side or not, it doesn't matter. You don't need to eliminate the excitation branch, since you know 1 voltage in the circuit. The key here is that you need to use the equations of the ideal transformer...

Suppose you have a cosinusoidal voltage V=Vmax*cos(wt). What would be the equation of the flux using this EMF in Faraday's law? What is the magnitude of this flux waveform in terms of V?

Well, I should have a problem at u=z1 or u=z2, since there's a singularity at log(0). However, since I am integrating over the real axis, I guess I only get the singularities if z1 and z2 are reals?

I think for the assumptions on k and m (Reals), the roots can be complex. Isn't there something special to do in that case? I think there is something about the complex plane and singularities, but it's been a while, can't remember very well.

Shouldn't I get: \int(k-mu)\log[(m^2+1)u^2-2mku+k^2]du Then, I can find the roots z1 and z2 of the polynomial and express the integral in the form: \int(k-mu)\log[(u-z_1)(u-z_2)]du=\int(k-mu)\log[u-z_1]du+\int(k-mu)\log[u-z_2]du and then I integrate by parts with f=k-m*u and dg=log[u-z]?