Short-circuited Resistance in Parallel with Reactance

[GoodPost]

Dears,

Please help me out with this. The attached figure is an equivalent circuit of an induction machine. I'm trying to find the current (Is). It's assumed that the core losses are neglected and hence Rc =0. If this is the case, then Rc (short-circuited) is in parallel with jXm. What's the resultant impedance for this parallel combination? Thx,

GoodPost.

 

[anorlunda]

Please fill out the template next time. At the minimum, you are required to show your own effort. Homework helpers are not allowed to help until you do that.

 

[GoodPost]

The homework problem is pretty long and has different parts and I can do it. My only issue is what happens when a resistor (short-circuited) is connected to a reactance in parallel. It's more to understand a concept than to get an answer for a homework.

As I know, if two resistors are connected in parallel and one of them is short-circuited, then the other one is short-circuited too. But i don't think this works here as well. I'm wondering how to deal with a reactance connected in parallel with a short circuit.

 

[anorlunda]

It's assumed that the core losses are neglected and hence Rc =0.

No, it means Rc is infinite. Not short circuit, but open circuit.

When an impedance (R or X or C) is short circuited, it can be neglected in the circuit.

 

[GoodPost]

It makes sense the way you interpreted it, but it's still confusing me since Rc=0 is actually assumed in the problem statement and if the resistance = 0 that means it's a short circuit, right? Plz check the attached photo.

Thx.

 

[anorlunda]

In your schematic, if Rc=0, then it short circuits everything except Rs and Xs. Does that make sense to you?

 

[cnh1995]

but it's still confusing me since Rc=0 is actually assumed in the problem statement

I believe it's a typo.

Have you performed 'no load' and 'blocked rotor' tests on an induction motor in your lab? The R2/s in the circuit is actually the sum of two components: 1) rotor resistance R₂ (slip independent) and 2) the electrical equivalent of the mechanical load i.e. R_m=R₂*(1-s)/s .

While analysing the blocked rotor circuit, we substitute s=1 in the above circuit. This gives Rm=0, and this corresponds to maximum copper loss. While analysing no load circuit, we put s=0 and get Rm=∞, which corresponds to zero (almost) copper loss.