Current Division Problem

Thread Starter

Connor Pupp

Joined Sep 29, 2019
5
I am trying to solve for Io using current division, but I can't figure out how to make it all in parallel to use the formula. Any help would be appreciated.
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WBahn

Joined Mar 31, 2012
24,692
Have you don't anything recently with Thevenin or Norton equivalent circuits?
 

WBahn

Joined Mar 31, 2012
24,692
Not in the frequency domain, is it the same as in the time domain?
Yep.

That's the primary motivation for working in the complex frequency domain -- we get to turn hideous time-domain differential equations for time-varying signals into familiar tame equations analogous to our friendly DC circuit equations.
 

Thread Starter

Connor Pupp

Joined Sep 29, 2019
5
Yep.

That's the primary motivation for working in the complex frequency domain -- we get to turn hideous time-domain differential equations for time-varying signals into familiar tame equations analogous to our friendly DC circuit equations.
How would you end up finding Io though? You would instantly loose it
 

Thread Starter

Connor Pupp

Joined Sep 29, 2019
5
Why would you instantly lose it?

Treat the component with I0 as your load and find the Norton equivalent of everything else.
How would you start to combine the elements without first doing Io with -j2 ohm? Even using a delta - wye transform wouldn't help that much. Unless I'm missing something?
 

Thread Starter

Connor Pupp

Joined Sep 29, 2019
5
Never mind, I figured it out. I combined the Io resistor with the -j2 ohm in parallel, then I did current division to find the current through the 4 ohm inductor and then did current division again to find Io
 

WBahn

Joined Mar 31, 2012
24,692
How would you start to combine the elements without first doing Io with -j2 ohm? Even using a delta - wye transform wouldn't help that much. Unless I'm missing something?
Fall back on what the entire meaning of a Thevenin or Norton equivalent circuit is.

Remove the "load" and leave it open circuited. Since there are no dependent sources, set all of the independent sources to zero. Find the impedance as seen between the load terminals. In this case, the result is immediately seen to be infinite impedance, meaning that it will be an ideal current source (and, consequently, an undefined voltage source).

Now short circuit the load and determine the current in it. That's the Norton equivalent current. This immediately turns out to be -Is (the negative of the current source output).

So you have an ideal current source of -5 A driving a 2 Ω resistor, which yields a current of -5 A (so 5 A flowing upward) and a voltage across the resistor of -10 V (top relative to bottom).
 

MrAl

Joined Jun 17, 2014
6,471
Never mind, I figured it out. I combined the Io resistor with the -j2 ohm in parallel, then I did current division to find the current through the 4 ohm inductor and then did current division again to find Io
Hello there,

You would do well to also look up Thevenin and Norton source transformations. I cant begin to tell you how much this can simplify things sometimes if not always.
 
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