hi people ...
In the attached circuit, if we want to find Ix,we cannot say (V1-50)/0 because this yields infinity,and we know that V1 mainly = 50V, so this expression becomes (50-50)/0, namely 0/0, which is an undeterminate quantity, so how can such a current be evaluated ...?

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Here's how I solved it: Convert the current source and the 75 ohm resistor to the Thevenin equivalent. Write the two loop current equations. You have two equations and two unknowns. You don't even have to use determinants!

__________________
Ron

 

The thevenin equivalent that Ron refers to is that you can take any part out of the circuit and look at back into the resulting terminals.

Calculate the voltage you see at the terminals. This voltage becomes the thevenin or open circuit voltage.

Open current sources and short voltage sources and calculate the thevenin reistance.

The equivalent circuit then is an ideal voltage source equal to the thevenin voltage with the thevenin resistance in series.

If you then put your load back in you will know what voltage and current it will run at having simplified whatever mess of impedances and sources are in the circuit to a single source and impedance.

D.

 

thanks everybody for your solution, but it's well known that all methods of solution should imply the same results, however, if we say that Ix = (V1-50)/0 where V1 = 50 volts, this becomes 0/0 ...??????????!!!!!!!!!!!

 

"all method of solution should imply the same results.." yes, that's true.

0/0 is not a method of solution (indeterminate is not a solution)
You asked how and the answer was given. So what's the problem?

 

Both ends of your "zero ohm resistor" are the same node!

__________________
Ron