KVL Analysis for Solving Vx: Why Can We Ignore the 6 Amp Source?

[Jeremy Burke]

the question is to solve Vx. chegg did a kvl in the middle mesh and it turned out to be -2+Vx+8=0, i was wondering why they just ignored the 6 amp source in the kvl. i understand no voltage can be dropped across an ideal current source so why can we just skip over the 6 amp source and go to the 2 ohm resistor

 

[gneill]

the question is to solve Vx. chegg did a kvl in the middle mesh and it turned out to be -2+Vx+8=0, i was wondering why they just ignored the 6 amp source in the kvl. i understand no voltage can be dropped across an ideal current source so why can we just skip over the 6 amp source and go to the 2 ohm resistor

The 6 A current source and the 2 Ω resistor are in parallel, so they MUST have the same potential drop.

An ideal current source will produce any potential difference necessary in order that it maintains its specified current.

 

[Jeremy Burke]

then i don't understand why the solution was showing -2. since the current is point upward, wouldn't that make the positive end on the bottom, making it +2 and not -2

 

[gneill]

then i don't understand why the solution was showing -2. since the current is point upward, wouldn't that make the positive end on the bottom, making it +2 and not -2

Current sources don't care about voltage polarity. They simply maintain the required current in the specified direction. Again, an ideal current source will produce any potential difference necessary in order that it maintains its specified current. Even if that potential change is negative in the direction of the current.

 

[Jeremy Burke]

then how do we decide whether or not the 2 is positive or negative

 

[gneill]

then how do we decide whether or not the 2 is positive or negative

In this case, taking your "KVL walk" around the loop including the 2 Ω resistor does the trick. In the figure below, the first potential change is -2 V as you "walk" through that resistor: