S1E7: KCL-0

While further poking around in his experiment, Joe found an exposed interconnect. It connected the positive output of the V power supply, a resistor, and two unknown 2-terminal elements, as shown:

He couldn't trace where the other ends of the elements went. However, he had a nifty (very expensive!) Hall-effect clamp-on ammeter that he used to measure the three of the four currents entering the elements. He didn't have enough space to measure the current entering the fourth element. He found that A, A, A.

That is i1 = -0.7A, i3 = 3.0A, i4 = 1.3A

What was the current (in Amperes) into the fourth element?

So this is a KCL problem; all of the currents entering a junction/node must sum to 0.
Again, we need to establish a convention for our current equations:

• Following a current arrow from tail to head will be a "positive" term in the equation.
• Following a current arrow from head to tail will be a "negative" term in the equation.

(-i2) + (-i1) + (-i5) = 0


Hmmm, we don't yet have a value or an equation for i5. So let's solve the next problem and come back to this one...


... and we're back with the answer that i5 = 4.3 A


So we solve in terms of i2:


(-i2) = i1 + i5


i2 = -i1 - i5


i2 = -(-0.7) - 4.3 = 0.7 - 4.3 = -3.6 A

Joe managed to get his ammeter probe around a wire and measured . What current (in Amperes) did he measure?

Again, we follow the KCL convention we established before:


(+i5) + (-i4) + (-i3) = 0


i5 = i3 + i4


i5 = 3.0 + 1.3 = 4.3 A