# Circuits - KCL Question

1. May 4, 2013

### GreenPrint

1. The problem statement, all variables and given/known data

[Broken]

2. Relevant equations

3. The attempt at a solution

For the node labeled Vload, why is the current in the 10 ohm resistor 100 Ix. I understand that there's a total of 100 Ix going into the node and so there must be 100 Ix going out of the node, however the dependent source makes it so that there's both 99Ix going into the node and 99Ix leaving the node, so there's 99Ix going through the 10 ohm resistor. How can there be two different currents going through the same wire?

I don't understand. The answer key solves the problem as if there's 100Ix going through the resistor.

Thanks for any help.

Last edited by a moderator: May 6, 2017
2. May 4, 2013

### Staff: Mentor

Don't ignore the ground connection on the Vload loop... current can "escape" there...

3. May 6, 2013

### GreenPrint

But why can current just escape to ground like that? I remember being told that current can be lost to ground, but why? I was never told why. Do you know?

4. May 6, 2013

### Staff: Mentor

Current isn't "lost to ground", so much as the ground path provides another path back to the source.

Remove all the ground symbols from your diagram and join those points with a wire. See any new loops?

5. May 7, 2013

### CWatters

Perhaps take a look at the wiring diagram for a car. The metal chassis is frequently used as a ground return path. A typical circuit loop would be ..

battery +ve....switch....light bulb....chassis....battery -ve

Most cars are "-ve earth" meaning that the chassis is connected to the -ve terminal of the battery. Some years back there were cars made that used a +ve earth system. These used the chassis to distribute the battery +ve. In that case the switching was done in the return path wiring. Take care if you are ever asked to help jump start a vintage car.