# Applying KCL problem

1. Feb 21, 2010

### mike41

i have been trouble with this. iunderstand the concept but i dont know how to set up the current directions onto a circuit when i apply KCL. Im not sure which way the arrows should be going when doing kcl. for example like in this problem.

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2. Feb 21, 2010

### Okefenokee

Current flows from the "+" side to the "-" side. So, for the dependent current source you have +V3 - V2 = 5i.

You can draw the current arrows through a resistor anyway you like. The arrow tells us to take the voltage at the start of the arrow and minus the voltage at the end of the arrow. Sources have an explicit polarity so you should choose an arrow that goes from it's positive to negative.

3. Feb 21, 2010

### mike41

ok but lets say for Node v1 how do i draw the currents

would this be correct. this is the part that is holding me back

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4. Feb 21, 2010

### Okefenokee

The 10V source doesn't need a current arrow. V1 will always be exactly 10V higher than V2. The equation for that branch will by +V1 - V2 = 10V. The rest looks fine.

Look at the direction of the arrow that you drew for the 3 Ohm resistor. It implies this formula: i3 = (0V - V3)/(3Ohm)= -V3/(3Ohm). The math will still work out just fine.

5. Feb 21, 2010

### mike41

ok so i can choose what ever i want for my reference at first, but i must stick to the current flow for the whole circuit and the math will work then?

6. Feb 21, 2010

### Okefenokee

exactly. You can flip the arrows and repeat the problem. You'll get negative currents where you had positive currents the first time and vice versa. A negative current implies that it flows in the opposite direction of the reference arrow that you drew.

7. Feb 21, 2010

### mike41

how about choosing your reference nodes. is there something special to that? choosing the node above the 60v wouldnt be the best choice would it?