- #1

- 8

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter madchiller
- Start date

- #1

- 8

- 0

- #2

berkeman

Mentor

- 60,567

- 10,885

http://en.wikipedia.org/wiki/Kirchhoff's_current_law#Kirchhoff.27s_Current_Law_.28KCL.29

Draw the polarities on the circuit before you write the equations, and stay with those polarities as you write all of the simultaneous mesh equations. You'll get the hang of it as you do more problems.

- #3

berkeman

Mentor

- 60,567

- 10,885

Welcome to the PF!

- #4

- 8

- 0

Welcome to the PF!

thanks a lot. But then who is right? if i get a negative voltage or current do i take the absolute value because the polarity that i chose was wrong?

tx for the help

- #5

- 31,970

- 8,881

No, you most certainly don't do that. What is means is that the actual current is going in the opposite direction.thanks a lot. But then who is right? if i get a negative voltage or current do i take the absolute value because the polarity that i chose was wrong?

tx for the help

The easiest way to do node voltage analysis is as follows:

1) label all of the node voltages in your circuit

2) pick a node (note, do not pick a nose)

3) write expressions for currents

4) write a node equation setting the sum of all currents leaving the node = 0

5) repeat 2)-4) for all other nodes

6) solve the system of equations for the node voltages

That's it. Just do that consistently and don't worry about the fact that some of the currents in 3) are actually coming into the node instead of leaving the node.

In 3), since the convention is to write expressions for current leaving the node then, if your circuit element is a resistor and if you are working at node i then you will always have an expression of the form (Vi-Vj)/R. If you solve the system and get Vj>Vi, all that means is that the current is actually entering the node instead of leaving it. Do not try to guess if Vj>Vi or not, always use the convention of currents leaving the node. This applies also when you get around to node j, when you are at node j you assume again that the current is leaving the node, i.e. at node j you write (Vj-Vi)/R. Your assumption of currents leaving will therefore always be wrong exactly 50% of the time, but as long as you are consistent you will get the right voltages.

- #6

- 8

- 0

crystal clear tx a lot for the help : )

Share: