# Nodal analysis exercise,are my equations correct?

• Questions999
Remember to use consistent units and directions for currents and voltages when summing them. Good luck with the rest of your analysis!

## Homework Statement

We have the circuit in the figure.I have to find v using nodal analysis only.

## Homework Equations

Converted 1/6 F to -3j and 2sin2t to -2j.

## The Attempt at a Solution

In node v1 we have the equation
6= -12+ [v1/2] + [( v1-v)/1]
In node v we have the equation
2j+ [(v1-v)/1]= 12 + [ v/(-3j)]

Solving these equations to find v is very easy,the crucial part is finding out if the two main equations here are correct.Are they?http://i.imgur.com/nAEsUpo.jpg

Elaia06 said:

## Homework Statement

We have the circuit in the figure.I have to find v using nodal analysis only.

## Homework Equations

Converted 1/6 F to -3j and 2sin2t to -2j.

## The Attempt at a Solution

In node v1 we have the equation
6= -12+ [v1/2] + [( v1-v)/1]
In node v we have the equation
2j+ [(v1-v)/1]= 12 + [ v/(-3j)]

Solving these equations to find v is very easy,the crucial part is finding out if the two main equations here are correct.Are they?http://i.imgur.com/nAEsUpo.jpg

They look basically right, but I usually prefer to just sum all of the currents leaving each node, and set each sum = 0. It's confusing having 2 sides to the equations, with some currents entering and some leaving each node. At least it is confusing for me. :-)

I would only question your handling of the sin() source. You've made the correct translation to a phasor value (-2j), then used +2j for it in your v-node equation on the left hand side where apparently you're summing currents flowing into the node.

I agree wholeheartedly with berkeman's suggestion about summing all currents and setting the sum to zero. Choose either in-flow or out-flow for the direction and be consistent. Do it the same way every time and you won't have to think about each current and the side of the equation to put it on and what sign to give it.

Many thanks to both of you :)
I guess finally I should write them like this?
6 +12- [v1/2] - [( v1-v)/1]=0
-2j+ [(v1-v)/1]- 12 - [ v/(-3j)]=0

Yes, looks good!