Solving a Circuit: Finding Voltage at Node 3

[pat666]

Homework Statement

I've been trying this Q for a week now and still can't get the answer correct. I've used a lab and multisim to know what the answer should be but my equations are wrong.

see attached for circuit

Homework Equations

The Attempt at a Solution

node 3 is 0v because its on the ground
0= (v_1-10)/4k7-v_1/10k-(v_2-v_1)/1k
0=(v_2-v_1)/1000-(v_2-v_4)/470-(v_2-v_4)/1000
0=(v_4-v_2)1k-(v_4-v_2)/470-(v_4-5)/10000-v_4/10k

That comes out to 9,8 and 7 volts respectively which is way off.

Thanks for any help.

 

[Jmf]

I haven't done the analysis myself (I will if you would like to see it, since you've obviously made a good attempt at the question) but I believe there are sign errors in your equations.

When doing node analysis, you're applying Kirchoff's Current Law, so for each node you can write, for example:

current into node 1 from node 2 + current into node 1 from node 3 + current into node 1 from node 4 = 0

or you can do the sum of current out of the node instead, if you wish. Note that there are no minus signs in the equation above, or it no longer holds.

What you seem to have done in your first equation is:

Current out of node 1 to node 2 - current out of node 1 to node 3 - current into node 1 from node 4 = 0

Notice the inconsistency? try to be more systematic about generating your equations and you should get there.

 

[pat666]

Hey,
I thought that there might have been something wrong with the signs. How do I tell the directions of the current? We have been taught current out of node = 0 so that's what I'm trying to do. and I wouldn't mind seeing how you do it so that I can try and see how you assign the current directions.

Thanks for the reply

 

[Jmf]

Sure; I'll just do the equation for node 1, then I can go into some detail. And I'll use the currents out of the node, since that's what you've learned:

by Ohm's law, the current going from node 1 to power supply 1 is:

\frac{v_1 - 10}{4 700}

note that I put v_1 first and 10V second. This makes sense, since it will be positive if v_1 is larger than 10: i.e. current will flow out of the node if the node voltage is larger than the battery voltage.

Similarily, the current from node 1 to node 3 is:

\frac{v_1 - 0}{10 000}

and from node 1 to node 2:

\frac{v_1 - v_2}{1 000}

So our equation for node 1 becomes:

\frac{v_1 - 10}{4 700} + \frac{v_1}{10 000} + \frac{v_1 - v_2}{1 000} = 0

Can you do the other two?

 

[pat666]

Hey I think I have it now but would you mind checking because it does differ a bit from multisim (not much) . I attached my solution.

Thanks