# Circuit analysis with nodal analysis

[EDIT] I'm going to give mesh analysis another go, and see where that gets me. [DOUBLE EDIT] Mesh analysis didn't get me anywhere :(

Im tearing my hair out over this problem, so any help would be greatly appreciated.

Im solving the circuit with nodal analysis, and following the steps of: 1: Ground node equation

2: Voltage source and current source equations

3: Equations for each resistor

4: KCL at each node apart from the ground node

5: Substituting resistor equations into the KCL equations and solving for the node voltages

6: Solve for branch currents

7: Find power consumed or supplied

The problem is, i get to step 5, find a node voltage, but apparently it's the wrong answer. It doesn't match any of the answers supplied by the teacher (it's a practice multiple choice exam).

Attached is my working, with a picture of the circuit.

I get to a voltage for node B (circled in red), however, when I sub this in to find the voltage at node A, i get the wrong answer (12.8).

Please please please tell me what I am doing wrong.

Thank you in advance

• Wow, people actually use whiteout? Get a pencil! Commented Apr 5, 2013 at 7:32
• Problems like this are what made me drop my EE degree, and go for a buisness administration degree...Good luck. I am not sure if if would help you or not, but you can simulate this circuit on circuitlab.com. That was a life saver for me when I was doing EE.
– Reid
Commented Apr 5, 2013 at 7:33
• @ConnorWolf, i did it in pen to make sure it was actually readable when I took a photo of it. I do use a pencil most of the time ;) Commented Apr 5, 2013 at 7:37

First off, mesh analysis would be simpler because you would only have one unknown (the current in the right mesh).

As you know, in nodal analysis you are summing currents at the nodes. Current into the nodes positive and current out of the nodes negative. The way you have defined the problem you should have 3 equations and three unknowns. Your first equations seems to do this correctly for node B. It kind of falls apart after that. For example, node A equation should just be .005 = (Va-Vb)/3000. You already know that Vc = 16 volts.

So simplifying the node A equation gives you what Andy aka said, then just substitute it back into your node B equation.

V(A) must be 15v higher than V(B) because of the 5mA current flowing thru the 3k resistor.

So.... V(A) = V(B) + 15

I don't know if this helps in the way you want it to?

you've already obtained one equation relating v(a) and v(b), the other relation between v(a) and v(b) is to be obtained by using ohms law as you already have the knowledge of the current flowing and the resistance between nodes a and b. this way you will have 2 equations 2 variables and you can solve it.

for future problem solving you should avoid writing too much redundant variables (which, by observations can be eliminated)(writing too much variables and equations can also lead to confusion) like here you should've directly wrote 0.005A instead of I(1) and I(4); 16V instead of V(c).

WHy did you assume so many currents in the loop for mesh analysis?

Method 1 : Just use only one current that is in the BC branch because the BD branch current will be 5mA +/- BC branch current . So there will be only 1 variable in the 2nd loop.

Method 2 : Take only one node : point B & apply nodal analysis method .There will be only one equation with only one variable.