# How to solve for node voltages using mesh analysis?

I am given the following circuit to analyze:

The problem asks me to solve for the unknown currents and then find the node voltages at N1 and N2.
Here are the mesh equations I came up with:
$$Mesh1$$ $$24000i_1-20000i_2-4000i_3=20$$ $$Mesh2$$ $$-20000i_1+22200i_2-200i_3=0$$ $$Mesh3$$ $$-4000i_1-200i_2+610i_3=0$$
Solving for the system of equations, I get the following:

The thing I'm stuck with now is calculating the node voltages N1 and N2. Do I calculate the individual voltage drops on the resistors first?

• Yes, and you then work from the known voltage (20 V) at the V2/R6/R8 node.
– Chu
May 2 '19 at 5:53
• @Chu I just subtract the voltage drops from 20V right? May 2 '19 at 6:12
• e.g. if the voltage across R6 is, say, 5 V, with the polarity (+) on its left and (-) on its right (+ sign where the current enters the resistor), then N1 voltage is: 20-5 = 15 V. A neat way of doing this, generally, is: start from ground and do what the polarities tell you as you travel to your destination (node N1 in this case), thus: 0V at ground, then go up in voltage through the battery (0 +20 = 20V), then go down in voltage through R6 (20-5 = 15V). You can go on any path to the destination.
– Chu
May 2 '19 at 6:47
• ... btw, your current calculations aren't correct; i1 is the source current and must be +ve in this circuit as it's the only source, (assuming this is a complete circuit diagram and there isn't a V1 lurking somewhere!)
– Chu
May 2 '19 at 7:16
• I would quickly like to say that this is how homework questions should be asked on this site. It is the second one by OP, and both questions have shown clear effort, OP has shared their attempt, and reading through comments has a willingness to learn and understand where they went wrong. +1 from me!
– MCG
May 2 '19 at 7:48

It's often best to redraw the schematic. Sometimes, this works well in your favor. This is an unbalanced wheatstone bridge:

simulate this circuit – Schematic created using CircuitLab

This would be very easy with nodal. But you want mesh. So:

\begin{align*} 20\:\text{V}-\left(I_1+I_3\right)\cdot R_8-\left(I_1+I_2\right)\cdot R_7-I_1\cdot R_6&=20\:\text{V}\\\\ 0\:\text{V}-\left(I_2-I_3\right)\cdot R_9-\left(I_1+I_2\right)\cdot R_7-I_2\cdot R_{10}&=0\:\text{V}\\\\ 20\:\text{V}-\left(I_1+I_3\right)\cdot R_8-\left(I_3-I_2\right)\cdot R_9&=0\:\text{V} \end{align*}

You should be able to solve for $$\I_1\$$, $$\I_2\$$, and $$\I_3\$$ with ease.

Once you have the currents, it's very easy to solve for $$\V_x\$$ and $$\V_y\$$. For example, $$\V_x=20\:\text{V}-\left(I_1+I_3\right)\cdot R_8\$$.

Should you have wanted nodal, then it would be these:

\begin{align*} \frac{V_x}{R_7}+\frac{V_x}{R_8}+\frac{V_x}{R_9}&=\frac{V_y}{R_7}+\frac{20\:\text{V}}{R_8}+\frac{0\:\text{V}}{R_9}\\\\ \frac{V_y}{R_6}+\frac{V_y}{R_7}+\frac{V_y}{R_{10}}&=\frac{20\:\text{V}}{R_6}+\frac{V_x}{R_7}+\frac{0\:\text{V}}{R_{10}} \end{align*}

Here, you'd solve for just two voltages. And from there, you could get the currents if you wanted them.

Either way works.

• Just a question, are my mesh calculations correct? May 2 '19 at 6:10
• @Augue, No, they don't appear to be correct to me. Compare them and see what you think.
– jonk
May 2 '19 at 6:17
• Ok. Using your approach, I got Vx as 2.71V. Simulations show that Vx is 0.29V. May 2 '19 at 6:21
• @AugieJavax98, using nodal, I get Vx as .291. How do you get your value??
– jonk
May 2 '19 at 6:23
• @AugieJavax98 For example, $V_x=0.291116727511\:\text{V}$ and $V_y=0.1085837312\:\text{V}$. That's what I get with the nodal equations, anyway.
– jonk
May 2 '19 at 6:25