# Solve a circuit with the nodal analysis (3)

I have the following circuit, and I have to find Vo:

I solved it in this way:

$$\\ \begin{cases} \frac{v_1 - 40}{1} + \frac{v_1 - v_0}{2} + 5 = 0 \\ \frac{v_1 - v_0}{2} + 5 = \frac{v_0 - (-20)}{8} + \frac{v_0}{4} \end{cases} \\ \begin{cases} 2v_1 - 80 + v_1 - v_0 + 10 = 0 \\ 4v_1 - 4v_0 + 40 - v_0 -20 - 2v_0 = 0 \end{cases} \\ \begin{cases} 3v_1 - v_0 = 70 \\ 4v_1 - 7v_0 = - 20 \end{cases} \\ v_0 = 3v_1 - 70 \\ 4v_1 - 21v_1 + 490 = -20 \\ 17v_1 = 510 \\ v_1 = \frac{510}{17} = 30 V \\ v_0 = 3v_1 - 70 = 20 \\ \begin{cases} v_1 = 30 V \\ v_0 = 20 V \end{cases}$$

But the solution is this:

I think that the error is in the solution, because in the first equation there is:

$$\frac{40 - v_0}{1}$$

$$\frac{40 - v_1}{1}$$

but I'm not sure. Is my solution right or wrong?

• Wrong, I can see without doing any sums that V0 will be positive (+40V from 3ohm source vs -20V from 8ohms source and you have that current source pumping it up too) so you must expect a positive answer. – user1582568 Feb 12 '16 at 15:19
• Yes, it looks like there's an error in the given solution, but you also have an error in your own setup of the problem, in the second equation. – Dave Tweed Feb 12 '16 at 15:24
• Its a badly drawn circuit, I presume that the top of the 4ohm is intended to be connected to the 2 and 8 ohm resistors, not just to the current source. – user1582568 Feb 12 '16 at 15:27
• @Andyaka, normally I agree 100% with your posts here, but on this one, I can't see where the OP has edited, the answer and the question look consistent to me – user1582568 Feb 12 '16 at 15:41
• Your updated solution is now correct. – Dave Tweed Feb 12 '16 at 15:46