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

circuit

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:

solution

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

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

instead of:

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

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

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    \$\begingroup\$ 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. \$\endgroup\$ Commented Feb 12, 2016 at 15:19
  • \$\begingroup\$ 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. \$\endgroup\$ Commented Feb 12, 2016 at 15:24
  • \$\begingroup\$ 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. \$\endgroup\$ Commented Feb 12, 2016 at 15:27
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    \$\begingroup\$ @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 \$\endgroup\$ Commented Feb 12, 2016 at 15:41
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    \$\begingroup\$ Your updated solution is now correct. \$\endgroup\$ Commented Feb 12, 2016 at 15:46

1 Answer 1

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In your solution the second equation should have 8 in the denominator instead of 2: enter image description here

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  • \$\begingroup\$ It is a transcription error, the following calculations are not affected by this. However, there was another mistake, which now I have corrected. \$\endgroup\$ Commented Feb 12, 2016 at 15:34

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