# Determining voltages for a network

I am solving some problems from my text book, and in this question (image below) I don't understand why Va is 13V and not 27V. It makes sense to me if Va were 27V so that Vab would be 21V. But this is wrong, why? This question is from the textbook 'Introductory Circuit Analysis'.

• This is strange. Can you include the full problem text? Mar 20, 2017 at 11:29
• Yeah, it looks like a trap. Mar 20, 2017 at 11:32
• Doesn't look like GND is 0V in this circuit. Mar 20, 2017 at 11:33
• The question says 'determine Va, Vb and Vab' that's it. But here's a screenshot anyway. prntscr.com/em7467 Mar 20, 2017 at 11:36

What you show is inconsistent with itself. If we consider the batteries to be perfect voltage sources, then this circuit can't exist.

You are explicitly told the left node is at 10 V. Since no reference is given, that can only be assumed to be relative to ground. The left battery adds 3 V to that, putting Va at 13 V.

However, the right battery puts Vb at 6 V. The middle battery then puts Va at 27 V.

Both can't be right, so this circuit is impossible as you presented it here.

Perhaps the text mentions some assumed internal resistances of the batteries? If so, you model the batteries as perfect voltage sources in series with their internal resistance. With some finite resistance in series with the batteries, this circuit is possible.

Either the textbook contains a major mistake, or there is some other implied information about this circuit you're not telling us.

As stated by Olin that circuit is inconsistent. The inconsistency has been corrected in a newer edition of the book. I checked the 12th edition, 2014, and the 6 V battery has been substituted with an 8 V one, with the polarity reversed, as represented in the circuit below.

simulate this circuit – Schematic created using CircuitLab

With the amended circuit, you get $V_a = 13\,\mathrm{V}$ from both sides.