Circuit analysis, simple KVL

Question

I have come across this example problem and can't determine the why behind why this comes out the way it does:

So in figure 3.9 part a) they get 4 volts for \$v_{R2}\$, however the negative voltage is throwing me off for example:

KVL around that first loop should be:

$$-12V + v_{R2} + (-8V) = 0$$

This is clearly wrong but I have no idea why and am seaking an explanation.

The answer is messed up. You are right. The author salted signs all over the place and failed to account for them. I think the author's mistake was the \$-8\:\text{V}\$ across a resistor marked with signs around it that determine the supposedly intended polarity as it relates to the \$-8\:\text{V}\$ denotation. But if you take both the sign marks around that resistor, and also the sign of the magnitude given, then you are forced to conclude that the author messed up with their own answer. Your brain is fine and you do not need to send out for a replacement. — jonk, Commented May 29, 2018 at 14:19

johnger · Accepted Answer · 2018-05-29 16:03:49Z

1

i think both answer are wrong, "-12+Vr2+(-8)=0" gives Vr2=20, and from errata : Errata link : http://docdro.id/nufs1QS

answered May 29, 2018 at 15:53

johnger

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\$\begingroup\$ But the answer "-12V + Vr2 +(-8) = 0" gives Vr2 = 20 which is what is given in p. 44 practice 3.3 answer (a) from errata? \$\endgroup\$
– Anonymous3.1415
Commented May 29, 2018 at 15:58
\$\begingroup\$ that is a picture from errata, getting 20 for Vr2 made me look into errata, which corrects wrong 4 to correct 20 \$\endgroup\$
– johnger
Commented May 29, 2018 at 15:59
1

\$\begingroup\$ Oh I see what you meant thank you that book has alot of errors I may be looking into another learning reasource \$\endgroup\$
– Anonymous3.1415
Commented May 29, 2018 at 16:11

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