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### Topic: Circuit Analysis (Read 597 times)previous topic - next topic

#### Bennington ##### Feb 19, 2016, 07:37 am
Hello,

I'm reading a book to teach myself electronics. There's a small section devoted to circuit analysis from which I have two questions.

I1 then reaches node A and flows through R3 downwards into node B and thus completes its loop. Now we can write -V+Vr1+Vr3=0. We don't want voltages, so we substitute Vr1=I1R1 and Vr3=(I1-I2)R3.

Why is it in the first equation the author places the voltage in the negative? Does this have anything to do with voltage drop?

Why is it in the last equation the voltage of R3 is calculated by multiplying the difference of the currents in the two loops?

I've attached an image of the schematic.

#### MorganS #1
##### Feb 19, 2016, 08:10 am
Are you sure you've copied that correctly? not -V1 + Vr1 + Vr3 = 0 ?

It's negative because if you follow the current around the loop, the different voltages are pointing different directions. The + on Vr1 is facing the + on V1.
"The problem is in the code you didn't post."

#### Hutkikz #2
##### Feb 19, 2016, 01:14 pmLast Edit: Feb 19, 2016, 01:27 pm by Hutkikz
in both questions it is because the voltages/currents are in opposite directions.

VR3 = ( I1 + (-I2) ) R3

if you rearrange it this way it becomes clear that he is adding currents that are flowing in opposite directions.

#### MarkT #3
##### Feb 19, 2016, 08:02 pm
voltages and currents have a sign, voltages are always the difference between two points in
a circuit, and sum to zero round a loop (if there's no induction).  You have to sum all the way
round the loop consistently, either always clockwise or always anticlockwise.
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#### tmd3 #4
##### Feb 19, 2016, 08:20 pmLast Edit: Feb 19, 2016, 08:21 pm by tmd3
Kirhoff's voltage law is sometimes rendered as, "the sum of the voltage drops around a loop is zero."  For a bit of charge starting at point B and going around loop 1, the voltage rises through the power supply, drops through resistor R1, and drops through resistor R3, if the reference direction for R3 is seen as positive at point A and negative at point B.  The voltage rise through the power supply V1 can be described as a negative voltage drop, so V1 appears as a negative term.

I note that there's a discrepancy between the drawing and the equation you posted,
-V+Vr1+Vr3=0
Each resistor in the drawing has a polarity marker indicating the reference direction for the voltage across it.  R3's polarity marker is at point B.  Traveling aroung loop 1, the voltage across R3 comes up as a voltage rise rather than a voltage drop, so I'd expect to see its appearance in the equation as negative, like this:
-V1 + VR2 - VR3 = 0
The sign of VR3 works itself out by how the current is represented:  The voltage across the resistor is the resistance multiplied by  the current entering the polarity mark.  In this case, that's R3 * (I2 - I1).  I1 appears as a negative term because the reference direction for I1 shown in the drawing enters the other end of the resistor.  That's the opposite of the signs of the two loop currents shown in your post:
Quote
Vr3=(I1-I2)R3
so, the math in your post is consistent.  It's just not entirely consistent with the reference directions for voltage shown in the drawing.

Edit: clarity

#### MarkT #5
##### Feb 19, 2016, 10:52 pm
Of course to be complete I will emphasize that Kirchoff's voltage law ignores magnetic induction, so it
doesn't for instance apply to the winding on a transformer.  At DC and for most purposes its perfectly
fine of course, but its an approximation, not a law really.
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