# KVL On A Circuit With 2 Voltage Sources

I want to apply Kirchhoff's Voltage Law on the circuit showed below. When I apply it, I form this equation : $$-20 + Vx - 3Vx - 6i = 0$$

However, my book forms this equation:

$$Vx + 3Vx + 6i - 20 = 0$$

What mistake am I doing here ? I'm assuming clockwise current flow.

simulate this circuit – Schematic created using CircuitLab

• Define how current i runs by adding arrows on every trace between every component (maybe bit overdone) and assign the corresponding + and - to each resistor. Commented Apr 3, 2019 at 13:09
• Remove + Vx - for now. And apply your own + and - according to this picture Commented Apr 3, 2019 at 13:18
• Your signs for R1 are not correct. Plain said: where the current enters the resistor should be +. Do remove the + Vx - from R2 (we need it later) and call this voltage $V_{R2}$ for now apply the correct signs here as well. Commented Apr 3, 2019 at 13:24
• After having defined $V_{R2}$ check whether it has the same polarity as how Vx is defined. (It has). Note that the signs around Vx determine how Vx is measured. Commented Apr 3, 2019 at 13:34

## 2 Answers

The sign of the source is drawn incorrectly or the polarity of how $$\V_x\$$ is defined should be reversed.

And most likely the equation is correct.

$$Vx + 3Vx + 6i - 20 = 0$$

together with the relation $$V_x=3\cdot i$$

yields

$$18i - 20 = 0$$

If it were $$Vx - 3Vx + 6i - 20 = 0$$ it would yield $$-20=0$$

• I have suspected the same thing but I simultaneously made the resistor polarity mistake. This book has many mistakes. You all have been really helpful. Commented Apr 3, 2019 at 13:54
• @NickDelta The book title is?
– G36
Commented Apr 3, 2019 at 13:55
• Fundamentals of Electric Circuits 4th edition - Charles K. Alexander , Matthew N.O Sadiku . But maybe it isn't their fault because the book is translated in Greek and maybe these mistakes have occurred in the translation process. Commented Apr 3, 2019 at 13:57
• How would you translate + and - ?? :-D Commented Apr 3, 2019 at 13:59
• I think the polarity of Vx should have been reversed. Because that way, the additional difficulty is the student should take into account that $V_{R2} = - V_x$ and the equation becomes $$3i - (-9i) + 6i - 20 = 0$$ Commented Apr 3, 2019 at 14:00

Resistors do not decide about the polarity. The current direction determines the resistor voltage drop polarity.

simulate this circuit – Schematic created using CircuitLab

• Very good answer, made a crucial mistake there but there's got to be something more as still that's not exactly what my book says. With your fix the equation becomes: $$-20+Vx-3Vx+6i=0$$ which is still not correct. Am I missing something in the polarity of the sources ? Commented Apr 3, 2019 at 13:39
• I think the book is wrong. The sign of $3V_x$ is incorrect. Commented Apr 3, 2019 at 13:43
• It seems that this is the case. The book shows the wrong polarity of a VCVS.
– G36
Commented Apr 3, 2019 at 13:54
• Your answer is misleading. What you state is true only if the value of the current and the value of the voltage are positive. When analyzing circuits we usually have to assume some particular direction of current flow and/or assume a direction of voltage polarity, and then solve the circuit. If any current or voltage turns out to be negative then the actual direction of the current or voltage is the opposite of what was assumed. But it is perfectly legal to assume that current enters the negative end of a resistor voltage. Commented Apr 3, 2019 at 14:07
• @ElliotAlderson Your remark confuses me. I thought the direction of (conventional) current is defined as the same direction as positive charges flow. So, drawing an arrow indicates the direction of a positive current. Is this correct? Or can the current have any sign independent of the direction of the drawn arrow? Commented Apr 4, 2019 at 10:09