# Simple KVL Problem Not So Simple

We're only a week in to the first EE class here at Uni and I'm already stumped. I can build a power supply in my sleep but anything theory-related turns me upside down. Now, I've done KVL on the following circuit six times now, and have been working at it for nearly an hour, and something just isn't making sense here.

simulate this circuit – Schematic created using CircuitLab

I'm trying to find the voltage across the 25 ohm and 10 ohm resistors, as well as the power absorbed by the 25 ohm. It really seems like I'm missing something extremely simple here, but my calculations just aren't working out properly.

Here's how I approached KVL the most recent time. V3 is the 15 ohm resistor's voltage drop and the sign (from my understanding) shouldn't matter as it will work itself out when solve. Starting from the bottom-left:

-20V + V3 + 30V + V1 - V2 = 0

*simplified*

10V + V3 + V1 = V2


That's easy. Now since the circuit elements are all in series the current should be the same for all elements. I applied ohm's law as follows:

V1 = IR1 = 25I
V2 = IR2 = 10I
V3 = IR3 = 15I


Looking good, narrowing down to one variable lets me solve for current.

10V + 15I + 25I = 10I
10V + 40I = 10I
10V = -30I
-1/3 A = I


Here I assume that current is just flowing counter-clockwise when we approached KVL in the clockwise direction, which is where the negative sign comes from.

So finding voltages across the resistors should be as easy as plugging the current back in to ohm's law.

V1 = 25(1/3) = 8.333 V
V2 = 10(1/3) = 3.333 V
V3 = 15(1/3) = 5 V


This is where I hit a snag. These are not correct according to the answers in the book. I don't know where I'm going wrong here. This is the last in a series of KVL/KCL problems and I had no issues with any of the other circuits. For some reason this one is just throwing me way off.

If I'm missing something very fundamental here, please tell me. If my assumptions stated above are dead wrong, please tell me. I like to think I know a decent amount about hands-on electrical topics, but this theory stuff really makes me struggle.

• Since R1 and R2 are in series, and hence the currrent through them is the same and have the same direction, their voltage must have the same polarity. Also, specify the voltage polarity on R3, too (the same as R1 and R2). Aug 31 '16 at 2:43
• Will changing the voltage polarity affect the numerical answer other than the signs? I need to give the voltage using the polarity provided, which is the polarity I showed above. Also, R3 is not given a polarity in the problem so I gave it one in the KVL equation further down. Aug 31 '16 at 2:45
• supply polarity yes, measured polarity no.. Do you see 10V/50 Ohms? Aug 31 '16 at 2:57

Your mistake is at the line V2 = IR2 = 10I. That should be V2 = -IR2 = -10I because you've assumed a voltage opposite the direction of the current.

• I didn't even consider the fact that the current was flowing opposite of the voltage. Once again, the theoretical nonsense is just messing with my head. Aug 31 '16 at 3:19

That's why usually when analyzing circuits (if we are the ones assigning polarities) we go with the passive sign convention, where the currents flow into the terminal marked as positive (that way we don't have to use negative sign for Ohm's law).

Just to add a little more to the accepted answer, here a picture with the different directions we can assume for the current and the polarities and their corresponding Ohm's law relation:

Most of the time we stick to option (i) or (iv), although the other ones work, too, but you need to keep track of the signs.

It is simple to solve the circuit using Kirchhoff's voltage law paying careful attention to the voltage directions.

Starting with $$\V_1\$$ and working along a current $$\I\$$, clockwise around the circuit:

$$+20~\textrm{V} - I\cdot 15~\Omega - 30~\textrm{V}- I\cdot 25~\Omega- I\cdot 10~\Omega=0.$$

(Ignore the $$\V_1\$$ and $$\V_2\$$ next to the resistors they will only add to the confusion!)

Therefore $$-10~\textrm{V} = I\cdot 50~\Omega$$ and $$I = \frac{-10~\textrm{V}}{ 50~\Omega} = -0.5~\textrm{A}.$$

Therefore the actual current is flowing anti-clockwise.

(You can imagine that the resistors "consumes" voltage along the current direction.)

series circuit so equivalent resistance is 50 ohms. two sources hooked up working against one another in series so subtract for effective voltage, 10 volts. series circuit so current is the same all over I=V/R= 10/50= 0.2A.

voltage used by 25 ohm resistor is V=IR= 0.2(25)= 5V.

voltage used by 10 ohm resistor is V=IR= 0.2(10)= 2V (the last resistor, 15 ohms, uses the left over 3V from the 10V total in the circuit).

power consumption of the 25 ohm resistor is P=IV= 0.2(5)= 1W.

Wonder if that's how to analyze the circuit?

• No, I don't think this is how the OP wants to solve the circuit. They were looking for guidance in using KVL specifically. Generally, we avoid providing answers to homework questions here unless the OP shows a significant effort and asks a specific question. As you can see from the other answers, we prefer to help the OP learn how to answer their own question. Apr 8 '20 at 20:32