Engineering Voltage drop w/ series/parallel circuit

[rms5643]

Homework Statement

Find the voltage drop (V0) across the 3 kΩ resistor
Circuit Pic: http://i.imgur.com/kX7hlQK.png

CS1= 14mA
CS2= 27mA
R1 = 7kΩ
R2 = 7kΩ
R3 = 3kΩ

Homework Equations

Ohms Law & Current Division

The Attempt at a Solution

1) First thing I did was combine the two current sources. Since they are pointing in opposite directions, 27mA -14mA = 13mA (I'll label this as CSE)

2) Next, I tried to find the current going through R2 & R3, the resistors in series, using current division by solving: R1/(R1+(R2+R3))*(CSE) -> 7000/(7000+(3000+7000))*0.013 = 5.3mA

3) Now that I know the current through R2 & R3, I used Ohms Law to find the voltage drop across R3, the resistor in question. V0=0.0053mA*3000=16.0588V, which is incorrect.

Note: The hint in this problem says to use voltage division after combing appropriate sources and resistors, but voltage division can only be used with elements in series, not parallel, right?

 

[NascentOxygen]

Hi rms5643. Welcome to Physics Forums!

Have you established whether Vo will be positive or negative?

 

[rms5643]

Hello Nascent!

That is one of my struggles with EE, I'm never really sure when things are negative or positive. Now that you mention it, since the 27 amp source is greater than the 14 amp source, the current would start at the bottom of the circuit, right? So, I'm not sure if this is the correct reasoning, but given the orientation of Vo in the problem, since the current is going from the negative terminal to the positive terminal, the voltage drop is negative? Is this the correct reasoning?

 

[NascentOxygen]

Current flows in an endless loop, neither starting somewhere nor ending anywhere.

Once you determine the direction of current "inside" R3 you can write + and - at either end of R3. How to tell? Use your knowledge that current always flows from a point of higher potential ("more positive") to a point of lower potential ("less positive" or "more negative"), both inside the resistor and in the circuit around it. Focus on R3. The end that current inside R3 flows from is marked + and the end it flows towards gets marked -.

If a voltmeter were to be connected across R3 with its red probe on the less-positive end of R3, the meter will read backwards, recording a negative reading. The arrow-head of Vo corresponds to where the red probe would be connected.

 

[rude man]

3) Now that I know the current through R2 & R3, I used Ohms Law to find the voltage drop across R3, the resistor in question. V0=0.0053mA*3000=16.0588V, which is incorrect.

Sure is. Try that multiplication again.
And pay attention to polarity as Nascent pointed out.

 

[rms5643]

Thanks for your help guys. Since the current would be entering the the negative terminal of R3 into the positive terminal (With respect to the picture), as Nascent said, the voltmeter would read the value as negative.

So, the correct answer was -16.058V!

 

[NascentOxygen]

It is easy to overlook the polarity.

It is best not to speak (or write) of the positive or negative terminals of a resistor, as resistors almost never have a polarity.