Why can't I add the voltage drop?

[flyingpig]

Why can't I add the voltage drop??

Homework Statement

What are the expected readings of the ammeter and voltmeter for the circuit in the figure below? (R = 6.00Ω , ΔV = 6.00 V.)

The Attempt at a Solution

I already solve for the ammeter, but the voltmeter is confusing me.

Sorry for the difference in pic size, Imageshack has gotten greedy and I can only attach some image files now

http://img156.imageshack.us/img156/692/p1863alt.th.gif

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[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP646619f154e92fe29514000016e30c7af875h3e4?MSPStoreType=image/gif&s=20&w=152&h=56

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP318819f1579c665h3gc900001dbf341h0bihh22e?MSPStoreType=image/gif&s=17&w=151&h=56


So I_A = 0.395A = I_C and I_B = 0.0315789A

Since we are only concerned with the magnitudes, I am going to leave out the signs for now.

So the current flowing into the 10Ω resistor is 0.395A - 0.0315789A = 0.363A and current down the 5.0Ω is 0.427A

So 5.0Ω * 0.427A = 2.133V and 10Ω * 0.363A = 3.63V

3.63V + 2.133V = 5.763V

Apparently the answer is 1.50V, which suggests it is 3.63V - 2.133V = 1.50V

Or that we could do 6.00V - 4.50V = 1.5V

Now the question, what is the real solution? And?

 

[gneill]

The voltmeter voltage is fixed by the two voltage sources. So V = 6.00V - 4.50V.

 

[SammyS]

Either way should work. They should give the same answer & they do.

 

[flyingpig]

No, but I am asking why. Because I thought I could get a equivalence voltage by adding the voltage drops, but I am not getting them

 

[gneill]

You have to pay attention to the directions of the net voltage drops on the resistors. The largest loop current flowing through a given resistor will tell you the polarity.

note: 3.63V - 2.133V ≈ 1.5V