Ignore voltage reference on resistors?

[rp1242]

I'm working on 3.19 from Schaum's Outlines: Basic Circuit Analysis (not homework).

The problem can be found at this http://books.google.com/books?id=EP...resnum=1&ved=0CBMQ6AEwAA#v=onepage&q&f=false".

I'm confused about why you're supposed to ignore the voltage reference of V3 in this case. I'm looking at my circuits textbook from school and it says to subtract voltages when negative signs are encountered first and add voltages when positive signs comes first (KVL).

Can anyone help me?

 

[Rainier9]

Its because that voltage reference ignored the direction of the current flow. In other words, Voltage V3= -I*6 = -1.8
Then, doing a Kirchhoff...
Vab= VR2+5-V3+V4-8
Vab= (15*0.3)+5-(-1.8)+(8*0.5)-8
Vab= 5.7 Volts

But that's not the best way to solve this kind of problem.Remember that n elements in series share the same current, so if you sum the ohmic value of all the resistors, multiply it by the current in the circuit and then doing the algebraic sum of the voltage sources you got your answer. This way the only reference you have to take in consideration is the one of the voltage you are after. Make sure your current 'enters' by the positive reference.

Lets do the same example using my logic, this time using the left side of the circuit since it has less components.
10 ohms + 11 ohms = 21 ohms
21ohms multiplied by -0.3Amps = -6.3V
-6.3 plus the voltage source 12v = 5.7V
The current in this case is -0.3 becase it was initially defined in the other direction, and the voltageAB is defined such as the current 'enters' by its positive reference.

If the voltage had been defined with opposite references, it would have been like:
21*0.3= 6.3V
6.3 - 12 = -5.7V
In this case the voltage is defined such as the current 'enters' by its negative reference. Which is the same direction it was initially defined.

Finally, you can always change the voltage references acording to you likings and then change it back as what the problem asked for. So in that same problem you can define V3 with the opposites references and simply solve the equation the book provided. Though keep in mind that if you were also asked for the voltage V3 and you answered it with the opposite reference it would have been a wrong answer. In the original circuit V3 = -1.8Volts
The equation used by the book defined V3 as 1.8 Volts.

 

[rp1242]

Thanks, Rainier. That helps a lot!

For some reason, I thought that the voltage reference symbols indicated whether an element supplied or consumed power. This clears up my confusion about "resistors supplying power."

Thanks again!

 

[vk6kro]

You could add up the voltages going clockwise around the circuit:

Voltage = 12 V - 5 V + 8 V = 15 volts

Then add up the resistances

Total resistance = 10 + 15 + 6 + 8 + 11 = 50 ohms.

So, the current = 15 volts / 50 ohms = 0.3 amps flowing clockwise around the loop.

Voltage drop across any resistor = 0.3 amps * R

So voltage drop across the 6 ohm R = 0.3 amps * 6 ohms = 1.8 volts,
but the positive end of the resistor is at the top in the diagram because the current is flowing clockwise and enters this end first.