Circuit Segment (Two sources with a resistor in between)

AI Thread Summary
The discussion focuses on calculating the voltage difference Va−Vb in a circuit with two voltage sources and a resistor. The current is given as 4 A, with resistors R1 and R2 both at 6 ohms, and voltage sources V1 at 16 V and V2 at 6 V. The solution involves using Ohm's Law (V=IR) and Kirchhoff's Voltage Law (KVL) to account for potential drops across the resistors and the effects of the voltage sources. The final answer is determined to be 58 V, with a hint suggesting a relationship between the contributions of the voltage sources. The discussion emphasizes the importance of correctly assigning potential drops and the direction of current flow in the calculations.
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Homework Statement


If a current I=4 A exists between points a and b, R1=6 [PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmr10/alpha/100/char0A.png, R2=6 [PLAIN]http://loncapa.mines.edu/adm/jsMath/fonts/cmr10/alpha/100/char0A.png, V1=16 V, and V2=6 V,the voltage difference Va−Vb is

Homework Equations


V=IR

The Attempt at a Solution


Well, I do V=IR for the first section up to source V1 (so V=4(6)), but once I get in between, I don't know how to deal with the inside having two voltage sources bombarding it. The answer is 58 V (it's multiple choice), but how?
 

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From the answer, you can guess the physics. Hint: 24 + 24 + x = 58?
What does the value of x imply the direction of the potential difference caused by the voltage sources?
 
You have the current and its direction. So you can assign potential drops across the resistors. Pencil them in on your diagram. Then do a "KVL walk" from b to a summing up the potential changes as you go.
 
gneill said:
You have the current and its direction. So you can assign potential drops across the resistors. Pencil them in on your diagram. Then do a "KVL walk" from b to a summing up the potential changes as you go.
There we go! I thought you had to do something with the two sources going towards the resistor.

I just went from A to B doing Kirchoff Law calcs (so going in the direction of the current was a negative, going from positive to negative terminals was negative, and the reverses were positive) and am getting it now. Thank ya!
 
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