Circuit Segment (Two sources with a resistor in between)

[Bluestribute]

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?

 

[td21]

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?

 

[gneill]

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.

 

[Bluestribute]

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!

 

[HalfLostAndSearching]

In order to solve this problem, you can use Kirchhoff's voltage law (KVL) which states that the sum of the voltage drops around a closed loop in a circuit must equal the sum of the voltage sources in that loop. In this case, the loop would be from point a to point b and back to point a.

You can start by labeling the voltage drop across R1 as V1 (since it is in the direction of the current flow). Similarly, the voltage drop across R2 can be labeled as V2. From KVL, we can write the following equation:

V1 + V2 = V1 + V2 + Vb - Va

Since we know the values of V1 and V2, we can rearrange the equation to solve for the voltage difference Va - Vb:

Va - Vb = V2 - V1

Substituting the values given in the problem, we get:

Va - Vb = 6 - 4(6) = -18 V

This means that Va is 18 V lower than Vb. However, the question asks for the voltage difference, which is a positive value. Therefore, the voltage difference Va - Vb is 18 V. To find the actual values of Va and Vb, we need to add or subtract this value from the given voltages. Since Va is lower than Vb, we can write:

Va = Vb - (Va - Vb) = 16 - 18 = -2 V

Vb = 16 V

Therefore, the voltage difference between Va and Vb is 18 V, and the actual values of Va and Vb are -2 V and 16 V, respectively. This means that the voltage at point b is 2 V higher than the voltage at point a. Adding this voltage difference to the voltage at point a (which is 16 V) will give us the voltage at point b, which is 18 V. Therefore, the answer is 18 V.