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- Thread starter pantera1441
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- #1

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- #2

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1. By KCL the current at the node is equal to 0

Since we don't know anything about the direction of the currents, we can safely let the node voltage determine the direction.

2. Let the [tex]V_{x}[/tex] denote the node voltage

Let [tex]I_{1}[/tex] denote the current from 28.3 V source

Let [tex]I_{2}[/tex] denote the current from 14.15 V source

Let [tex]I_{3}[/tex] denote the current in the 16.5Ohm resistance

3. Applying the nodal equations we get the following:

[tex]I_{1}+I_{2}+I_{3}=0[/tex]

Note that the current can take any direction. Now we replace currents with voltage drops divided by resistor values:

[tex]\frac{V_{x}-28.3}{30.6}+\frac{V_{x}-14.15}{4.32}+\frac{V_{x}}{16.5}=0[/tex]

Solving for [tex]V_{x}[/tex] we get 12.9

Now we can solve for the current in the leftmost resistor:

[tex]I_{3}=\frac{V_{x}}{16.5}=0.784A[/tex]

DONE!

- #3

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thanks a lot for the help...

much obliged!

much obliged!

- #4

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thanks a lot for the help...

much obliged!

No problem, I hope you understood the steps towards the solution.

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- #6

russ_watters

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- #7

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Correct and remember if the batteries were switched the other way around, then you would use [tex]V_{x}+V_{battery}[/tex] instead

I won't argue about that, but i thought that making the complete description of the steps including the description of what rules i applied would help him better, so he/she will know what to do next time he/she is faced with the problem.

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- #9

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do you think you can "push me in the right direction" on this one, I know they're looking for the difference between A and B, but i forget how to do the voltage drop and i'm clueless when there's more than one battery

Hehe, shure. First you have to determine the current flowing through the loop, note that the current will be the same at all places. When You've determined the current in the loop, start by writing the equation expressed as voltage drops , you should start at point a and move towards point b, note that the first voltage drop across the battery will be negative, since you're moving from -to +, and it will be positive across the resistors and positive across the battery that is near the b point, since the battery is going from + to -.

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- #11

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I tried 40.86V and that was wrong, is it 25V - 15.86V = 9.14V as the difference

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- #12

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This should make it easier to work out the circuit voltage and hence the current.

- #13

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Thanks,

You'll be hearing from me again this week about RC Circuits.....totally new to me

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LOL, ok ;)

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