# Circuit analysis potential difference

1. Jan 18, 2013

### hms.tech

1. The problem statement, all variables and given/known data

What is the Potential difference between X and Y ?

2. Relevant equations

KVL and KCL

3. The attempt at a solution

1. 9V (From general concepts)
2. 3V (after applying the KVL and KCL)

I think the 2nd answer looks correct. Any thoughts ?
Can someone give a description and an explanation for this answer .

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2. Jan 18, 2013

### Staff: Mentor

Can you explain the reasoning, or preferably the calculations, behind the two suggested answers? Where did they come from?

3. Jan 18, 2013

### hms.tech

Ofcourse !

using KVL :

9 - 5$I_{1}$ - 5$I_{2}$ = 0 >>>eq 1
9 - 5$I_{1}$ - 5$I_{3}$ = 0 >>>eq 2
$I_{1}$ = $I_{2}$ + $I_{3}$ >>>eq 3

By comparing eq 1 and eq 2 : $I_{2}$ = $I_{3}$

Now by substitution we find that : $V_{1}$ (voltage across the first resistor the one closest to the 9V rail) = 6 V

This leaves 3 V for the resistors in parallel and hence the terminals X and Y.

Is this reasoning sound for rejecting the answer 9V (across X and Y) and confirming the answer as 3V (across X and Y) ?

4. Jan 18, 2013

### Staff: Mentor

Yes, the method is fine and the result, once the stated equations are solved in detail, will give the correct result and allow you to reject the other one.

5. Jan 18, 2013

### hms.tech

Does that mean my answer is correct ?

6. Jan 19, 2013

### Staff: Mentor

I didn't see a complete answer... just some (correct) steps along the path to one

7. Jan 19, 2013

### hms.tech

3 V

There is no working left to show (excluding solving the linear equations)

8. Jan 19, 2013

### Staff: Mentor

All right, so 3 V is correct using KCL and KVL.

Now, what about this "9V (From general concepts)" option? What's the argument for that, and is it good enough to discount the KVL/KCL solution?

9. Jan 19, 2013

### hms.tech

I will actually have to upload another circuit to prove my point here

I used this circuit (only a little different than the first) as a reference that if no current flows through the circuit, the p.d across any component is always the same as the e.m.f of the source . Although, given that I worked out the math, I would disregard this flawed concept of mine.

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10. Jan 19, 2013

### Staff: Mentor

Yup. This new circuit lacks the resistance in series with the voltage source, so it's strictly a parallel circuit where all branches have the same potential which is the same as that of the voltage source. The lack of the series resistor is a critical difference.

11. Jan 19, 2013

### hms.tech

thank you for the explanation in post no.10
I was looking for something like this