Circuit analysis potential difference

[hms.tech]

Homework Statement

What is the Potential difference between X and Y ?

Homework Equations

KVL and KCL

The Attempt at a Solution

I am confused about the two possible answers :
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 .

 

[gneill]

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

 

[hms.tech]

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

Ofcourse !

using KVL :

9 - 5I_{1} - 5I_{2} = 0 >>>eq 1
9 - 5I_{1} - 5I_{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) ?

 

[gneill]

Ofcourse !

using KVL :

9 - 5I_{1} - 5I_{2} = 0 >>>eq 1
9 - 5I_{1} - 5I_{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) ?

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.

 

[hms.tech]

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.

Does that mean my answer is correct ?

(your reply is quite ambiguous )

 

[gneill]

Does that mean my answer is correct ?

(your reply is quite ambiguous )

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

 

[hms.tech]

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

3 V

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

 

[gneill]

3 V

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

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?

 

[hms.tech]

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?

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.

 

[gneill]

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.

 

[hms.tech]

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