What is the potential at point D in the circuit

[tots123450]

Homework Statement

I've attached a picture of the circuit below for reference, but the question is asking what the potential is at point D.

Homework Equations

V = IR[/B]

The Attempt at a Solution

I was told that the answer was 9V, but I'm a little confused because I thought if I looked at the battery on the left side that by the time the current starting at this battery reached point C, there would be 0 voltage and therefore 0 current. Going with this, I thought that if I kept going to point D it would still be 0 since there would be no voltage drop at resistor 5 if there was no longer any current. Therefore I thought I could find current I1 by taking 12V/3ohms = 4A

Looking at the battery on the right side I thought that by the same logic, the current starting at the right battery would reach 0 A by the time it reached point C. Likewise I thought current I2 would = 18V/3 ohms = 6A.

However when starting at point A and going around the left side path to point D, I would get 0V. But if I take the right side path using current 2 of 6A, I'm getting 18V - (6A*2ohms) = 6V.

I'm not sure what I'm doing wrong here; I know I'm definitely missing some conceptual bridge here.

 

[kuruman]

Hint: What is the potential difference between E and C?

 

[tots123450]

Is it not 0V; I was thinking that anywhere from A to E and A to C would be all 0 V, making the difference between E and C also 0, and that we could ignore the path with I3 as a whole also then.

 

[collinsmark]

Homework Statement

I've attached a picture of the circuit below for reference, but the question is asking what the potential is at point D.

Homework Equations

V = IR[/B]

The Attempt at a Solution

I was told that the answer was 9V, but I'm a little confused because I thought if I looked at the battery on the left side that by the time the current starting at this battery reached point C, there would be 0 voltage and therefore 0 current. Going with this, I thought that if I kept going to point D it would still be 0 since there would be no voltage drop at resistor 5 if there was no longer any current. Therefore I thought I could find current I1 by taking 12V/3ohms = 4A

Looking at the battery on the right side I thought that by the same logic, the current starting at the right battery would reach 0 A by the time it reached point C. Likewise I thought current I2 would = 18V/3 ohms = 6A.

However when starting at point A and going around the left side path to point D, I would get 0V. But if I take the right side path using current 2 of 6A, I'm getting 18V - (6A*2ohms) = 6V.

I'm not sure what I'm doing wrong here; I know I'm definitely missing some conceptual bridge here.

I think you are on the right track here. I also think that the person who told you that the answer is "9 V" is incorrect. (You might want to have them check their work.)

I'm assuming that I understand the diagram correctly in that there is a short between nodes A and C.

 

[kuruman]

Yes, it its zero. Straight lines in a circuit are equipotentials. so the potential is zero at points A, C and E. So you can draw a straight line joining C and E and disregard everything to the left of that line.

 

[kuruman]

I think you are on the right track here. I also think that the person who told you that the answer is "9 V" is incorrect. (You might want to have them check their work.)

I'm assuming that I understand the diagram correctly in that there is a short between nodes A and C.

I agree, 9 V is incorrect.

 

[tots123450]

Does that mean the voltage is 0 coming from the left battery? The thing I’m confused about is it doesn’t seem consistent coming from the right battery path

 

[kuruman]

Does that mean the voltage is 0 coming from the left battery? The thing I’m confused about is it doesn’t seem consistent coming from the right battery path

Voltage does not come out of a battery. The job of a battery is to maintain a constant potential difference across its terminals. Look at R₃. The potential difference across it is zero because its two ends are held at the same potential. What does Ohm's law say about the value of I₃? You said in post #3, I₃ = 0. This means that you can consider the circuit that you have as two separate circuits.

 

[tots123450]

You said that I could imagine the circuit with a straight line from C to E and everything to the left being ignored. I get that and I believe that would make the current I2 = 18/3 = 6A . Therefore the potential at D would be 18 - 2*6 = 12A. Is that correct ? If it is should I get the same value if I started with the battery on the left using I1 as my current value then ?

 

[kuruman]

Therefore the potential at D would be 18 - 2*6 = 12A.

No. First of all potential is expressed in Volts (V) not in Amps (A). Secondly, 18 - 2*6 = 12 is incorrect.

Consider this:
1. There is no current through I₃.
2. In branch EACD you have I₂ = 6 A.
3. In branch ABC you have I₁ = 4 A.
4. At A the two currents come together and you have 10 A flowing from A to C.
5. At C the currents split back into I₂ = 6 A and I₁ = 4 A and away they go in their own separate ways.