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What is the potential at point D in the circuit

  • Thread starter tots123450
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  • #1

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.
 

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  • #2
kuruman
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Hint: What is the potential difference between E and C?
 
  • #3
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.
 
  • #4
collinsmark
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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.
 
  • #5
kuruman
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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.
 
  • #6
kuruman
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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.
 
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  • #7
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
 
  • #8
kuruman
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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.
 
  • #9
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 ?
 
  • #10
kuruman
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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.
 

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