# Problem with open circuit

• inter060708
In summary, the conversation is discussing how to find the value of v1, v2, vab, vbc, and vca using Kirchoff's Law. The participants discuss how the current through the 3 and 4 ohm resistors is zero, and how this affects the voltage at different points in the circuit. They also mention removing the 3 ohm resistor and simplifying the circuit to solve for the voltages at different points.

## Homework Statement

Find the value of v1, v2, vab, vbc, vca. See picture

Kirchoff's Law

## The Attempt at a Solution

Is it true that both current through 3 and 4 ohm resistors are zero?
The way I think of this is if they aren't zero, there would be charge building up at a and b. Unless there are some way for them to neutralize, such as static electricity.

Thank You

Yes, it is true that the current through the 3 and 4 ohm resistors is zero.

Okay. In that case, there's no current in the bottom wire either? (Nodes connecting the negative ends of 7 and 2 ohm resistors)

inter060708 said:
Okay. In that case, there's no current in the bottom wire either? (Nodes connecting the negative ends of 7 and 2 ohm resistors)

Yes, I think the same logic applies. You can assign the potential at point c to be zero and work from there.

Thank you, Dick

The way I think of this is if they aren't zero, there would be charge building up at a and b. Unless there are some way for them to neutralize, such as static electricity.

Yes zero. Not sure why you think charge will build up.

With no current flowing through the 3Ohm there will be no voltage drop across the 3Ohm so both sides of the 3Ohm will be at the same voltage.

So how does one solve for Vab, vbc and vca? Since the current at a, b, and c are zero, that means the voltages at those points are zero too right? (V=IR --> a=0*3, b=0*4). Then wouldn't it just be zero minus zero for all them? (Vab = Va - Vb --> Vab = 0 - 0) Help please?

Doubledealer said:
So how does one solve for Vab, vbc and vca? Since the current at a, b, and c are zero, that means the voltages at those points are zero too right? (V=IR --> a=0*3, b=0*4). Then wouldn't it just be zero minus zero for all them? (Vab = Va - Vb --> Vab = 0 - 0) Help please?

No. Having no current through a resistor tells you the voltage at each terminal is the same. It doesn't tell you both are zero.

Dick said:
No. Having no current through a resistor tells you the voltage at each terminal is the same. It doesn't tell you both are zero.

So then how would you find the voltage? You can't use KVL because its not looped.

Doubledealer said:
So then how would you find the voltage? You can't use KVL because its not looped.
We are dealing here with a thread that is nearly 2 years old.

If you have a resistor, Ohm's law applies. E=IR. The potential difference (E) between the two terminals is equal to the current flowing (I) multiplied by the resistance of the resistor (R). If the current is zero, the voltage must be...

jbriggs444 said:
We are dealing here with a thread that is nearly 2 years old.

If you have a resistor, Ohm's law applies. E=IR. The potential difference (E) between the two terminals is equal to the current flowing (I) multiplied by the resistance of the resistor (R). If the current is zero, the voltage must be...

But the guy above me just said it's not zero.

Doubledealer said:
But the guy above me just said it's not zero.
The guy (@Dick) said that the potential difference between the two terminals is zero. And it is. That does not tell you the potential at both terminals, only that those two potentials are the same.

Edit: Since the potentials at either end of the 3 ohm resistor are the same and since that resistor carries no current, you can erase it from the drawing and move the label "a" to the black dot where the 3 ohm resistor had been attached. That is a key point that this exercise is driving at -- simplification.

jbriggs444 said:
The guy (@Dick) said that the potential difference between the two terminals is zero. And it is. That does not tell you the potential at both terminals, only that those two potentials are the same.

Edit: Since the potentials at either end of the 3 ohm resistor are the same and since that resistor carries no current, you can erase it from the drawing and move the label "a" to the black dot where the 3 ohm resistor had been attached. That is a key point that this exercise is driving at -- simplification.

oh I getcha. Sorry if I annoyed you for bringing up an old thread. Just still confused at how this open circuit works.

Doubledealer said:
oh I getcha. Sorry if I annoyed you for bringing up an old thread. Just still confused at how this open circuit works.
No problem. So are you comfortable with the simplification we made by removing the 3 ohm resistor? Can we proceed with some more simplification?

jbriggs444 said:
No problem. So are you comfortable with the simplification we made by removing the 3 ohm resistor? Can we proceed with some more simplification?

Oh yes you bet I am. I just want like the whole solution to the problem lol then discuss the parts I don't get. I got V1 = 14v and V2 = 6v but i don't get how to find the vab, vbc, vca ones.

Doubledealer said:
Oh yes you bet I am. I just want like the whole solution to the problem lol then discuss the parts I don't get. I got V1 = 14v and V2 = 6v but i don't get how to find the vab, vbc, vca ones.
So we agree that we can remove the 3 ohm resistor from the drawing without any problems.
Do you also agree that since there is a wire between "c" and the negative terminal of the 7 ohm resistor that those two points are at the same potential?

jbriggs444 said:
So we agree that we can remove the 3 ohm resistor from the drawing without any problems.
Do you also agree that since there is a wire between "c" and the negative terminal of the 7 ohm resistor that those two points are at the same potential?
Ok, but what about the negative terminal of the 2 ohm resistor?

Doubledealer said:
Ok, but what about the negative terminal of the 2 ohm resistor?
It too is connected by a wire leading to point c, so it too must be at that same potential.

jbriggs444 said:
It too is connected by a wire leading to point c, so it too must be at that same potential.
Oh ok, makes sense. But would that then mean the voltage from a to c is the same as v1? and does the same apply for b to c?

Doubledealer said:
Oh ok, makes sense. But would that then mean the voltage from a to c is the same as v1? and does the same apply for b to c?
BINGO!

jbriggs444 said:
BINGO!
Haha ok. So then Vca is -14V? Because Vc-Va. and is Vab then 14-6V=8V?

Doubledealer said:
Haha ok. So then Vca is -14V? Because Vc-Va. and is Vab then 14-6V=8V?
Yes, I agree with that result.

jbriggs444 said:
Yes, I agree with that result.
Ok then. Thanks so much! Oh and just to check Vbc is +6V?

Doubledealer said:
Ok then. Thanks so much! Oh and just to check Vbc is +6V?
Yes indeed.

inter060708 said:
The way I think of this is if they (currents through 3 and 4 ohm resistors) aren't zero, there would be charge building up at a and b. Unless there are some way for them to neutralize, such as static electricity.

There is no current flowing through the 3 and 4 ohm resistors, but there will be some electrostatic charge on terminals a and b because they are not at the same voltage.

David Lewis said:
There is no current flowing through the 3 and 4 ohm resistors, but there will be some electrostatic charge on terminals a and b because they are not at the same voltage.
inter060708 has not been around since April of 2015. There is little point responding to him.

In any case, if one is analyzing a circuit taking Kirchoff's laws as a given [see post #1] then electrostatic charge is not part of the model. It is idealized away.

davenn

## 1. What is an open circuit?

An open circuit is a type of electrical circuit in which the path for the flow of electricity is broken or interrupted by a gap or break in the circuit. This results in an incomplete circuit and prevents the flow of electricity.

## 2. What causes an open circuit?

An open circuit can be caused by a variety of factors, such as a loose connection, a broken wire, a blown fuse, or a faulty component. It can also be caused by physical damage to the circuit, such as corrosion or wear and tear.

## 3. How do I identify an open circuit?

An open circuit can be identified by using a multimeter to measure the resistance along the circuit. If there is an open circuit, the multimeter will show an infinite resistance, indicating that the circuit is not complete.

## 4. How do I fix an open circuit?

The first step in fixing an open circuit is to identify the location of the break or gap in the circuit. This can be done by visually inspecting the circuit or using a multimeter. Once the location is identified, the broken component or wire can be replaced or repaired to restore the circuit's continuity.

## 5. How can I prevent open circuits?

To prevent open circuits, it is important to use high-quality components and wires in the circuit. Regular maintenance and inspections can also help identify and address any potential issues before they become larger problems. Additionally, following proper safety precautions, such as turning off power before working on a circuit, can prevent physical damage that may lead to open circuits.