What is the potential difference across the resistor in this circuit?

[Shark 774]

Hi. In the attached document, is the potential difference across the resistor 0V or 15V? I have read two different books that say two different things. Personally I thought 0V seemed correct because no current is flowing hence how can there be a potential difference?

 

[Drakkith]

Are you asking for the voltage across the resistor?
This is from wikipedia:

Ohm's law states that the voltage (V) across a resistor is proportional to the current (I) passing through it, where the constant of proportionality is the resistance (R).

Equivalently, Ohm's law can be stated:


This formulation of Ohm's law states that, when a voltage (V) is present across a resistance (R), a current (I) will flow through the resistance. This is directly used in practical computations. For example, if a 300 ohm resistor is attached across the terminals of a 12 volt battery, then a current of 12 / 300 = 0.04 amperes (or 40 milliamperes) will flow through that resistor.

What did you read that said it was 0V?

 

[Shark 774]

Are you asking for the voltage across the resistor?
This is from wikipedia:



What did you read that said it was 0V?

In my school course the word "voltage" has been changed to "potential difference" for whatever reason, I don't know. That wikipedia piece seems to me to confirm that it should be 0V because if there is a voltage across a resistor than a current will flow through it. No current through it hence no voltage. They had an example in one of my textbooks like the one I gave and the answer was 0V but then another one of my texts books had a similar example that said 15V, hence my confusion.

 

[Drakkith]

What do you mean? If you are applying a voltage, then current will flow through the resistor. Your example has 15 volts being applied, correct?

 

[willem2]

What do you mean? If you are applying a voltage, then current will flow through the resistor. Your example has 15 volts being applied, correct?

It looks like those 2 points at the bottom aren't connected, so no current can flow.

 

[Evil Bunny]

The circuit, as drawn, is an open circuit. There is nothing connecting the the negative side to the resistor. You have zero volts across the resistor.

If you connected those two little circles together, you would complete the circuit, making a connection to both the positive and negative side, and you would measure 15V across the resistor.

 

[Shark 774]

The circuit, as drawn, is an open circuit. There is nothing connecting the the negative side to the resistor. You have zero volts across the resistor.

If you connected those two little circles together, you would complete the circuit, making a connection to both the positive and negative side, and you would measure 15V across the resistor.

Yes this is what I thought. However my teacher and one book said that the potential difference will be 15V across the resistor even when the circuit is open, which makes no sense to me.

 

[Evil Bunny]

If that's the case, then your teacher and one book are wrong.

 

[Studiot]

I suggest you discuss with your teacher the following addition to your circuit.

I have redrawn your circuit and included two switches 'S1' and 'S2 and a capacitor 'cap'.

Start with S1 and S2 both open circuit as in fig1.

Point A is directly connected to the battery and therefore at 15 volts potential.
Points B, C and D are all indeterminate, but would register near zero in a suitable voltmeter.

So in Fig 1 the voltage across the resistor (B-C) is zero.

Now close the first switch S1, as in fig2.

Our voltmeter would show that points A, B and C are all at 15 volts and only D is now indeterminate.
This change is explained by stating that a small current flows through the resistor R for just long enough to create the new potentials at B and C. Once the 15 volts is established the current ceases to flow.

So in Fig 2 the voltage across the resistor (B-C) is still zero.

Now close the second switch S2, as in fig3.


A larger current flows for a longer time until the capacitor is charged to 15 volts.
Whilst this current is flowing there is a measurable voltage drop across B-C the resistor.
Once the voltage at D has risen to 15 volts the current ceases and all A, B, C and D are now at 15 volts.

The capacitor, still charged to 15 volts, may be removed.

Edit: So in Fig 3 the voltage across the resistor (B-C) is still zero.


go well

 

[Drakkith]

The circuit, as drawn, is an open circuit. There is nothing connecting the the negative side to the resistor. You have zero volts across the resistor.

If you connected those two little circles together, you would complete the circuit, making a connection to both the positive and negative side, and you would measure 15V across the resistor.

Oh wow, I can't believe I didn't realize that lol. I guess that's what happens when you don't have much experience with circuits!

 

[Shark 774]

I suggest you discuss with your teacher the following addition to your circuit.

I have redrawn your circuit and included two switches 'S1' and 'S2 and a capacitor 'cap'.

Start with S1 and S2 both open circuit as in fig1.

Point A is directly connected to the battery and therefore at 15 volts potential.
Points B, C and D are all indeterminate, but would register near zero in a suitable voltmeter.

So in Fig 1 the voltage across the resistor (B-C) is zero.

Now close the first switch S1, as in fig2.

Our voltmeter would show that points A, B and C are all at 15 volts and only D is now indeterminate.
This change is explained by stating that a small current flows through the resistor R for just long enough to create the new potentials at B and C. Once the 15 volts is established the current ceases to flow.

So in Fig 2 the voltage across the resistor (B-C) is still zero.

Now close the second switch S2, as in fig3.


A larger current flows for a longer time until the capacitor is charged to 15 volts.
Whilst this current is flowing there is a measurable voltage drop across B-C the resistor.
Once the voltage at D has risen to 15 volts the current ceases and all A, B, C and D are now at 15 volts.

The capacitor, still charged to 15 volts, may be removed.

Edit: So in Fig 3 the voltage across the resistor (B-C) is still zero.


go well

Ok great, thanks for the help!