Potential Differences on an Electrical Circuit

In summary: So when the switch is open, the potential difference between A and B is the same as the potential difference between A and C, but when the switch is closed, the potential difference between A and C is 0, while the potential difference between A and B is 1/2*IR. In summary, when the switch is open, the potential difference across A is 2/3*emf, while when the switch is closed, the potential difference across A is 0.
  • #1
forestmine
203
0

Homework Statement



Just trying to understand some concepts regarding this particular circuit configuration.

20087282344136335288545316737501091.jpg


When the switch is open, I understand the bulbs B and C form a loop, and therefore the potential difference across each is the same. I don't understand why -- when the switch is closed -- the potential of both B and C drop to 0. Is it because there's no potential drop across a closed switch (essentially a wire) with no resistors or anything to pass through? And since it's in parallel with C, the potential difference across is C is 0, and similarly, across B? So if the potential difference is 0 across both B and C, then the emf is equal to the potential across A.
Please, correct me if any of that is wrong.

When the switch is open, the potential is split between A, B, and C, where B and C have the same potential. It turns out that the potential difference across A when the switch is open is 2/3*(emf), and I'm not sure at all where this comes from. The problem states in the beginning that R is the resistance of each bulb, so I would think the potential difference across each would be IR, so therefore 3*IR.

Not sure about how to arrive at 2/3*emf.

Hope this is clear enough.

Thank you!
 
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  • #2
forestmine said:

Homework Statement



Just trying to understand some concepts regarding this particular circuit configuration.

20087282344136335288545316737501091.jpg


When the switch is open, I understand the bulbs B and C form a loop, and therefore the potential difference across each is the same. I don't understand why -- when the switch is closed -- the potential of both B and C drop to 0. Is it because there's no potential drop across a closed switch (essentially a wire) with no resistors or anything to pass through? And since it's in parallel with C, the potential difference across is C is 0, and similarly, across B? So if the potential difference is 0 across both B and C, then the emf is equal to the potential across A.
Please, correct me if any of that is wrong.
That looks fine.
When the switch is open, the potential is split between A, B, and C, where B and C have the same potential. It turns out that the potential difference across A when the switch is open is 2/3*(emf), and I'm not sure at all where this comes from. The problem states in the beginning that R is the resistance of each bulb, so I would think the potential difference across each would be IR, so therefore 3*IR.

Not sure about how to arrive at 2/3*emf.
Thank you!
When the switch is open, the current passing through A divides at the junction to pass through B and C. It then rejoins at the next junction to head back to the battery. So how much current (compared to that through A) do each of B and C carry? What does that do to the potential drops across B and C? Hint: Ohm's law.
 
  • #3
So, in comparison to A, B and C each get 1/2 of the current that passes through A. If there is 1/2 I for each B and C, using Ohm's Law, the potential across both B and C would be 1/2 * IR, while the potential across A will be IR.

If we have total potential = 0 for a closed loop, then, emf - IR - 1/2IR - 1/2IR = 0. emf = 2IR...not sure if I'm going about this correctly or where do go from here at this point?
 
  • #4
forestmine said:
So, in comparison to A, B and C each get 1/2 of the current that passes through A. If there is 1/2 I for each B and C, using Ohm's Law, the potential across both B and C would be 1/2 * IR, while the potential across A will be IR.

If we have total potential = 0 for a closed loop, then, emf - IR - 1/2IR - 1/2IR = 0. emf = 2IR...not sure if I'm going about this correctly or where do go from here at this point?

Oops! You only pass through either B or C, not both when you travel around your loop. B & C are in parallel, not in series, so you don't add both of their potential drops.
 
  • #5
Hm, alright...in that case, the current that passes through A will be equal to the current that passes through either B or C.

So we have V= emf - IR - IR = 0.

emf = 2IR

I'm still a bit confused. I would have thought that when the switch is open, since there is no potential across C or B, it would have to be entirely across A, and since A is in parallel with the emf source, I would think that the potential difference across A would simply be equal to emf?

Thanks for the help!
 
  • #6
forestmine said:
Hm, alright...in that case, the current that passes through A will be equal to the current that passes through either B or C.
Nope. As discussed, each bulb B or C gets HALF of the current going through A; together (summed) they conduct the same current as A.
So we have V= emf - IR - IR = 0.
That second IR should be (1/2)IR.
I'm still a bit confused. I would have thought that when the switch is open, since there is no potential across C or B, it would have to be entirely across A, and since A is in parallel with the emf source, I would think that the potential difference across A would simply be equal to emf?
Do you mean when the switch is closed rather than open? When it's closed, bulbs B and C are effectively bypassed by a short circuit. In that case, yes, all of the battery's emf will be impressed across A alone.
 
  • #7
Let me try this again.

When the switch is open, we have V = emf - IR - (1/2)IR.

The 1/2 is the equivalent resistance of B and C, right?

From there:

emf = IR + 1/2 IR
emf = 2/3 (IR)

So V = 2/(3emf)

Somewhere along the way I'm messing up so that emf doesn't appear in the numerator, but I'm not sure where.

But conceptually, when the switch is open, the majority of the potential difference occurs across A.
 
  • #8
forestmine said:
Let me try this again.

When the switch is open, we have V = emf - IR - (1/2)IR.

The 1/2 is the equivalent resistance of B and C, right?

From there:

emf = IR + 1/2 IR
correct. So then emf = (3/2)IR (because 1 + 1/2 = 3/2)

and I = (2/3)(emf/R)

By Ohm's law, the the voltage across A is V = I*R, so...
But conceptually, when the switch is open, the majority of the potential difference occurs across A.
yup.
 
  • #9
Got it! Thanks so much for the help!
 

Related to Potential Differences on an Electrical Circuit

1. What is a potential difference on an electrical circuit?

A potential difference, also known as voltage, is the difference in electric potential energy between two points on an electrical circuit. It is measured in volts (V) and is responsible for the movement of electric charges in a circuit.

2. How is potential difference different from current?

Potential difference is the force that drives the flow of electric charges, while current is the rate of flow of these charges. In other words, potential difference is the cause and current is the effect.

3. What factors affect potential difference?

The main factors that affect potential difference are the resistance of the circuit, the type of material used in the circuit, and the length and thickness of the circuit. These factors can impact the flow of electric charges and therefore the potential difference.

4. How is potential difference measured?

Potential difference is measured using a voltmeter, which is connected in parallel to the circuit. The voltmeter measures the difference in electric potential energy between two points and displays it in volts.

5. Why is potential difference important in electrical circuits?

Potential difference is important in electrical circuits because it determines the flow of electric charges and the amount of current in the circuit. It also allows for the control and regulation of electricity in a circuit, making it a crucial aspect in the functioning of various electrical devices.

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