Solving potentiometer problem with addition of resistor

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Homework Help Overview

The discussion revolves around a potentiometer problem involving the emf of a battery and the effects of adding a resistor to the circuit. The original poster presents a scenario where the emf of battery A is determined using a potentiometer, and then questions arise regarding the impact of an additional resistor on the balance length.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the interpretation of the circuit configuration, particularly whether the added resistor is in series or parallel with the battery. There are discussions about the implications of current flow and voltage drops across components in the circuit.

Discussion Status

The discussion is ongoing, with participants questioning assumptions about the circuit setup and the behavior of current in the system. Some guidance has been offered regarding the interpretation of the circuit, but no consensus has been reached on the implications of these interpretations for solving the problem.

Contextual Notes

Participants note the complexity introduced by the internal resistance of the battery and the additional resistor, as well as the potential for confusion regarding the circuit's configuration. There is also mention of the need for a circuit diagram to clarify the discussion.

toforfiltum
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Homework Statement


The emf of a battery A is balanced by a length of 75 cm on a potentiometer wire. The emf of a standard cell, 1.02V, is balanced by a length of 50 cm. What is the emf of A?
Calculate the new balance length if A has an internal resistance of 2 Ω and a resistor of 8 Ω is joined to its terminals.

Homework Equations


E= I(R+r)

The Attempt at a Solution


I can answer the first part of this question. Emf of A is 1.53 V. It's the second part of the question that I don't know.
With the addition of the 8 Ω resistor, the voltage across the potentiometer wire must be equal to the voltage from A after voltage drop across 8Ω resistor, right?

And if so, I don't know how to find it, for I don't know how to find the value of current. If my reasoning is correct, I will end up with E= I(R+r) + V', where V' is the potential difference across potentiometer wire.

I don't know if I'm totally wrong.
 
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toforfiltum said:

Homework Statement


The emf of a battery A is balanced by a length of 75 cm on a potentiometer wire. The emf of a standard cell, 1.02V, is balanced by a length of 50 cm. What is the emf of A?
Calculate the new balance length if A has an internal resistance of 2 Ω and a resistor of 8 Ω is joined to its terminals.

Homework Equations


E= I(R+r)

The Attempt at a Solution


I can answer the first part of this question. Emf of A is 1.53 V. It's the second part of the question that I don't know.
With the addition of the 8 Ω resistor, the voltage across the potentiometer wire must be equal to the voltage from A after voltage drop across 8Ω resistor, right?
That depends upon your interpretation of the circuit description. Do you consider the 8 Ω resistor to be in series with the A cell (and its internal resistance of 2 Ω), or in parallel with it? The description does say "terminals" (plural)...

What's the current drawn through the potentiometer connection when the system is balanced? Perhaps sketching your interpretation of the circuit diagram and labeling all known values would be helpful.
 
gneill said:
That depends upon your interpretation of the circuit description. Do you consider the 8 Ω resistor to be in series with the A cell (and its internal resistance of 2 Ω), or in parallel with it? The description does say "terminals" (plural)...

What's the current drawn through the potentiometer connection when the system is balanced? Perhaps sketching your interpretation of the circuit diagram and labeling all known values would be helpful.
upload_2015-8-13_23-19-34.png

Ok, haha this is my interpretation. I don't know how to provide those nice circuits some users here do, so I just use paint:-p. But I think it is clear, I hope.

And it is with this interpretation that I'm stuck. Since you might me wondering what the circuit might be, the answer is 60 cm, if it helps.
And its precisely the current that I don't know how to find.
 
Okay, so the problem is with your interpretation of the circuit description. The 8 Ω resistor is connected to BOTH terminals of the A cell, not just one. That is, it is in parallel with the cell. See here (I use MS Visio for drawings, by the way):

Fig1.gif


From your knowledge of how a potentiometer circuit works, what is the current ##I_s## when the circuit is balanced?
 
gneill said:
Okay, so the problem is with your interpretation of the circuit description. The 8 Ω resistor is connected to BOTH terminals of the A cell, not just one. That is, it is in parallel with the cell. See here (I use MS Visio for drawings, by the way):

View attachment 87276

From your knowledge of how a potentiometer circuit works, what is the current ##I_s## when the circuit is balanced?
It's zero. But other than that, I still don't know how to proceed. Is the voltage drop across the potentiometer equal to the voltage across 8 Ω resistor and cell A?
And how did you get 2.04 V?
 
toforfiltum said:
It's zero. But other than that, I still don't know how to proceed.
If that current is zero then the cell + resistor is essentially an isolated circuit (no current enters or leaves this circuit "island". You can determine the local current and all potential drops in this subcircuit.
Is the voltage drop across the potentiometer equal to the voltage across 8 Ω resistor and cell A?
Yup.
And how did you get 2.04 V?
From the standard cell test conditions described in the problem statement. 50 cm is half the 100 cm of the full potentiometer wire.
 
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gneill said:
If that current is zero then the cell + resistor is essentially an isolated circuit (no current enters or leaves this circuit "island". You can determine the local current and all potential drops in this subcircuit.

Yup.

From the standard cell test conditions described in the problem statement. 50 cm is half the 100 cm of the full potentiometer wire.
Ok, thanks. One last question. Actually, if I put an ammeter at the 8 Ω resistor and at cell A, there would be no reading right? Since there's no net potential difference. You are just isolating the circuit to solve the question right?
 
toforfiltum said:
Ok, thanks. One last question. Actually, if I put an ammeter at the 8 Ω resistor and at cell A, there would be no reading right? Since there's no net potential difference. You are just isolating the circuit to solve the question right?
No , you would , in both cases .
 
Qwertywerty said:
No , you would , in both cases .
Why? I thought there's no potential difference across resistor and cell A?
 
  • #10
toforfiltum said:
You are just isolating the circuit to solve the question right?
We're doing this because there will be current in this loop , which will , however , not move out of this loop ( as galvanometer shows zero reading ) .

We then just write Kirchoff ' s loop law , for this particular loop .

Hope this helps .
 
  • #11
toforfiltum said:
Why? I thought there's no potential difference across resistor and cell A?
If there wasn't , then how would the potentiometer work ?
 
  • #12
Qwertywerty said:
If there wasn't , then how would the potentiometer work ?
I don't understand. I thought that the potential drop across potentiometer is equal to potential drop across resistor and cell A? If so, shouldn't that potential drop be balanced by potential difference across terminals of cell A and resistor? This potential difference is supplied by cell A. Then, how can there be current?
 
  • #13
toforfiltum said:
I thought that the potential drop across potentiometer is equal to potential drop across resistor and cell A?
Note : Or ; not And .
toforfiltum said:
If so, shouldn't that potential drop be balanced by potential difference across terminals of cell A and resistor? This potential difference is supplied by cell A. Then, how can there be current?
The previous part of the post should answer this .
 
  • #14
It might be beneficial to have a labeled circuit diagram to refer to in order to avoid ambiguities during discussions. Here's an attempt that I hope will help:

Fig2.gif
 
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