# Solving potentiometer problem with addition of resistor

• toforfiltum
In summary, the conversation discusses the use of a potentiometer wire to balance the emf of a battery A and a standard cell. The emf of A is found to be 1.53 V, but the addition of an 8 Ω resistor complicates the calculation. By interpreting the circuit correctly and using knowledge of potentiometer circuits, it is determined that the current in the circuit is zero and the voltage drop across the potentiometer is equal to the voltage across the 8 Ω resistor and cell A. The final answer is found to be 2.04 V.
toforfiltum

## 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.

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.

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.

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.

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. 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):

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.

Qwertywerty
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?

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 .

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 ?

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?

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 .

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:

Qwertywerty and SammyS

## 1. What is a potentiometer and how does it work?

A potentiometer, also known as a variable resistor, is an electronic component that can vary its resistance according to the position of a sliding contact. It consists of a resistive track with three terminals, where the first and third terminals are connected to the ends of the track and the middle terminal is connected to the sliding contact. By moving the sliding contact, the resistance between the first and third terminals can be adjusted.

## 2. Why would I need to add a resistor to a potentiometer?

Adding a resistor to a potentiometer can help in situations where the range of resistance provided by the potentiometer is not enough for the desired application. By adding a resistor in series with the potentiometer, the total resistance can be increased, allowing for a wider range of control.

## 3. How do I calculate the value of the resistor to add to my potentiometer?

The value of the resistor to add to a potentiometer can be calculated using the formula R = (V/Rp) - 1, where R is the desired resistance, V is the total voltage applied, and Rp is the resistance of the potentiometer. Make sure to choose a resistor with a value closest to the calculated value.

## 4. Can I add multiple resistors to a potentiometer?

Yes, you can add multiple resistors in series with a potentiometer to increase the range of resistance. However, keep in mind that the total resistance of the circuit will be the sum of all the resistors, so make sure to calculate and choose the appropriate values for your application.

## 5. How do I troubleshoot a potentiometer problem with the addition of a resistor?

If you are experiencing issues with your potentiometer after adding a resistor, first check that the connections are correct and secure. Then, make sure the values of the resistor and potentiometer are appropriate for your application. If the problem persists, it could be due to a faulty component or a wiring issue, and further troubleshooting may be necessary.

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