# Potentiometer and EMFs with internal resistance

• Tanishq Nandan
In summary: The unknown battery may have internal resistance, but it will still produce 0V of deflection when connected to the pot.
Tanishq Nandan

## Homework Statement

In a potentiometer 99cm of wire of resistance 99ohms is connected with a battery of 50V and 1ohm internal resistance.Find emf of battery giving zero deflection for a length of 13cm.

## Homework Equations

Sadly,the formula I know is only for ideal emfs,which states that:
If two batteries with emf E1 and E2 obtain null point at L1 and L2 lengths of the potentiometer wire respectively,then : E1/E2=L1/L2

## The Attempt at a Solution

The question doesn't mention connecting 2 emfs,just 1 case,1setup,hence the formula I know can't be used here.

Hence,my problem.

A potentiometer is supposed to compare emfs based on where their null points are obtained,right?Then,how am I supposed to find the emf with just 1 setup(that too batteries with internal resistance)?
(Maybe I misunderstood the language of the question.Any other interpretations would be useful)

It says:

In a potentiometer 99cm of wire of resistance 99ohms is connected with a battery of 50V and 1ohm internal resistance. Find emf of battery giving zero deflection for a length of 13cm.
So you have a known battery of 50 volts, and a battery of unknown EMF. The question is, do we assume the unknown battery also has 1 ohm internal resistance?
That would be a decent assumption to go on (and should be stated in your solution steps.
Since 99cm is 99 ohms, you have 1 ohm/cm so you could just add 1 cm of virtual wire to simulate the internal resistance. You could do this for both sides so that the potentiometer is now 101 cm long, and the setting is at 14 cm.

scottdave said:
The question is, do we assume the unknown battery also has 1 ohm internal resistance?
I doubt it.The unknown emf needs to be ideal.

Tanishq Nandan said:
A potentiometer is supposed to compare emfs based on where their null points are obtained,right?Then,how am I supposed to find the emf with just 1 setup(that too batteries with internal resistance)?
(Maybe I misunderstood the language of the question.Any other interpretations would be useful
I think there are two batteries. One with emf of 50V and internal resistance 1 ohm and the other with unknown emf.
You are supposed to find the emf of the unknown battery for which the null point is obtained at 13cm.

scottdave
I just re-read your problem and formulas, and realize now how the setup is: 50 volt battery (1 ohm internal) across the entire pot, then the unknown battery is connected via galvometer or ammeter to the tap point, so when there is no current flowing to the tap, the deflection is zero (and it won't matter how much internal resistance).

Last edited:

## 1. What is a potentiometer?

A potentiometer, also known as a variable resistor, is an electronic component used to adjust and control the flow of electric current in a circuit. It consists of a resistive element and a sliding contact, which can be moved along the resistive element to change the resistance and thus the amount of current that can pass through it.

## 2. How does a potentiometer measure EMFs?

A potentiometer measures EMFs (electromotive forces) by comparing the potential difference between two points in a circuit. The sliding contact of the potentiometer is connected to one point, while the other point is connected to the resistive element. By adjusting the position of the sliding contact, the resistance can be changed until the potential difference between the two points is equal. This value can then be used to calculate the EMF.

## 3. What is internal resistance in a potentiometer?

Internal resistance in a potentiometer refers to the resistance of the potentiometer itself. This resistance is caused by the material and construction of the potentiometer and can affect the accuracy of the measurements. It is important to consider and account for the internal resistance when using a potentiometer to measure EMFs.

## 4. How can internal resistance affect the accuracy of potentiometer measurements?

Internal resistance can affect the accuracy of potentiometer measurements as it adds to the overall resistance in the circuit, leading to a lower potential difference between the two points being measured. This can result in a lower calculated EMF value, leading to an inaccurate measurement. To minimize the impact of internal resistance, it is important to use a potentiometer with a low internal resistance and to account for it in the calculations.

## 5. Can potentiometers be used to measure all types of EMFs?

Yes, potentiometers can be used to measure all types of EMFs as long as the EMF can be expressed as a potential difference between two points in a circuit. However, the accuracy and precision of the measurements may vary depending on the type and strength of the EMF being measured and the internal resistance of the potentiometer. It is important to choose a potentiometer with the appropriate range and sensitivity for the specific type of EMF being measured.

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