In a potentiometer experiment, if the balance point is found at 4 m with a 2 V battery, what is the EMF of the cell being measured?

1 question

1 item

Q: In a potentiometer experiment, if the balance point is found at 4 m with a 2 V battery, what is the EMF of the cell being measured?

Solution: The EMF of the cell is equal to the potential gradient multiplied by the length to the balance point. Here, it is 2 V.

Steps: 7

Step 1: Understand that a potentiometer is used to measure the EMF (Electromotive Force) of a cell.
Step 2: Identify that the balance point is where the potential difference across the potentiometer wire equals the EMF of the cell being measured.
Step 3: Note that the balance point is found at 4 meters with a 2 V battery.
Step 4: Recognize that the potential gradient (voltage per meter) can be calculated using the 2 V battery and the length of the wire (4 m).
Step 5: Calculate the potential gradient: 2 V / 4 m = 0.5 V/m.
Step 6: Use the potential gradient to find the EMF of the cell: EMF = potential gradient × length to balance point.
Step 7: Since the length to the balance point is 4 m, the EMF = 0.5 V/m × 4 m = 2 V.