In a potentiometer experiment, if the balance point is found at 4 m with a 6 V battery, what is the EMF of the cell being measured?
Practice Questions
1 question
Q1
In a potentiometer experiment, if the balance point is found at 4 m with a 6 V battery, what is the EMF of the cell being measured?
2 V
3 V
4.5 V
6 V
Using the potential gradient, the EMF of the cell can be calculated as (6 V / 10 m) * 4 m = 2.4 V.
Questions & Step-by-step Solutions
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Q
Q: In a potentiometer experiment, if the balance point is found at 4 m with a 6 V battery, what is the EMF of the cell being measured?
Solution: Using the potential gradient, the EMF of the cell can be calculated as (6 V / 10 m) * 4 m = 2.4 V.
Steps: 8
Step 1: Understand that a potentiometer measures the electromotive force (EMF) of a cell by comparing it to a known voltage.
Step 2: Identify the known voltage from the battery, which is 6 V.
Step 3: Note the distance where the balance point is found, which is 4 m.
Step 4: Recognize that the total length of the potentiometer wire is 10 m (this is a common assumption unless stated otherwise).
Step 5: Calculate the potential gradient (voltage per meter) using the formula: Potential Gradient = Voltage / Length of wire = 6 V / 10 m.
Step 6: Multiply the potential gradient by the distance to the balance point to find the EMF of the cell: EMF = Potential Gradient * Distance to balance point = (6 V / 10 m) * 4 m.
Step 7: Perform the calculation: (6 V / 10 m) * 4 m = 2.4 V.
Step 8: Conclude that the EMF of the cell being measured is 2.4 V.
