Question
State the working principle of a potentiometer with the help of a circuit diagram. Describe a method to find the internal resistance of a primary cell.
In a potentiometer arrangement, a cell of emf 1.20 volt gives a balance point at 30 cm length of the wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emfs of the two cells is 1.5. Calculate the difference in the balancing length of the potentiometer wire in the two cases.
Solution
Potentiometer basically allows us to compare the emf's of any two sources where, one of the cell is chosen as the standard cell whose emf is known. The other emf can be calculated.
The basic equation involving is
The jockey is moved along the wire at a distance l1 where the balance point is obtained and, similarly another emf E2 is balanced against l2 .
We can use a potentiometer to find the internal resistance of a cell. For this, the cell (E) whose internal resistance has to be determined is connected across a resistance box through a key K.
The connections are made as shown in the figure.
The key k’ is closed and k is kept open. The balance point is found and let, the balancing length be l1.
Then,
E ∝ l1 ...(i)
Now, a suitable resistance R is introduced in the resistance box R and with the key k’ closed, the balancing length l2 is found out.
When the circuit is closed, the potential difference across the cell falls to
Then,
...(ii)
Dividing equation (i) by (ii), we get
Given,
The basic equation involving is
The jockey is moved along the wire at a distance l1 where the balance point is obtained and, similarly another emf E2 is balanced against l2 .
We can use a potentiometer to find the internal resistance of a cell. For this, the cell (E) whose internal resistance has to be determined is connected across a resistance box through a key K.
The connections are made as shown in the figure.
The key k’ is closed and k is kept open. The balance point is found and let, the balancing length be l1.
Then,
E ∝ l1 ...(i)
Now, a suitable resistance R is introduced in the resistance box R and with the key k’ closed, the balancing length l2 is found out.
When the circuit is closed, the potential difference across the cell falls to
Then,
...(ii)
Dividing equation (i) by (ii), we get
Given,
Difference in the balancing length of potentiometer wire is ,
l1 – l2 = 30 – 20
= 10 cm.
