How would me measure the int. resistance of a cell using poentiometer? Will still no current be derived from the cell?
The necessary circuit for determining the internal resistance of the cell is given below. The primary circuit consists of a battery, rheostat and key k1. The positive terminal of the battery is fixed to the end A of the potentiometer wire AB while negative terminal of the cell is fixed to the point B. The positive terminal of the cell of emf E with internal resistance r is attached to the higher potential point A while its negative terminal to the jockey J through a galvanometer. A resistance box and a key k2 are transversely connected to the cell.
Initially the key k1 is closed and the key k2 is kept open. We obtain the null point by sliding the jockey J on the wire of the potentiometer. Let the distance of this null point from end A is l1. In this case, the cell is in the open circuit, therefore, the potential difference between the terminals is equal to the emf E of the cell.
E = x l1 (1)
Where x is the potential gradient of the wire
Now, a suitable resistance R is taken in the resistance box and the key k2 is closed. Again, we find the null position on the wire with the help of jockey J. Let the distance of this null point from the point A is l2. If V is the potential difference across the resistance R, we have
V = x l2 (2)
The internal resistance of the cell is given by r = (E V) R/ V,
so by putting the values of E and V from equation (1) and (2), we get
Therefore, by substituting the values of l1 and l2 we obtain the internal resistance of the cell.
Yes, it draws current from the cell due to internal resistance of the cell. The current drawn from a cell of emf E when an external resistance R is connected across it is given by Ohm's Law;
where r=internal resistance of the cell.