Question

explain comparison of EMF of two cells by using potentiometer with necessary diagram??
plzz help ​

Answer 1

Answer:

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Explanation:

Comparison of EMFs of two given cells using potentiometer : The potentiometer wire AB is connected in series with a battery (Bt), Key (K), rheostat (Rh) as shown in figure. This forms the primary circuit. The end A of potentiometer is connected to the terminal C of a DPDT switch (six-way key-double pole double throw). The terminal D is connected to the jockey (J) through a galvanometer (G) and high resistance (HR). The cell of emf E1 is connected between terminals C1 and D1 and the cell of emf E2 is connected between C2 and D2 of the DPDT switch.

Let I be the current flowing through the primary circuit and r be the resistance of the potentiometer wire per metre length.

The DPDT switch is pressed towards C1,D1 so that cell E1 is included in the secondary circuit. The jockey is moved on the wire and adjusted for zero deflection in galvanometer. The balancing length is l1. The potential difference across the balancing length l1=Irl1. Then, by the principle of potentiometer,

E1=Irl1....(1)

The DPDT switch is pressed towards E2. The balancing length l2 for zero deflection in galvanometer is determined. The potential difference across the balancing length is l2=Irl2, then

E2=Irl2....(2)

Dividing (1) and (2) we get,

E2E1

Answer 2

   Potentiometer is a device used to compare the emf of the two cells,the cells measure the internal resistance of the cells ,and potential diffrence across the resistor.

   The potentiometer works on the principles that when a constant current flows through a wire of uniform cross sectional area, potential difference between its two points is directly proportional to the length of the wire between the two points.

   Electromotive force is a measurement of the energy that causes current to flow through the current to flow through a circuit. It is a measurement of energy per unit charge.

   Using a potentiometer, we can determine the emf of a cell by obtaining the balancing length I.Here the fall of potential along the length I of the potentiometer wire is equal to the emf of the cell.

                                              E = KL

         where K is the potential gradient along the wire.

    Thus it is possible to compare the emf's of the two given cells by measuring the respective balancing length l1 and l2.

 ie;

     E1=Kl1           and

     E2=Kl2

or

             E1\E2=l1/l2