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How the emf of two cells are compared using potentiometer ? |
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Answer» Solution :Comparison of emf of two cells with a potentiometer:To compare the emf of two cells, the circuit connections are made as shown in figure. Potentiometer WIRE CD is connected to a battery Bt and a key K in series. This is the primary circuit. The end C of the wire is connected to the terminal M of a DPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR. The cells whose emf `xi_(1)` and `xi_(2)` to be compared are connected to the terminals `M_(1), N_(1) and M_(2), N_(2)` of the DPDT switch. The positive terminals of Bt, `xi_(1) and xi_(2)` should be connected to the same end C. The DPDT switch is pressedtowards `M_(1), N_(1)` so that cell `xi_(1)` is included in the secondarycircuit and the balancing length `l_(1)` is found by adjusting the jockey for zero deflection. Then the second cells `xi_(2)` is included in the circuit and the balancing length `l_(2) ` is determined. Let r be the resistance per UNIT length of the potentiometer wire and I be the current flowing through the wire.  we have `xi_(1)=Irl_(1)""...(1)` `xi_(2)=Irl_(2)""...(2)` By dividing equation (1) by (2) `(xi_(1))/(xi_(2))=(l_(1))/(l_(2)) ""...(3)` By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it. |
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