A battery `epsilon_(1)` of 4 V and a variable resistance Rh are connected in series with the wire AB of the potentiometer. The length of the wire of the potentiometer of 1 meter. When a cell `epsilon_(2)` of emf 1.5 V is connected between points A and C, no currents flows through `epsilon_(2)` Length AC=60 cm. (i) Find the potetial difference between the ends A and B of the potentiometer. (ii) Would the method work, if the battery `epsilon_(1)` is replaced by a cell of emf of 1 V ?

A battery `epsilon_(1)` of 4 V and a variable resistance Rh are connec

1.	A battery `epsilon_(1)` of 4 V and a variable resistance Rh are connected in series with the wire AB of the potentiometer. The length of the wire of the potentiometer of 1 meter. When a cell `epsilon_(2)` of emf 1.5 V is connected between points A and C, no currents flows through `epsilon_(2)` Length AC=60 cm. (i) Find the potetial difference between the ends A and B of the potentiometer. (ii) Would the method work, if the battery `epsilon_(1)` is replaced by a cell of emf of 1 V ?
Answer» (i) Let V be the pot. Diff. between the ends A and B of the potentiometer wire. Then `V/100 = epsilon_(2)/60` or `V = epsilon_(2) xx 100/60 = 1.5 xx 100/60 = 2.5V` (ii) If battery `epsilon_(1)` is replaced by a cell of emf 1 V, then method would not work. As `epsilon_(1) lt epsilon_(2)`, the balance point cannot be obtained on the potentiometer wire.

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