The length of a wire of a potentiometer is 100 cm, and the emf of its standard cell is E volt. It is employed to measure the e.m.f of a battery whose internal resistance is 0.5 Ω Ω . If the balance point is obtained at l = 30 cm from the positive end, the e.m.f. of the battery is (where i is the current in the potentiometer)
$(a) \frac{30E}{100} (b) \frac{30E}{100.5} (c) \frac{30E}{(100 - 0.5)} (d) \frac{30(E - 0.5i)}{100}$
Text Solution
Verified by ExpertsA
Let V be the potential across balance point and one end of wire. Hence according to the principle of potentiometer $v \propto l$
Also if a cell of emf E is employed in the circuit between the ends of potentiometer wire of length L then $E \propto L$ .
Therefore, $\frac{V}{E} = \frac{1}{L}$ $V = \frac{1}{L} E = \frac{30}{100} E = \frac{30E}{100}$
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