Ab is a potentiometer wire of length 100 cm and resistance 10 ohm it is connected in series with a resistance r =40 ohms and a battery of emf 2v if a source of unknown emf e is balanced by 40cm of potentiometer the value of e

Question

Arun · Answer

Dear student 2/100=e/40 1/50=e/40 40/50=e 0.8v=e Hope it helps Regards Arun (askIITians forum expert)

Khimraj · Answer

given, resistance of potentiometer wire, R = 10 ohm length of potentiometer wire , l = 100cm emf of cell, connected in series with potentiometer wire, E = 2V. balancing length is obtained at 40cm. so, I = E/(R + r) where I is the current through potentiometer . I = 2/(10 + r) ....(1) now resistance of 40cm wire = 40/100 × 10 = 4 ohms. given, emf of the cell is balanced by a length of wire, E' = 10mV = 10^-2 volts using Ohm'slaw, V = IR 10^-2 = I × 4 ohms or, I = 2.5 × 10^-3 A now putting the value of I in equation (1), 2.5 × 10^-3 = 2/(10 + r) 10 + r = 2/(2.5 × 10^-3) = 2000/2.5 = 800 or, r = 790 ohms hence, answer is 790ohms.

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Ab is a potentiometer wire of length 100 cm and resistance 10 ohm it is connected in series with a resistance r =40 ohms and a battery of emf 2v if a source of unknown emf e is balanced by 40cm of potentiometer the value of e

Ab is a potentiometer wire of length 100 cm and resistance 10 ohm it is connected in series with a resistance r =40 ohms and a battery of emf 2v if a source of unknown emf e is balanced by 40cm of potentiometer the value of e

Ab is a potentiometer wire of length 100 cm and resistance 10 ohm it is connected in series with a resistance r =40 ohms and a battery of emf 2v if a source of unknown emf e is balanced by 40cm of potentiometer the value of e

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