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Grade: 12th pass
        

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

7 months ago

Answers : (2)

Arun
23483 Points
							
Dear student
 
2/100=e/40
1/50=e/40
40/50=e
0.8v=e
 
Hope it helps
 
Regards
Arun (askIITians forum expert)
7 months ago
Khimraj
3008 Points
							
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.
 
7 months ago
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