Question

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

Answer 1

2/100=e/40
1/50=e/40
40/50=e
0.8v=e

Answer 2

Answer: The correct answer is 0.16 A.

Explanation:

The expression of the current is as follows;

$I=\frac{E}{R+R'}$

Here, I is the current,R' is the internal resistance, R is the resistance of the wire and E is the EMF of the cell.

Put R= 10 ohm , R'= 40 ohm and E= 2 V.

$I=\frac{2}{10+40}$

$I=0.04 A$

Calculate the resistance of 40 cm length of the potentiometer.

$R= \frac{10}{100}(40)$

$R= 4 ohm$

Calculate the unknown emf e by using Ohm's law.

e=V=IR

Put R= 4 ohm and I= 0.04 A.

e=V= (0.04)(4)

e=V= 0.16 A

Therefore, the correct value of e is  0.16 A.