NTA Abhyas JEE Main2020PhysicsCurrent ElectricityPractice
In a potentiometer experiment, the balancing length obtained with a cell is 240 cm . On shunting the cell with a resistance of 2 Ω , the balancing length becomes 120 cm . The internal resistance of the cell is
Options
- A4   Ω
- B1   Ω
- C0.5   Ω
- D2 Ω
Correct answer
D. 2 Ω
Step-by-step solution
In the first case, the emf of the battery is E = Kl 1 , where K is the potential gradient and l 1 is the first balancing length. In the second case the potential difference across the battery is ER r + R = Kl 2 , where R is the shunt resistance used. Dividing the above two equations, we get r R + 1 = l 1 l 2 ⇒r = 2 240 - 120 120 = 2   Ω