A potentiometer has a uniform wire of length 5 m. A battery of emf 10 V and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at 200 cm. The emf of the secondary cell is

Question

A potentiometer has a uniform wire of length 5 m. A battery of emf 10 V and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at 200 cm. The emf of the secondary cell is (A) 4V (B) 6V (C) 2V (D) 8V

Tardigrade Admin · Accepted Answer

Emf of cell in the secondary circuit =potential gradient x balancing length ⇒ (10 / 5) × 2=4 V

Q. A potentiometer has a uniform wire of length $5 m$ . A battery of emf $10 V$ and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at $200 c m$ . The emf of the secondary cell is

4V

6V

2V

8V

Emf of cell in the secondary circuit =potential gradient $x$ balancing length
$\Rightarrow (10/5) \times 2 = 4 V$

Q. A potentiometer has a uniform wire of length 5m. A battery of emf 10V and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at 200cm. The emf of the secondary cell is

4V

6V

2V

8V

Emf of cell in the secondary circuit =potential gradient x balancing length ⇒(10/5)×2=4V

Q. A potentiometer has a uniform wire of length $5 m$ . A battery of emf $10 V$ and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at $200 c m$ . The emf of the secondary cell is

Emf of cell in the secondary circuit =potential gradient $x$ balancing length
$\Rightarrow (10/5) \times 2 = 4 V$