Question

A potentiometer wire of length 'L' and a resistance 'r' are connected in series with a battery of E.M.F. 'E₀' and a resistance 'r₁'. A cell of unknown E.M.F, 'E' is balanced at a length 'ℓ' of the potentiometer wire. The unknown E.M.F. E is given by ______ 

विकल्प

• `(E_0ℓ)/L`

• `(E_0r)/(r + r_1) . ℓ/L`

• `(LE_0r)/(ℓr_1)`

• `(LE_0r)/((r + r_1)ℓ)`

Answer 1

A potentiometer wire of length 'L' and a resistance 'r' are connected in series with a battery of E.M.F. 'E₀' and a resistance 'r₁'. A cell of unknown E.M.F, 'E' is balanced at a length 'ℓ' of the potentiometer wire. The unknown E.M.F. E is given by `underline((E_0r)/(r + r_1) . ℓ/L)`.

Explanation:

Current through the circuit I = `E_0/(r + r_1)`

P.d. across the wire V = Ir = `(E_0r)/(r + r_1)`

Potential gradient = `(E_0r)/((r + r_1)L)`

If balancing length for cell of emf E is ℓ then

E = `(E_0r)/((r + r_1)) . ℓ/L`