Question

Two cells having unknown emfs E₁ and E₂ (E₁ > E₂) are connected in potentiometer circuit, so as to assist each other. The null point obtained is at 490 cm from the higher potential end. When cell E₂ is connected, so as to oppose cell E₁, the null point is obtained at 90 cm from the same end. The ratio of the emfs of two cells `("E"_1/"E"_2)` is ______.

पर्याय

• 0.689

• 5.33

• 1.45

• 0.182

Answer 1

Two cells having unknown emfs E₁ and E₂ (E₁ > E₂) are connected in potentiometer circuit, so as to assist each other. The null point obtained is at 490 cm from the higher potential end. When cell E₂ is connected, so as to oppose cell E₁, the null point is obtained at 90 cm from the same end. The ratio of the emfs of two cells `("E"_1/"E"_2)` is 1.45.

Explanation:

The emf of a cell in a potentiometer is

E = `"V"/"L"`l

where, I = length of wire at null point,

V = voltage of source

and L = total length of potentiometer wire.

∴ E ∝ I

When two cell of emf E₁ and E₂ (E₁ > E₂) are connected to assist each other, then

E₁ + E₂  = 490   ...(i)

When the cell of emf E₂ Is connected, so as to oppose cell E₁, then

E₁ - E₂  = 90   ...(ii)

From Eqs. (I) and (Ii), we get

E₁ = 290 V and E₂ = 200 V

∴ `"E"_1/"E"_2 = 290/200 = 29/20` = 1.45