Question

The length of potentiometer wire is 4m and its resistance is 18ohm a battery of emf 2V and internal resistance is 2ohm is connected across the wire calculate the potential gradient if the balancing length for a cell of emf E1 is 200cm calculate E1

Answer 1

Given that,

Length of potentiometer = 4 m

Resistance = 18 Ω

Emf = 2 V

Internal resistance = 2 Ω

We need to calculate the potential gradient

Using formula of potential gradient

$K=\dfrac{ER}{(R+r)L}$

Where, E = emf

R = resistance

r = internal resistance

L = length

Put the value into the formula

$K=\dfrac{2\times18}{(18+2)\times4}$

$K=0.45\ V/m$

If the balancing length for a cell of emf E₁ is 200 cm.

We need to calculate the value of emf E₁

Using formula of cell of emf

$E_{1}=Kl_{1}$

Where, K = potential gradient

l = balancing length

Put the value into the formula

$E_{1}=0.45\times200\times10^{-2}$

$E_{1}=0.9\ V$

Hence, The potential gradient is 0.45 V/m.

The value of emf E₁ is 0.9 V.

Answer 2

the potential gradient is o.45v/m,

the value of emf E1 is 0.9v