Question

The emf of a cell is balanced by a length of 120 cm of potentiometer wire. When the cell is shunted by a resistance of 10 Ω, the balancing length is reduced by 20 cm. Find the internal resistance of the cell.

Answer 1

Given that,

Original length l₁ = 120 cm

Resistance = 10 Ω

Reduced length l₂ = 20 cm

We need to calculate the internal resistance of the cell

Using formula of internal resistance

$r=(\dfrac{l_{1}-l_{2}}{l_{2}})\times R$

Where, $l_{1}$ = original length

$l_{2}$ = reduced length

R = resistance

Put the value into the formula

$l_{2}=\dfrac{120-20}{20}\times 10$

$l_{2}=50\ \Omega$

Hence, The internal resistance of the cell is 50 Ω.

Answer 2

Answer:

Explanation:

Given data in this question is:

balancing length = l1 = 120 cm

balancing length = l2 = 100 cm (since original length is reduced by 20 cm).

R = 10 Ω

To find: internal resistance = r =?

using the formula of internal resistance we get,

$r = R(\frac{l1}{l2} -1)$

$r = 10(\frac{120}{100} -1)$

and thus r = 2 Ω.

Hence the internal resistance comes out to be 2Ω.