Question

In the given circuit potentiometer wire is of length 5m and resistance is 5. Resistance of 4 is connected across the potentiometer wire with a cell of emf 10 v and internal resistance 1. If the balance length is found at 300 cm, then emf e of the cell is

Answer 1

Explanation:

Resistance = resistivity x L/A

Resistivity = (RxA) / L

Area = 1 x 10^-6 m^2

Resistivity = (8x 1x10^-6)/1

Resistivity = 8 x10^-6

Answer 2

Concept:

Resistance is a unit of measurement for the restriction of current flow in an electrical circuit.

Given:

A length of 5m and internal resistance of 5Ω and resistance of 4Ω is connected with a cell of emf 10V, 1Ω.

The balance length is 300 cm.

Find:

The emf of the cell.

Solution:

In the potentiometer circuit,

A wire with an internal resistance of 5Ω, a resistance of 4Ω, and an emf with an internal resistance of 1Ω are connected in series.

So, the equivalent resistance is:

$R_1=5+4+1=10$ Ω

The emf of a cell is:

$E_1=10V$

The balance length when the two cells are connected in parallel is $3m$.

The equivalent emf of any cells is E.

The resistance at the balance point of the wire with 5m length and 5Ω resistance is:

$R_2=\frac{5}{5}\times 3=3$

Now,

at balance length, both the circuits will have equal lengths.

Therefore,

$\frac{E_1}{R_1}=\frac{E_2}{R_2}$

$\frac{10}{10}=\frac{E}{3}$

$E=3V$

Hence, the emf of the cell is 3V.

#SPJ2