Question

A cell is connected to a voltmeter with a very high resistance. The voltmeter reads 1.49v. The cell is then connected to a fixed resistor with a resistance of 3.25 Ohms and the curreby flow is 0.27A. Calculate internal resistance of the cell.​

Answer 1

The internal resistance of the cell is, r = 2.27 Ω

Step-by-step explanation:

Given data,

The voltage of the battery, v = 1.49 V

The external resistor of resistance, R = 3.2 Ω

The current through the circuit after connecting resistor, I = 0.27 A

The theoretical value of current through the circuit,

Iₓ = v / R

= 1.49 / 3.2

= 0.4585 A

But, the actual reading of the current through the circuit is,

I = 0.27 A

Therefore, the total resistance Rₓ = v / I

= 1.49 / 0.27

= 5.52 Ω

Therefore, the internal resistance of the cell,

r = Rₓ - R

= 5.52 Ω - 3.2 Ω

∴ r = 2.27 Ω