Question

A resistance of 9 ohm is connected to the terminal of a cell .a voltmeter connected across the cell reads is 1.8 volt on. when the resistance of 10 ohm in a series with 9 ohm the voltmeter reading changes to 1.9 volt calculate the EMF of the cell and its internal resistance

Answer 1

the EMF of the cell is 2 volts and the internal resistance is 1Ω.

EXPLANATION :

Let the EMF of the cell = E and the internal resistance be r,

Case -I

current in the circuit = E/(r + 9)

voltage drop in the internal resistance = r x E/(r + 9) = Er/(r + 9)

since the voltmeter is connected across the cell, hence

E - Er/(r + 9) = 1.8

=> Er + 9E - Er = 1.8(r + 9)

=> 9E = 1.8(r + 9)

=> 5E = r + 9.......................eq1

Case -II

current in the circuit = E/(r + 19)

voltage drop in the internal resistance = r x E/(r + 19) = Er/(r + 19)

since the voltmeter is connected across the cell, hence

E - Er/(r + 19) = 1.9

=> Er + 19E - Er = 1.9(r + 19)

=> 19E = 1.9(r + 19)

=> 10E = r + 19 .......................eq2

subtracting eqn2 from eq1 we get

5E = 10

=> E = 2v

putting the value of E in eqn1 we get

5 x 2 = r + 9

=> r = 1Ω

Hence the EMF of the cell is 2 volts and the internal resistance is 1Ω.