Question

A resistance of 9 ohm is connected to the terminals of a cell. A voltmeter connected across the cell reads 1.8V . When a resistance of 10 ohm is connected in series with 9 ohm the voltmeter reading changes to 1.9V . Calculate the emf of the cell and its internal resistance

Answer 1

Hey dear,

● Answer-
r = 1 ohm
E = 2 V

● Explanation-
R1 = 9 ohm
R2 = 10 ohm
V1 = 1.8 V
V2 = 1.9 V

# Solution-
Let E be emf of cell and r be internal resistance.

In first situation,
I1 = V1 / R1
I1 = 1.8 / 9
I1 = 0.2 A

But, we know
I1 = E / (R1+r)
0.2 = E / (9+r) ...(1)

In second situation,
R' = R1 + R2
R' = 9 + 10
R' = 19 ohm

In this situation,
I2 = V2 / R'
I2 = 1.9 / 19
I2 = 0.1 A

But, we know
I2 = E / (R'+r)
0.1 = E / (19+r) ...(2)

Dividing (1) by (2),
2 = (19+r) / (9+r)
18 + 2r = 19 + r
r = 1 ohm

Therefore,
E = I1 (R1+r)
E = 0.2 (9+1)
E = 2 V

Hence, emf of cell is 2 V and internal resistance is 1 ohm.

Hope this helps...