Question

As illustrated in the given circuit diagram, an ammeter with a resistance of 0.8 and two resistors of 4.5 Ω and 9 Ω are connected to a cell with an emf of 2 V and internal resistance of 1.2 Ω. Determine the ammeter reading and the potential difference between the terminals of the cell.

Answer 1

Given:

emf of the cell, E = 2 V

Internal resistance, r = 1.2 Ω

Ammeter resistance = 0.8 Ω

Resistors = 4.5 Ω and 9 Ω

1. Calculate the equivalent resistance of the parallel combination:

`1/R = 1/4.5 + 1/9`

`1/R = 2/9 + 1/9`

`1/R = 3/9`

`1/R = 1/3`

R = 3 Ω

2. Calculate the total resistance of the circuit:

The ammeter resistance, equivalent resistance, and internal resistance are in series.

`R_"total"` ​= 0.8 + 3 + 1.2 = 5 Ω

3. Calculate the current in the circuit (ammeter reading):

Using Ohm's law,

I = `E/R_"total"`

= `2/5`

= 0.4 A

Therefore, the ammeter reading is 0.4 A​.

4. Calculate the terminal potential difference:

V = E − Ir

= 2 − (0.4 × 1.2)

= 2 − 0.48

= 1.52 V