Question

A cell of EMF 1.8 V and internal resistance $2\Omega$ is connected in a series with an ammeter of resistance $0.7\Omega$ and a resistor of $4.5\Omega$ as shown below.

(a) What would be the reading of the ammeter?
(b) What is the potential difference across the terminals of the cell?
[4 MARKS]

Answer 1

Each subpart: 2 Marks

External resistance of circuit,
$R=4.5+0.7=5.2\Omega$
(a) Current through ammeter (A),
$I=\frac{E}{R+r}=\frac{1.8}{5.2+2}=\frac{1.8}{7.2}=0.25A$
(b) Potential difference across the internal resistance $=Ir=0.25$$\times$ $2=0.5V$
$\therefore$ Potential difference across the terminals of cells, $V=ϵ-Ir$
$=1.8-0.5=1.3V$