Question

A cell of emf $2V$ and internal resistance of $1.2\Omega$ is connected with an ammeter of resistance $0.8\Omega$ and two resistors of $4.5\Omega$ and $9\Omega$ as shown in the diagram:
$\left(i\right)$ What would be the reading on the Ammeter?
$\left(ii\right)$ What is the potential difference across the terminals of the cell?

Answer 1

Given: $E=2V$, $r=1.2\Omega$, $R=$ (external resistance)

Let $4.5\Omega$ and $9\Omega$ connected in parallel, then equivalent resistance

$\frac{1}{R_p}=\frac{1}{4.5}+\frac{1}{9}$

$\frac{1}{R_p}=\frac{2+1}{9}=\frac{3}{9}=\frac{1}{3}$

$R_p=3\Omega$

Now, $0.8\Omega$ and $R_p$ resistances are in series. Then total resistance

$R=3+0.8=3.8\Omega$

$\left(i\right)$ Current in $0.8\Omega$ resistance $I=\frac{E}{R+r}=\frac{2}{3.8+1.2}=\frac{2}{5}=0.4A$

$\left(ii\right)$ Terminal Voltage $\left(V\right)=E-Ir=2-(0.4\times 1.2)$

$=2-0.48$

$=1.52V$