Question

A cell of emf $1.8V$ and internal resistance $2ohm$ is connected in series with an ammeter of resistance $0.7ohm$ and resistance of $4.5ohm$ as shown in figure.
A. What would be the reading of the ammeter?
B. What is the potential difference across the terminals of the cell?

Answer 1

The effective resistance of the circuit:
Given:
The internal resistance of the cell, $r=2\Omega$
Ammeter resistance, $RA=0.7\Omega$
External resistance, $R=4.5\Omega$
The resistances are effectively in series.
Therefore, the total Resistance,
$R_{series}=r+R_A+R$
$=2+0.7+4.5$
$=7.2\Omega$

(a) Finding the current in the circuit:
The e. m. f. of the cell $\epsilon =1.8V$ can be
assumed to appear across the three resistances.
By applying Ohm’s law for the $R_{series}$ , we get
$\epsilon =I{R}_{series}$
So, the current in the circuit
$I=\frac{\epsilon }{R_{series}}=\frac{1.8V}{7.2\Omega }=0.25A$
The current in the circuit $I=0.25A$, is the reading of the ammeter.
(b) Potential difference across the terminals of
the cell
The terminal voltage is the sum of the voltage
drop across the cell of emf ε and across the
internal resistance r.
$V=\epsilon -Ir$
Given,
e. m. f. of the cell, $\epsilon =1.8V$
Current (as calculated in part (a)),$I=0.25A$
Internal resistance, $r=2\Omega$
So,
The potential difference across the terminals of the cell,
$V=1.8V-(0.25\times 2)V=1.3V$