Question

In the circuit diagram shown below, three resistors of resistance $1\Omega ,2\Omega and3\Omega$ are connected to a cell of EMF 2V and internal resistance $0.5\Omega$.

(a) Calculate the total resistance of the circuit.
(b) What is the reading of the ammeter?
(c) What will be the ammeter reading if an exactly similar cell is connected in series with the given cell?
[5 MARKS]

Answer 1

part a: 2 Marks
part b: 1 Mark
part c: 2 Marks

(a) Resistor of $1\Omega$ and $2\Omega$ are in series and give $3\Omega$ as equivalent resistance which is parallel to $3\Omega$ resistor.
$\therefore \frac{1}{R}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\Rightarrow R=\frac{3}{2}=1.5\Omega$ Let r be internal resistance of the cell.
Then, total resitance of the circuit $=R+r=1.5+0.5=2.0\Omega$
(b) Reading of ammeter will be the current in the circuit, i.e.
$I=\frac{EMF}{Totalresistance}=\frac{2}{2}=1A$
(c) New total resistance = R + r + r
$=1.5+0.5+0.5=2.5\Omega$
$\therefore$ New current, $I=\frac{TotalEMFoftwocell}{Totalresistance}=\frac{2+2}{2.5}=\frac{4}{2.5}=1.6A$