Question

The given figure shows three resistors of $2\Omega$, $4\Omega$ and $R\Omega$ connected to a battery of EMF 2V and internal resistance $3\Omega$.A main current of 0.25 A flows through the circuit.

(a) What is the potential difference across the $4\Omega$ resistor?
(b) Calculate the potential difference across the internal resistance of the cell.
(c) What is the potential difference across the $R\Omega$ or $2\Omega$ resistors?
(d) Calculate the value of R.
[5 MARKS]

Answer 1

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

Given that current in the circuit = 0.25A
$\therefore$ Current through $4\Omega$ resistor = 0.25A
(a) Potential difference across $4\Omega$ resistor
$=R\times I=4\times 0.25=1V$
(b) Potential difference across $3\Omega$ (internal resistance of the cell) = R $\times$ I
$=3\times 0.25=0.75V$
(c) the potential difference accross the $R\Omega$ or $2\Omega$ resistors = $2-0.75-1=0.25V$

(d) Current through the $2\Omega$ resistance $=\frac{0.25}{2}=0.125A$
Now ,$V=IR\Rightarrow R=\frac{V}{I}$
On substituting the values we get:
$R=\frac{0.25}{0.125}=2\Omega$