Question

A cell of $e.m.f.1.5V$, records a p.d. of $1.35V$ when connected to an external resistance $R$, such that current flowing through the circuit is$0.75A$. Calculate the value of $R$ and internal resistance of the cell.

Answer 1

Step 1: To find the value of external resistance $R$:

Given data :

$\text{E}=1.5\text{volts},\text{V}=1.35\text{volts, I}=0.75\text{A}$

By using the external resistance $\text{R}$(Ohm's law) formula,

$\text{R}=\frac{\text{V}}{\text{I}}$

$R=\frac{1.35}{0.75}$

$R=1.8\Omega$

Thus, the value of the external resistance $\text{R}$ for the given cell is $1.8\Omega$.

Step 2: To find the value of internal resistance $r$:

By using , $I=\frac{E}{R+r}$

$0.75=\frac{1.5}{1.8+r}$

$1.8+r=\frac{1.5}{0.75}$

$1.8+r=2$

$r=2-1.8$

$r=0.2\Omega$

Thus, the internal resistance $r$ for the given cell is $0.2\Omega$.