Answer:
The internal resistance of the cell is 0.051 ohm.
Explanation:
Given;
emf of the battery, E = 12 V
terminal voltage of the cell, V = 8.2 V
current in the circuit, I = 75 A
let the potential drop of the cell due to internal resistance (r) = Ir
The internal resistance of the cell is calculated from the equation below;
E = V + Ir
where;
r is the internal resistance of the cell
[tex]Ir = E - V\\\\r = \frac{E-V}{I} \\\\r = \frac{12-8.2}{75} \\\\r = 0.051 \ ohm[/tex]
Therefore, the internal resistance of the cell is 0.051 ohm.
