Answer :
Answer:
1.77V
Explanation:
The electromotive force voltage (E) in a cell, is related to the lost voltage ([tex]V_{L}[/tex]) and the terminal voltage ([tex]V_{T}[/tex]) as follows;
E = [tex]V_{T}[/tex] - [tex]V_{L}[/tex]
Where;
The lost voltage ([tex]V_{L}[/tex]) is the product of the internal resistance (r) of the cell and current (I) in the cell. i.e
[tex]V_{L}[/tex] = I x r
Substitute [tex]V_{L}[/tex] = I x r into equation (i) as follows;
E = [tex]V_{T}[/tex] - (I x r) ----------------------(ii)
According to the question;
E = 1.54V
I = 2.15A
r = 0.105Ω
Substitute these values into equation(ii) as follows;
1.54 = [tex]V_{T}[/tex] - (2.15 x 0.105)
1.54 = [tex]V_{T}[/tex] - (0.22575)
1.54 = [tex]V_{T}[/tex] - 0.22575
Solve for [tex]V_{T}[/tex];
[tex]V_{T}[/tex] = 1.54 + 0.22575
[tex]V_{T}[/tex] = 1.54 + 0.22575
[tex]V_{T}[/tex] = 1.77V
Therefore, the terminal voltage of the cell is 1.77V