Consider a large 1.54 V carbon-zinc dry cell used in a physics lab to supply 2.15 A to a circuit. The internal resistance of the cell is 0.105 Ω. What is the terminal voltage, V, of the cell, in volts?

1.77V

The electromotive force voltage (E) in a cell, is related to the lost voltage () and the terminal voltage () as follows;

E = -

Where;

The lost voltage () is the product of the internal resistance (r) of the cell and current (I) in the cell. i.e

=  I x r

Substitute =  I x r into equation (i) as follows;

E = - (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 = - (2.15 x 0.105)

1.54 = - (0.22575)

1.54 = - 0.22575

Solve for ;

= 1.54 + 0.22575

= 1.54 + 0.22575

= 1.77V

Therefore, the terminal voltage of the cell is 1.77V

not too sure thinking all accelerations are pointed down