Question

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?

Answer 1

Answer:

1.77V

Explanation:

The electromotive force voltage (E) in a cell, is related to the lost voltage ($V_{L}$) and the terminal voltage ($V_{T}$) as follows;

E = $V_{T}$ - $V_{L}$

Where;

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

$V_{L}$ =  I x r

Substitute $V_{L}$ =  I x r into equation (i) as follows;

E = $V_{T}$ - (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 = $V_{T}$ - (2.15 x 0.105)

1.54 = $V_{T}$ - (0.22575)

1.54 = $V_{T}$ - 0.22575

Solve for $V_{T}$;

$V_{T}$ = 1.54 + 0.22575

$V_{T}$ = 1.54 + 0.22575

$V_{T}$ = 1.77V

Therefore, the terminal voltage of the cell is 1.77V