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

Question

2 years ago

10

cell is 0.105 Ω. What is the terminal voltage, V, of the cell, in volts?

1 answer:

andrew-mc [135] · Answer 1 · 2022-07-18T00:11:08+03:00

<h2>Answer:</h2>

1.77V

<h2>Explanation:</h2>

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

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

E = $V_{T}$ - (I x r) ----------------------(ii)

<em>According to the question;</em>

E = 1.54V

I = 2.15A

r = 0.105Ω

<em>Substitute these values into equation(ii) as follows;</em>

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

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

1.54 = $V_{T}$ - 0.22575

<em>Solve for </em> $V_{T}$ <em>;</em>

$V_{T}$ = 1.54 + 0.22575

$V_{T}$ = 1.77V

Therefore, the terminal voltage of the cell is 1.77V