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lisov135 [29]
3 years ago
10

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?
Physics
1 answer:
andrew-mc [135]3 years ago
8 0
<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} = 1.54 + 0.22575

V_{T} = 1.77V

Therefore, the terminal voltage of the cell is 1.77V

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3 years ago
A block weighing 10 newtons is resting on a plane inclined 30° to the horizontal. What is the magnitude of the normal force acti
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