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Correct answer: 250 watt

Power dissipated in a resistor can be calculated using:

\[ P = I^2 R \]

Given:

• \(I = 500\ \mathrm{mA} = 0.5\ \mathrm{A}\)
• \(R = 1000\ \Omega\)

Substituting:

\[ P = (0.5)^2 \times 1000 = 0.25 \times 1000 = 250\ \mathrm{W} \]

• 0.25 W would result from a much smaller current.
• 2.5 W would require a lower resistance.
• 25 W would require either lower current or resistance.

Therefore, the power dissipated is 250 watt.

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Correct answer: 1.25 watt

Power dissipated in a resistor can be calculated using:

\[ P = I^2 R \]

Given:

• \(I = 0.25\ \mathrm{A}\)
• \(R = 20\ \Omega\)

Substituting:

\[ P = (0.25)^2 \times 20 = 0.0625 \times 20 = 1.25\ \mathrm{W} \]

• 5 W would require a higher current.
• 2.50 W would require either more current or resistance.
• 10 W would require significantly more current.

Therefore, the power dissipated is 1.25 watt.

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Correct answer: 20 watt

Power dissipated in a resistor can be calculated using:

\[ P = \frac{V^2}{R} \]

Given:

• \(V = 200\ \mathrm{V}\)
• \(R = 2000\ \Omega\)

Substituting:

\[ P = \frac{(200)^2}{2000} = \frac{40{,}000}{2000} = 20\ \mathrm{W} \]

• 30 W is not supported by the calculation.
• 10 W would result from a lower applied voltage.
• 40 W would require either higher voltage or lower resistance.

Therefore, the resistor will dissipate 20 watt.

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Correct answer: 5 amp

Power dissipated in a resistive load is given by:

\[ P = I^2 R \]

Solving for current:

\[ I = \sqrt{\frac{P}{R}} \]

Given:

• \(P = 500\ \mathrm{W}\)
• \(R = 20\ \Omega\)

Substituting:

\[ I = \sqrt{\frac{500}{20}} = \sqrt{25} = 5\ \mathrm{A} \]

• 25 amp would correspond to a much higher power for a 20 \(\Omega\) load.
• 2.5 amp would only deliver \(P = (2.5)^2 \times 20 = 125\ \mathrm{W}\).
• 10 amp would deliver \(P = (10)^2 \times 20 = 2000\ \mathrm{W}\).

Therefore, the antenna current is 5 amp.

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Correct answer: 5 watt

Two 10 \(\Omega\) resistors in series give a total resistance of:

\[ R_{\text{total}} = 10 + 10 = 20\ \Omega \]

Using:

\[ P = \frac{V^2}{R} \]

Given:

• \(V = 10\ \mathrm{V}\)
• \(R = 20\ \Omega\)

Substituting:

\[ P = \frac{10^2}{20} = \frac{100}{20} = 5\ \mathrm{W} \]

• 10 W would require a lower resistance.
• 20 W or 100 W would require much higher current.

Therefore, the battery load is 5 watt.

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Correct answer: 24 watt

Three equal resistors in parallel share the load, so the equivalent resistance is:

\[ R_{eq} = \frac{18\ \Omega}{3} = 6\ \Omega \]

Using the power formula:

\[ P = \frac{V^2}{R} \]

Substituting \(V = 12\ \mathrm{V}\) and \(R = 6\ \Omega\):

\[ P = \frac{12^2}{6} = \frac{144}{6} = 24\ \mathrm{W} \]

• 3 watt is far too small for a 12 V supply across a low resistance.
• 18 watt would correspond to a higher equivalent resistance than calculated.
• 36 watt would require an even lower equivalent resistance.

Therefore, the total power dissipation of the resistor load is 24 watt.

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By convention, \(P\) is power, \(I\) is current, \(R\) is resistance, and \(E\) is Voltage (Electro-motive force).

The Power Law states: \(P = I \times E\). Ohm's law states that \(E = I \times R\). Combining those to remove E, we get a restatement of the Power Law as \(P = I \times (I \times R)\), or \(P = I^2 \times R\)

So in this case:

\[P = (.02)^2 \times 10,000\] \[P = 0.0004 \times 10,000\]

Thus

\[P = 4 \text{ watts}\]

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Correct answer: power

A resistor becomes hot when it dissipates electrical energy as heat.

The power dissipated in a resistor is given by:

\[ P = I^2 R \quad \text{or} \quad P = \frac{V^2}{R} \]

If too much power is dissipated, the resistor’s temperature rises and it may overheat or burn.

• Current and voltage contribute to power, but are not dissipated themselves.
• Resistance is a property, not something that is dissipated.

Therefore, the resistor is dissipating too much power.

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Correct answer: 10,000 watt

For a resistive load, power dissipation is given by:

\[ P = I^2 R \]

Given:

• \(I = 10\ \mathrm{A_{rms}}\)
• \(R = 100\ \Omega\)

Substituting:

\[ P = (10)^2 \times 100 = 100 \times 100 = 10{,}000\ \mathrm{W} \]

The frequency (50 Hz) does not affect the power dissipated in a pure resistor when RMS values are used.

• 500 watt would correspond to a much smaller current.
• 707 watt is a common trap related to peak vs RMS values, but RMS current is already given.
• 50,000 watt would require a significantly higher current.

Therefore, the power dissipated is 10,000 watt.

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Correct answer: increase by four times

For a fixed total resistance, electrical power is related to voltage by:

\[ P = \frac{V^2}{R} \]

If the applied voltage is doubled from \(V\) to \(2V\), the new power becomes:

\[ P_{new} = \frac{(2V)^2}{R} = \frac{4V^2}{R} = 4P \]

So the total power dissipated increases by a factor of four.

• decrease to half is incorrect because increasing voltage increases power.
• double would only occur if power were directly proportional to voltage, which it is not.
• not change is incorrect because power depends on the square of the voltage.

Therefore, doubling the applied voltage causes the total power dissipated to increase by four times.

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