Correct answer: 9 watt

DC input power is given by:

\[ P = VI \]

Given:

• \(V = 300\ \mathrm{V}\)
• \(I = 30\ \mathrm{mA} = 0.03\ \mathrm{A}\)

Substituting:

\[ P = 300 \times 0.03 = 9\ \mathrm{W} \]

• 300 W and 9000 W are far too high.
• 6 W would require a lower current or voltage.

Therefore, the DC input power is 9 watt.

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Correct answer: 6 watt

DC input power is given by:

\[ P = VI \]

Given:

• \(V = 12\ \mathrm{V}\)
• \(I = 500\ \mathrm{mA} = 0.5\ \mathrm{A}\)

So:

\[ P = 12 \times 0.5 = 6\ \mathrm{W} \]

• 12 W and 20 W are too high.
• 500 W is far too high.

Therefore, the DC input power is 6 watt.

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Correct answer: 2 watt

Each resistor is rated at:

\[ 1\ \mathrm{W} \]

When two resistors are connected in series, the same current flows through both, and each resistor can safely dissipate up to its rated power.

So the total maximum power that can be dissipated without exceeding either resistor’s rating is:

\[ P_{\text{total}} = 1\ \mathrm{W} + 1\ \mathrm{W} = 2\ \mathrm{W} \]

• 4 W would exceed the rating of each resistor.
• 1 W would be the limit for only one resistor.
• 1/2 W is below the combined rating.

Therefore, the maximum total power that can be dissipated is 2 watt.

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Correct answer: C — 10 watt

When two identical resistors are connected in parallel, each resistor retains its own individual power rating — it can still dissipate up to 5 W. Because both resistors share the load equally, the total maximum power the combination can safely dissipate is simply the sum of the two individual ratings.

\[ P_{\text{total}} = P_1 + P_2 = 5\ \mathrm{W} + 5\ \mathrm{W} = 10\ \mathrm{W} \]

We can verify this using the parallel resistance and the voltage at maximum power. Two 1000 Ω resistors in parallel give 500 Ω. Each 1000 Ω resistor dissipates 5 W at:

\[ V = \sqrt{P \times R} = \sqrt{5 \times 1000} \approx 70.7\ \mathrm{V} \]

At 70.7 V across the parallel combination, total power is:

\[ P_{\text{total}} = \frac{V^2}{R_{\text{parallel}}} = \frac{70.7^2}{500} = \frac{5000}{500} = 10\ \mathrm{W} \]

• A. 40 watt — Incorrect; this would only be possible if the power rating of each resistor also doubled when paralleled, which it does not.
• B. 20 watt — Incorrect; connecting resistors in parallel does not double each resistor's individual power rating.
• D. 5 watt — Incorrect; this is the rating of only one resistor. With two resistors sharing the dissipation, the combined limit is higher.

Therefore, two 1000 Ω 5 W resistors in parallel can safely dissipate a maximum total of 10 W.

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Correct answer: B — 10 watt

Power dissipated in a resistor can be calculated using the formula relating power, current, and resistance:

\[ P = I^2 R \]

Given:

• Current I = 10 mA = 0.01 A
• Resistance R = 100 kΩ = 100,000 Ω

\[ P = (0.01)^2 \times 100{,}000 = 0.0001 \times 100{,}000 = 10\ \mathrm{W} \]

• A. 1 watt — This would result from an arithmetic error, such as using I = 0.1 A or R = 10 kΩ incorrectly.
• C. 100 watt — This results from failing to convert mA to A correctly (e.g., using I = 0.1 A instead of 0.01 A).
• D. 10,000 watt — This results from using the current in milliamps (10) directly without converting to amps, giving a value 1,000,000 times too large.

Therefore, the power dissipated in a 100 kΩ resistor carrying 10 mA is 10 watts.

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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: B — watt

The watt (symbol W) is the SI unit of power, defined as the rate at which energy is transferred or work is done. One watt equals one joule of energy per second.

\[ P = VI \]

Where P is power in watts, V is voltage in volts, and I is current in amperes. For example, a circuit with 12 V and 2 A delivers:

\[ P = 12 \times 2 = 24\ \mathrm{W} \]

• Ohm — the unit of electrical resistance, not power.
• Ampere — the unit of electrical current, not power.
• Volt — the unit of electromotive force (voltage), not power.

Therefore, the correct unit for power is the watt.

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Correct answer: B — voltage and current

Electrical power is the rate at which energy is transferred in a circuit. It is found by multiplying the voltage (potential difference) across a component by the current flowing through it.

\[ P = V \times I \]

For example, if a circuit has 12 V across a load and 2 A flowing through it:

\[ P = 12 \times 2 = 24\ \mathrm{W} \]

• A. resistance and capacitance — These are two different types of opposition to current flow; multiplying them does not give power (R × C gives a time constant in seconds).
• C. voltage and inductance — Inductance is a property of a coil that opposes changes in current; multiplying voltage by inductance does not yield power.
• D. inductance and capacitance — Multiplying these two reactive components together gives no meaningful power result; LC combinations relate to resonant frequency, not power.

Therefore, power in watts is always found by multiplying voltage in volts by current in amperes.

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Correct answer: A — volt and ampere

Electric power is the product of voltage (volts) and current (amperes). One watt is defined as one volt multiplied by one ampere, expressing the rate at which electrical energy is transferred or converted.

\[ P = V \times I \]

For example, a 12 V supply delivering 2 A produces:

\[ P = 12 \times 2 = 24\ \mathrm{W} \]

• Volt and farad: Farads measure capacitance, not current flow. Volts × farads gives units of charge (coulombs), not power.
• Farad and henry: Farads measure capacitance and henries measure inductance. Their product has no direct relation to power in watts.
• Ampere and henry: Henries measure inductance. Amperes × henries gives units of magnetic flux linkage (weber), not power.

Therefore, the only pair of units that multiplies to give watts is volts and amperes.

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Correct answer: B — 0.1 watt

Power dissipated in a resistor can be calculated directly from the voltage across it and the current through it using the power formula. With 10 mA (0.01 A) of current and 10 V across the resistor:

\[ P = V \times I \]

Substituting the values:

\[ P = 10\ \mathrm{V} \times 0.01\ \mathrm{A} = 0.1\ \mathrm{W} \]

• A (0.01 watt) — Incorrect; this would result from forgetting to convert milliamps to amps and multiplying 10 × 0.001, or from a similar arithmetic error.
• C (1 watt) — Incorrect; this would arise from mistakenly using 100 mA (0.1 A) instead of 10 mA, giving 10 × 0.1 = 1 W.
• D (10 watt) — Incorrect; this would result from treating the current as 1 A rather than 10 mA, or simply multiplying the two numbers without unit conversion.

Therefore, the power dissipated is 0.1 W, found by converting 10 mA to 0.01 A and applying P = V × I.

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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: D — 3 ampere

The total power dissipated by all nine resistors must first be found, then the supply current is calculated using the relationship between power, voltage, and current.

Step 1 – Total power:

\[ P_{\text{total}} = 9 \times 4\ \mathrm{W} = 36\ \mathrm{W} \]

Step 2 – Total current from the supply:

\[ P = VI \implies I = \frac{P}{V} \]

\[ I = \frac{36\ \mathrm{W}}{12\ \mathrm{V}} = 3\ \mathrm{A} \]

• A — 48 ampere: Incorrect; this would result from multiplying 12 V × 4 W, which is not a valid operation for finding current.
• B — 36 ampere: Incorrect; 36 is the total power in watts, not the current in amperes.
• C — 9 ampere: Incorrect; this is simply the number of resistors, not a current value.

Therefore, with 36 W of total dissipation from a 12 V supply, the total circuit current is 3 amperes.

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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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