Electrical Power and Energy Lesson Test

Answer: C

Correct answer: One joule of energy transferred each second. The watt is the SI unit of power, and power is the rate of energy transfer. One watt therefore means one joule of energy transferred every second.

• A: Volts divided by amperes gives resistance in ohms, not power.
• B: Power is measured per second; joules per hour would be a much smaller rate and is not the definition of a watt.
• C (correct): A watt is defined as a rate of one joule per second, which is why power describes how fast energy moves.
• D: One coulomb per second is one ampere, the unit of current rather than power.

Answer: A

Correct answer: kilowatt-hours. Utilities charge for energy, which is power multiplied by the time it is used. One kilowatt-hour is the energy of a 1,000 W load running for one hour, or a 100 W load running for ten hours. The most tempting wrong choice is kilowatts, but that is a rate, like speed, while kilowatt-hours are the total, like distance traveled. A meter tallies kilowatt-hours over the month, and the bill multiplies that total by a price per kWh.

• A (correct): kWh measures energy accumulated over time.
• B: Ampere-hours measure electric charge, as on battery ratings, not billed energy.
• C: V measures electrical potential difference.
• D: Kilovolt-amperes rate apparent power for equipment such as transformers, not energy used over time.

Answer: B

Correct answer: 72 W. Use

because voltage and current are given.

• A: This subtracts the current from the voltage, which has no physical meaning.
• B (correct): Multiplying 24 V by 3 A with $P=VI$ gives 72 W.
• C: This adds the voltage and current instead of multiplying them.
• D: This comes from dividing voltage by current, which gives resistance in ohms, not power.

Answer: D

Correct answer: 60 Wh. Energy equals power multiplied by time, so a watt-hour is simply one watt sustained for one hour. Multiplying watts by hours keeps the units together as watt-hours. The key distinction is between power, measured in watts, which is how fast energy is used, and energy, measured in watt-hours, which is how much has been used in total. An answer in plain watts has the right number but describes a rate, not an amount.

• A: 23 Wh adds unlike quantities.
• B: 6 Wh miscalculates the product.
• C: W alone is power, not energy.
• D (correct): E = 20 W × 3 h = 60 Wh. $E=20\times 3=60\mathrm{Wh}$

Answer: C

Correct answer: 2,500 W. The prefix kilo means one thousand, so converting kilowatts to watts means multiplying by 1,000, which moves the decimal point three places to the right. Since a watt is the smaller unit, the number in watts must be larger. Errors usually come from moving the decimal only one or two places. The same factor converts kilometers to meters or kilograms to grams, so the rule carries over across all metric units.

• A: This is too small by a factor of 10.
• B: This is too large by a factor of 10.
• C (correct): 2.5 × 1,000 = 2,500 W.
• D: This is too small by a factor of 100.

Answer: A

Correct answer: 4 A. Rearrange the power formula to find current.

• A (correct): Dividing 480 W by 120 V with $I=P/V$ gives 4 A.
• B: This divides voltage by power, which flips the formula upside down.
• C: This is a decimal slip; 480 divided by 120 is 4, not 40.
• D: This subtracts the voltage from the power, which mixes two different units.

Answer: A

Correct answer: An energy quantity. A watt-hour is power multiplied by time, so it measures how much energy has been or can be delivered. A label of 40 Wh on a battery pack means it can supply, for example, 40 W for one hour or 10 W for four hours. The tempting confusion is with watts: a 40 W rating describes how fast a device uses energy, while 40 Wh describes a total amount stored or consumed.

• A (correct): A watt-hour is an energy unit.
• B: Wh does not measure potential difference.
• C: W and Wh are distinct quantities.
• D: Wh does not measure current.

Answer: D

Correct answer: Electric charge. An ampere-hour equals one ampere of current flowing for one hour, so it measures a quantity of charge. It becomes energy only after multiplying by the battery voltage.

• A: Power is a rate measured in watts, not a quantity measured in ampere-hours.
• B: Potential, or voltage, is measured in volts and is a separate value on the label.
• C: Energy needs the voltage too; volts times ampere-hours gives watt-hours of energy.
• D (correct): An ampere-hour is one ampere flowing for one hour, which is an amount of charge.

Answer: B

Correct answer: 0.6 h. Watt-hours need time in hours. Divide minutes by 60:

h.

• A: This divides 60 by 36, which flips the conversion.
• B (correct): Dividing 36 minutes by 60 gives 0.6 hours, the time unit that pairs with watts to give watt-hours.
• C: This simply moves the decimal point, but an hour has 60 minutes, not 100.
• D: This multiplies by 60, which converts minutes to seconds, and then labels the result as hours.

Answer: A

Correct answer: 2 kWh. Energy equals power times time, and keeping power in kilowatts and time in hours gives kilowatt-hours directly. A kilowatt-hour is defined as one kilowatt running for one hour, so two hours at that rate is twice as much. The tempting answer 2 kW has the right number but the wrong unit: kilowatts describe the heater's rate of use, while kilowatt-hours describe the total energy that accumulates while it runs.

• A (correct): 1 × 2 = 2 kWh.
• B: kW alone measures power, not accumulated energy.
• C: This accounts for only one hour.
• D: This would be for half an hour at 1 kW.

Answer: D

Correct answer: 80 W. With current and resistance given, use

• A: This squares the resistance instead of the current, using the incorrect form $I{R}^2$.
• B: This multiplies current by resistance, which gives the voltage across the resistor in volts, not power.
• C: This squares the voltage drop of 20 V, which squares the resistance as well as the current.
• D (correct): Squaring 4 A gives 16, and 16 times 5 $\Omega$ gives 80 W.

Answer: B

Correct answer: 48 W. With voltage and resistance given, use

• A: This squares the resistance as well as the voltage, using $V^2/R^2$.
• B (correct): Squaring 24 V gives 576, and 576 divided by 12 $\Omega$ gives 48 W.
• C: This multiplies voltage by resistance, which does not match any power formula.
• D: This divides voltage by resistance without squaring, which gives the current of 2 A.

Answer: C

Correct answer: 12 V. Rearrange the power formula to find voltage.

Check:

• A: This divides current by power, which flips the formula.
• B: This subtracts the current from the power, which mixes units.
• C (correct): Dividing 90 W by 7.5 A with $V=P/I$ gives 12 V.
• D: This multiplies power by current instead of dividing.

Answer: A

Correct answer: 85 percent. Efficiency is useful output divided by input.

• A (correct): Dividing the 340 W output by the 400 W input gives 0.85, or 85 percent.
• B: This divides input by output; an efficiency above 100 percent signals the ratio was flipped.
• C: This treats the 60 W of lost power as if it were a percentage.
• D: This is the fraction lost as heat and friction, 60 W out of 400 W, not the useful fraction.

Answer: C

Correct answer: It becomes four times as large. For a fixed resistor,

Since voltage is squared, doubling it makes the power

times larger. Another view: doubling V doubles I, and

then grows by

• A: Doubling applies only when current is held fixed, but in a resistor the current also doubles.
• B: Raising the voltage on a fixed resistor raises the current, so power cannot drop.
• C (correct): With $P=V^2/R$ and R fixed, doubling V multiplies power by $2^2=4$.
• D: The resistance stays the same, but the power depends on voltage squared, so it must change.

Answer: B

Correct answer: $21.60. Daily energy:

kWh. Monthly energy:

kWh.

Cost:

• A: This treats 12 cents as $0.012, a decimal slip in the rate.
• B (correct): The heater uses 180 kWh in the month, and 180 times $0.12 is $21.60.
• C: This treats 12 cents as $1.20 per kWh, a decimal slip in the other direction.
• D: This is the cost for only one day, 6 kWh at 12 cents.

Answer: A

Correct answer: 0.2 kWh. To get kilowatt-hours, both inputs must be in matching units: power in kilowatts and time in hours. Divide watts by 1,000 to get 0.4 kW, and divide minutes by 60 to get 0.5 h, then multiply. The most common error is skipping the time conversion, either treating 30 minutes as 30 hours or ignoring that half an hour is less than a full hour. Converting first and multiplying second avoids both mistakes.

• A (correct): 0.4 kW × 0.5 h = 0.2 kWh. $E=0.4\times 0.5=0.2\mathrm{kWh}$
• B: This ignores that it runs only half an hour.
• C: This multiplies incorrectly.
• D: This treats minutes as hours.

Answer: B

Correct answer: 7,200 J. A joule is one watt for one second, and a watt-hour is one watt for one hour. Since an hour contains

each watt-hour equals 3,600 J. Multiply 2 Wh by 3,600 to get the answer. A frequent mistake is using 60 instead of 3,600, converting only minutes rather than seconds. The same logic shows that one kilowatt-hour equals 3.6 million joules.

• A: This uses an incorrect seconds-per-hour factor.
• B (correct): 2 × 3,600 = 7,200 J.
• C: This is too small by a factor of ten.
• D: This is too large by a factor of ten.

Answer: C

Correct answer: $0.18. Electricity cost is energy multiplied by price per unit of energy, so first find kilowatt-hours. Convert 500 W to 0.5 kW, then multiply by 2 hours to get 1 kWh. At $0.18 for each kilowatt-hour, the charge for exactly one is $0.18. A common error is multiplying the price by the hours or by the watts without converting, which produces a cost too large. Always reduce power and time to kWh before applying the rate.

• A: This doubles the 1-kWh charge.
• B: The load uses 1 kWh, not 0.5 kWh.
• C (correct): 0.5 kW × 2 h = 1 kWh; cost = $0.18.
• D: This is ten times the correct charge.

Answer: A

Correct answer: 18,000 J. Joules require time in seconds:

• A (correct): Five minutes is 300 seconds, and 60 W times 300 s gives 18,000 J.
• B: This multiplies 60 W by 5 without converting minutes to seconds.
• C: This divides power by time, which gives a meaningless unit instead of energy.
• D: This treats 5 minutes as 5 hours before converting to seconds.

Answer: D

Correct answer: 7.4 hours. Convert capacity: 5,000 mAh is 5 Ah. Energy:

Wh.

Runtime:

• A: This divides ampere-hours by watts and leaves out the voltage, so it does not use energy.
• B: This multiplies stored energy by load power instead of dividing.
• C: This uses 5,000 instead of 5 Ah, forgetting to convert milliampere-hours to ampere-hours.
• D (correct): The pack stores 3.7 V times 5 Ah, or 18.5 Wh, and 18.5 Wh divided by 2.5 W is 7.4 hours.

Answer: C

Correct answer: 36 Wh. Amp-hours measure charge, and multiplying charge by voltage gives energy, so watt-hours equal volts times amp-hours. The battery could, ideally, deliver 3 A for one hour at about 12 V, which is 36 W for an hour. The estimate is rough because the actual voltage sags as the battery discharges. A common mistake is treating amp-hours as energy by themselves, but two batteries with equal Ah and different voltages store different energy.

• A: 12/3 is not the energy estimate.
• B: This is ten times the nominal estimate.
• C (correct): Nominal Wh ≈ 12 × 3 = 36 Wh. $E\approx 12\times 3=36\mathrm{Wh}$
• D: 12 + 3 is not the energy calculation.

Answer: C

Correct answer: 4 h. Runtime is stored energy divided by the rate at which it is used, just as travel time is distance divided by speed. Watt-hours divided by watts leaves hours, which confirms the setup. A quick check: 15 W for 4 hours uses 60 Wh, exactly the battery's capacity. Multiplying instead of dividing gives a huge, unreasonable figure, and real batteries run somewhat shorter than this ideal estimate because of conversion losses.

• A: Multiplying energy by power is not runtime.
• B: 60/15 is not 1.
• C (correct): t = 60 Wh / 15 W = 4 h. $t=\frac{60}{15}=4\mathrm{h}$
• D: 15 is the wattage, not duration.

Answer: A

Correct answer: 72 W. Series parts share the same current, so use

for each:

W and

Total:

which matches

• A (correct): The 2 $\Omega$ resistor dissipates 18 W and the 6 $\Omega$ resistor 54 W, for 72 W in total.
• B: This squares the resistance instead of the current, using $I{R}^2$ with 8 $\Omega$.
• C: This multiplies current by total resistance, which gives the total voltage of 24 V, not power.
• D: This is the power in the 6 $\Omega$ resistor alone and leaves out the 2 $\Omega$ resistor.

Answer: D

Correct answer: 60 W. Parallel branches share the same voltage, so use

for each:

W and

Total:

• A: This adds the branch currents of 3 A and 2 A and labels the result as power.
• B: This is the power in the 6 $\Omega$ branch only and leaves out the 4 $\Omega$ branch.
• C: This adds the resistances as if they were in series, which does not apply to parallel branches.
• D (correct): Each branch has 12 V across it, giving 36 W and 24 W, for 60 W in total.

Answer: A

Correct answer: They use equal energy. Energy depends on both power and time,

The 30 W lamp uses

Wh, and the 60 W lamp uses

Wh. Doubling the power while halving the time leaves the product unchanged. The tempting answer is the 60 W lamp, because a higher wattage feels like more energy, but wattage is only the rate; the time it runs matters just as much.

• A (correct): Each uses 120 Wh.
• B: 30 × 4 equals 120 Wh, not more than the other.
• C: 60 × 2 also equals 120 Wh.
• D: Operating electrical lamps transfer energy.

Answer: D

Correct answer: 75 W. Daily energy:

Wh.

Average power is energy divided by total time:

• A: This is the instantaneous power while the compressor runs, not the average over the whole day.
• B: This is the daily energy in watt-hours mislabeled as power.
• C: This divides running power by running hours, which does not give average power.
• D (correct): The freezer uses 180 W times 10 h, or 1,800 Wh, and spreading that over 24 hours gives 75 W.

Answer: A

Correct answer: 15 A. Loads running at once add their powers:

• A (correct): The combined load is 1,800 W, and 1,800 W divided by 120 V is 15 A.
• B: This is the toaster's current alone and leaves out the microwave.
• C: This divides voltage by total power, which flips the current formula.
• D: This is the microwave's current alone and leaves out the toaster.

Answer: D

Correct answer: 2.88 kWh. Hours in the month:

h. Energy:

Wh.

Divide by 1,000 to get 2.88 kWh.

• A: This is the energy for a single day, 96 Wh, not the whole month.
• B: This is the watt-hour value labeled as kilowatt-hours, missing the division by 1,000.
• C: This multiplies 4 by 30 days, skipping the 24 hours per day and the watt-to-kilowatt step.
• D (correct): It runs 720 hours, and 4 W times 720 h is 2,880 Wh, or 2.88 kWh.

Answer: B

Correct answer: 15 W. Source power:

In an ideal circuit the loads absorb all of it.

Third load:

• A: This subtracts only the 25 W load from the 60 W source and forgets the 20 W load.
• B (correct): The source delivers 60 W, and subtracting the 45 W used by the other loads leaves 15 W.
• C: This is the combined power of the two known loads, not the missing one.
• D: This adds the source power to the known loads instead of subtracting.

Answer: C

Correct answer: 1,215 W. Find the resistance from the rating:

At 108 V:

This equals $1500\times 0.9^2$.

• A: This scales the power by 0.9 in proportion to voltage, but power in a resistor depends on voltage squared.
• B: The rating applies only at 120 V; at a lower voltage a fixed resistance makes less power.
• C (correct): The element is 9.6 $\Omega$, and $108^2/9.6$ gives 1,215 W.
• D: This divides by 0.9, assuming a lower voltage makes more power, which is backward for a resistor.

Answer: C

Correct answer: 4 hours. Stored energy:

Wh. Battery output needed:

Runtime:

• A: This multiplies the load by 0.8, making the battery supply less than the load uses.
• B: This applies the 80 percent efficiency twice, once to the stored energy and once to the load.
• C (correct): The battery must supply 48 W divided by 0.8, or 60 W, and 240 Wh divided by 60 W is 4 hours.
• D: This divides 240 Wh by the 48 W load and ignores the inverter loss.

Answer: B

Correct answer: $0.51. Preheat:

kWh. Baking:

kWh.

Total: 2.55 kWh. Cost:

• A: This counts only the baking stage and leaves out the preheat energy.
• B (correct): The two stages use 0.75 kWh and 1.8 kWh, a total of 2.55 kWh, which costs $0.51 at 20 cents.
• C: This treats the 15-minute preheat as 15 hours instead of 0.25 hours.
• D: This averages the two power levels over the full 1.75 hours, but the stages last different times.

Answer: D

Correct answer: 18 W. For a fixed resistor,

so power scales with the square of current.

• A: This cubes the factor of 3, but current is squared in the power formula.
• B: This assumes a fixed resistance fixes the power, but power grows with current squared.
• C: This scales power directly with current, which holds only at fixed voltage, not for a fixed resistor.
• D (correct): With $P=I^{2}R$, tripling the current multiplies power by 9, and 2 W times 9 is 18 W.

Answer: B

Correct answer: 1 W. Expected dissipation:

Twice that is 0.9 W, so the smallest listed rating that meets it is 1 W.

• A: This rating is safe but larger than needed, since 1 W already meets the guideline.
• B (correct): The resistor dissipates 0.45 W, twice that is 0.9 W, and 1 W is the smallest listed rating that covers it.
• C: This covers the 0.45 W dissipation with almost no margin, falling short of the twice-dissipation guideline.
• D: This comes from using $V/R=0.05$ as the power; the actual dissipation is 0.45 W, which would overheat this part.

Answer: B

Correct answer: 10 A. Required input:

• A: This divides by the efficiency twice, overstating the input power.
• B (correct): Input power is 2,160 W divided by 0.9, or 2,400 W, and 2,400 W divided by 240 V is 10 A.
• C: This multiplies the output by the efficiency, which makes the input smaller than the output.
• D: This uses the output power as the input and ignores the motor's losses.

Answer: C

Correct answer: The 10 Ω resistor dissipates three times as much. Parallel branches have the same voltage, so use

Power is inversely proportional to resistance:

so the 10 $\Omega$ resistor dissipates three times as much.

• A: This squares the resistance ratio, but only voltage is squared in $V^2/R$.
• B: This is the series result, where equal current makes the larger resistor dissipate more.
• C (correct): With equal voltage, $P=V^2/R$, so the resistor with one-third the resistance dissipates three times the power.
• D: Equal voltage does not mean equal power; the branch currents differ, so the powers differ.

Answer: B

Correct answer: The 12 V, 2 Ah battery, about 24 Wh. Energy in watt-hours is volts times ampere-hours. The first: 6,000 mAh is 6 Ah, so

Wh.

The second:

Wh, so the 12 V battery stores more despite its smaller ampere-hour number.

• A: This value is right for the 3.7 V pack, but 22.2 Wh is less than the 24 Wh of the other battery.
• B (correct): Multiplying 12 V by 2 Ah gives 24 Wh, which beats the 22.2 Wh of the 3.7 V pack.
• C: This divides voltage by ampere-hours instead of multiplying them.
• D: This multiplies 3.7 V by 6,000 without converting milliampere-hours to ampere-hours.

Answer: D

Correct answer: 15 A. Time in seconds:

Power:

• A: This divides the energy by 2 without converting minutes to seconds.
• B: This converts 2 minutes to hours and divides joules by hours, which is a unit mismatch.
• C: This divides energy by voltage and leaves out the time altogether.
• D (correct): Power is 216,000 J divided by 120 s, or 1,800 W, and 1,800 W divided by 120 V is 15 A.

Answer: D

Correct answer: $0.015. Work in three steps: power, energy, cost. Power is voltage times current, so 20 V × 2 A = 40 W, which is 0.04 kW. Ninety minutes is 1.5 hours, so the energy is 0.06 kWh. Multiplying by the price per kWh gives the cost. The most common errors are leaving power in watts instead of kilowatts, which inflates the answer a thousandfold, or treating 90 minutes as 0.9 hours.

• A: This is one third of the correct charge.
• B: This is ten times the energy-only charge.
• C: This is one hundred times the correct charge.
• D (correct): P = 20 V × 2 A = 40 W = 0.04 kW; E = 0.04 kW × 1.5 h = 0.06 kWh; cost = 0.06 × $0.25 = $0.015.

Answer: C

Correct answer: 18 Wh. Stored energy is about volts times ampere-hours. Convert 2,500 mAh to 2.5 Ah, then

Wh.

• A: This multiplies by 2,500 without first converting milliampere-hours to ampere-hours.
• B: This divides the voltage by the capacity instead of multiplying them.
• C (correct): Converting 2,500 mAh to 2.5 Ah and multiplying by 7.2 V gives 18 Wh.
• D: This divides 2,500 by 7.2, which mixes up the units and the operation.

Answer: C

Correct answer: 180 W. Use

Here

• A: This divides voltage by current, which gives resistance in ohms, not power.
• B: This adds the voltage and current, which are different quantities.
• C (correct): Power is voltage times current: 120 times 1.5 is 180 W.
• D: This misplaces the decimal point when multiplying by 1.5.

Answer: A

Correct answer: 1.25 kW. One kilowatt is 1,000 W, so divide by 1,000:

• A (correct): Dividing 1,250 W by 1,000 gives 1.25 kW.
• B: This divides by 100 instead of 1,000.
• C: This divides by 10,000, moving the decimal one place too far.
• D: This divides by 10 instead of 1,000.

Answer: B

Correct answer: 2,160 J. Joules come from watts times seconds. Convert the time:

Then

• A: This multiplies by 8 without converting minutes to seconds.
• B (correct): Eight minutes is 480 s, and 4.5 W times 480 s gives 2,160 J.
• C: This treats the 8 minutes as 8 hours and then multiplies by 3,600.
• D: This uses only one minute, 60 s, instead of the full 8 minutes.

Answer: B

Correct answer: 30 W. Input power equals output divided by efficiency:

Check: 90 percent of 30 W is 27 W, and the other 3 W is lost as heat.

• A: This multiplies by 0.90, which would make the input smaller than the output.
• B (correct): Dividing the useful output by the efficiency, 27 divided by 0.90, gives 30 W of input.
• C: This adds 10 percent of the output instead of dividing by the efficiency.
• D: This is the power lost as heat, not the input power.

Answer: D

Correct answer: $15.12. Convert to kilowatts: 300 W is 0.3 kW. Hours:

h, so energy is

kWh.

Cost:

• A: This is the cost of a single day; the 30 days were left out.
• B: This assumes the unit runs 24 hours a day instead of 12.
• C: This uses 300 W directly as if it were kilowatts, skipping the divide-by-1,000 step.
• D (correct): The month uses 0.3 kW times 360 h, or 108 kWh, and 108 times $0.14 is $15.12.

Answer: A

Correct answer: 48 W. Total resistance is

so the current is

A through both resistors.

With a shared current, use

• A (correct): The series current is 32 divided by 16, or 2 A, and $2^2\times 12=48$ W.
• B: This applies the full 32 V to the 12 Ω resistor alone, but it only has part of the source voltage.
• C: This is the total power delivered by the source, not the share in the 12 Ω resistor.
• D: This is the power in the 4 Ω resistor, not the 12 Ω resistor.

Answer: D

Correct answer: 294 Wh. Find each stage separately. Warm-up: 6 minutes is 0.1 h, so

Wh. Printing:

Wh.

Total:

Wh.

• A: This multiplies 240 W by 6 without converting the minutes to hours.
• B: This averages the two powers even though the stages last very different times.
• C: This simply adds the two power ratings, which gives watts, not energy.
• D (correct): The warm-up uses 240 W times 0.1 h, or 24 Wh, and the printing uses 270 Wh, for 294 Wh in all.

Answer: D

Correct answer: From 12 W to 36 W. With voltage fixed, use

Before:

After:

Cutting the resistance to one-third triples the power.

• A: This assumes power rises and falls with resistance, which is only true at fixed current.
• B: This squares the threefold resistance change, but resistance is not squared in the formula.
• C: These are the currents in amperes, 24 divided by each resistance, not the powers.
• D (correct): At fixed voltage, $P=V^2/R$: 576 divided by 48 is 12 W, and 576 divided by 16 is 36 W.

Answer: A

Correct answer: 9.6 Ω. Convert 1.5 kW to 1,500 W. From

Check: the current is

• A (correct): Rearranging $P=V^2/R$ gives R = 14,400 divided by 1,500, or 9.6 Ω.
• B: This is 1,500 divided by 120, which is the current in amperes, not a resistance.
• C: This divides the voltage by the power, which does not give ohms.
• D: This divides the power by the voltage squared, which is the formula turned upside down.