Electric Power
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AP Physics C: Electricity and Magnetism › Electric Power
A space heater is modeled as a single resistor of $R=24,\Omega$ connected across a $120,\text{V}$ outlet. The heater is the only load, so the voltage across the resistor equals the source voltage. Use $P=\frac{V^2}{R}$ to compute the electrical power converted to heat. Report the power in watts (W). Determine the power consumed by the appliance.
300 W
5.0 W
600 J
600 W
Explanation
This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes a 24 Ω heater connected across 120 V, requiring calculation of power using P=V²/R. Choice A (600 W) is correct because it applies the formula P=V²/R using values V=120 V and R=24 Ω, resulting in P=(120 V)²/(24 Ω)=14400/24=600 W. Choice D (600 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize that space heaters typically consume hundreds of watts and always check units. Practice with common household appliances to build intuition for realistic power values.
A household night-light is modeled as a single resistor of $R=240,\Omega$ connected directly across a $120,\text{V}$ outlet. The circuit is steady-state and the resistor is the only load. Use $P=\frac{V^2}{R}$ to find the electrical power converted to thermal energy and light. Report the result in watts (W). Determine the power consumed by the appliance.
60 W
0.50 W
60 J
30 W
Explanation
This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes a 240 Ω resistor connected across 120 V, requiring calculation of power using P=V²/R. Choice B (60 W) is correct because it applies the formula P=V²/R using values V=120 V and R=240 Ω, resulting in P=(120 V)²/(240 Ω)=14400/240=60 W. Choice A (0.50 W) is incorrect due to likely calculating I=V/R=0.5 A without completing the power calculation, often a result of stopping midway through the problem. To help students, emphasize completing all steps and choosing the appropriate power formula based on given quantities. Practice with household appliance examples to build intuition for typical power values.
A long extension cord is modeled as a $R=0.20,\Omega$ resistor in series with a load, and the current through the cord is $I=8.0,\text{A}$. The cord warms during operation, indicating resistive heating in the wire. Use $P=I^2R$ to determine the power lost as heat in the cord. Give your answer in watts (W). How much power is lost in the transmission line?
12.8 W
40 W
1.6 W
12.8 J
Explanation
This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes an extension cord with R=0.20 Ω carrying I=8.0 A, requiring calculation of resistive power loss. Choice B (12.8 W) is correct because it applies the formula P=I²R using values I=8.0 A and R=0.20 Ω, resulting in P=(8.0 A)²(0.20 Ω)=64×0.20=12.8 W. Choice D (12.8 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize that power loss in wires is continuous dissipation measured in watts. Practice calculating wire losses to understand why thick wires are used for high currents.
A 9.0 V battery powers a wearable device modeled as two resistors in series: $R_1=3.0,\Omega$ and $R_2=6.0,\Omega$, so the total resistance is $9.0,\Omega$. The same current flows through both resistors and equals $1.0,\text{A}$. Determine the power dissipated by $R_2$ using $P=I^2R$. Give your answer in watts (W). What is the power dissipated by the resistor in the circuit?
6.0 J
9.0 W
3.0 W
6.0 W
Explanation
This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes two resistors in series with R₂=6.0 Ω and current I=1.0 A, requiring calculation of power dissipated by R₂. Choice A (6.0 W) is correct because it applies the formula P=I²R using values I=1.0 A and R=6.0 Ω, resulting in P=(1.0 A)²(6.0 Ω)=6.0 W. Choice D (6.0 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize unit consistency and that power dissipation in a resistor is always P=I²R when current is known. Practice recognizing series circuits where the same current flows through all components.
A desk lamp uses a bulb modeled as a resistor of $R=144,\Omega$ connected across a $12.0,\text{V}$ DC supply. The lamp is used in a low-voltage lighting circuit with negligible wire resistance. Use $P=\frac{V^2}{R}$ to determine the bulb’s power usage. Report the answer in watts (W). What is the power usage of the light bulb?
12 W
1.0 W
144 W
0.083 W
Explanation
This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes a 144 Ω bulb connected across 12.0 V DC, requiring calculation of power using P=V²/R. Choice A (1.0 W) is correct because it applies the formula P=V²/R using values V=12.0 V and R=144 Ω, resulting in P=(12.0 V)²/(144 Ω)=144/144=1.0 W. Choice D (144 W) is incorrect due to using V² without dividing by R, often a result of incomplete formula application. To help students, emphasize careful formula application and unit checking. Practice with low-voltage DC circuits to build familiarity with typical power values.
A transmission line is modeled as a resistor of $R=0.80,\Omega$ delivering power to a remote load, and the line current is $I=15,\text{A}$. The line’s heating is a safety concern, so you calculate resistive loss. Use $P=I^2R$ with current in amperes and resistance in ohms. Give the power loss in watts (W). How much power is lost in the transmission line?
12 W
180 J
180 W
225 W
Explanation
This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes a transmission line with R=0.80 Ω carrying I=15 A, requiring calculation of resistive power loss. Choice B (180 W) is correct because it applies the formula P=I²R using values I=15 A and R=0.80 Ω, resulting in P=(15 A)²(0.80 Ω)=225×0.80=180 W. Choice D (180 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize that transmission line losses are calculated using P=I²R and represent continuous power dissipation. Practice with real-world examples to understand why minimizing transmission losses is important.
A reading lamp uses a bulb modeled as a $R=60,\Omega$ resistor connected to a $12,\text{V}$ supply in a simple lighting circuit. The voltage across the bulb remains $12,\text{V}$ during operation. Use $P=\frac{V^2}{R}$ to compute the bulb’s power usage. Report the result in watts (W). What is the power usage of the light bulb?
2.4 J
2.4 W
24 W
0.20 W
Explanation
This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes a 60 Ω bulb connected across 12 V, requiring calculation of power using P=V²/R. Choice A (2.4 W) is correct because it applies the formula P=V²/R using values V=12 V and R=60 Ω, resulting in P=(12 V)²/(60 Ω)=144/60=2.4 W. Choice D (2.4 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize consistent unit usage and that low-voltage bulbs typically consume a few watts. Practice with various voltage and resistance combinations to build computational fluency.
A 18.0 V battery powers a portable circuit with two resistors in parallel, $R_1=9.0,\Omega$ and $R_2=18.0,\Omega$, giving total resistance $R_\text{eq}=6.0,\Omega$. The total current drawn from the battery is $I=3.0,\text{A}$. Use $P=IV$ to compute the power supplied by the battery. Report power in watts (W). Calculate the power supplied by the battery.
54 W
54 J
162 W
6.0 W
Explanation
This question assesses understanding of electric power calculations in circuits (AP Physics C: Electricity and Magnetism). Electric power in circuits is determined by the product of current and voltage (P=IV), or equivalently by P=I²R or P=V²/R, depending on known values. In this scenario, the circuit includes an 18.0 V battery with total current I=3.0 A, requiring calculation of power supplied by the battery. Choice B (54 W) is correct because it applies the formula P=IV using values I=3.0 A and V=18.0 V, resulting in P=(3.0 A)(18.0 V)=54 W. Choice D (54 J) is incorrect due to using joules instead of watts, often a result of confusing power units with energy units. To help students, emphasize that battery power output is always P=IV where I is total current drawn and V is battery voltage. Practice with parallel circuits to understand how total current relates to individual branch currents.