AP Physics 2 Flashcards: Electric Power

Study Electric Power in AP Physics 2 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

AP Physics 2

Electric Power

0 mastered0 still learning

0% Complete

QUESTION
1/ 57

How is power related to current and voltage in an AC circuit?

Tap card or press Space to flip

ANSWER

P=VIcosϕP = VI \cos \phi. AC power includes the phase angle factor.

How well did you know it?

Card 1 / 57

What this deck covers

This deck focuses on Electric Power, giving you a quick way to review the definitions, rules, and examples that matter most for AP Physics 2.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

All flashcards

Flashcard 1: How is power related to current and voltage in an AC circuit?

Answer: P=VIcosϕP = VI \cos \phi. AC power includes the phase angle factor.

Flashcard 2: State the expression for power factor in terms of real and apparent power.

Answer: Power Factor=PS\text{Power Factor} = \frac{P}{S}. Power factor equals real power over apparent power.

Flashcard 3: Calculate the power if V=12VV = 12 \, V and I=0.5AI = 0.5 \, A.

Answer: P=6WP = 6 \, W. Using P=VI=12×0.5=6P = VI = 12 \times 0.5 = 6 W.

Flashcard 4: Find the power factor if the power is 400W400 \, W, I=5AI = 5 \, A, and V=100VV = 100 \, V.

Answer: cosϕ=0.8\cos \phi = 0.8. Power factor = PVI=4005×100=0.8\frac{P}{VI} = \frac{400}{5 \times 100} = 0.8.

Flashcard 5: How is reactive power denoted in an electrical circuit?

Answer: QQ. QQ represents reactive power in circuits.

Flashcard 6: What does a power factor of 1 signify in an AC circuit?

Answer: Purely resistive load. Unit power factor means no reactive components.

Flashcard 7: Identify the formula for efficiency in terms of power.

Answer: η=PoutPin×100%\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%. Efficiency is output power divided by input power.

Flashcard 8: Identify the formula for power loss in terms of voltage drop and current.

Answer: Ploss=IΔVP_{\text{loss}} = I \cdot \Delta V. Power loss equals current times voltage drop.

Flashcard 9: Convert 10kW10 \, \text{kW} to watts.

Answer: 10,000W10,000 \, W. 11 kW =1000= 1000 W, so 1010 kW =10,000= 10,000 W.

Flashcard 10: What is the role of a transformer in power transmission?

Answer: Adjusts voltage levels. Transformers step voltage up or down efficiently.

Flashcard 11: What is the power dissipation if V=10VV = 10 \, V and R=5ΩR = 5 \, \text{Ω}?

Answer: P=20WP = 20 \, W. Using P=V2R=1025=20P = \frac{V^2}{R} = \frac{10^2}{5} = 20 W.

Flashcard 12: What is the power output if E=500JE = 500 \, \text{J} and t=10st = 10 \, \text{s}?

Answer: P=50WP = 50 \, W. Using P=Et=50010=50P = \frac{E}{t} = \frac{500}{10} = 50 W.

Flashcard 13: What is the unit of reactive power?

Answer: Volt-ampere reactive (VAR). VAR is the unit for reactive power.

Flashcard 14: Define apparent power in an AC circuit.

Answer: S=VIS = VI. Apparent power is the total power in AC circuits.

Flashcard 15: Calculate power if V=120VV = 120 \, V, I=3AI = 3 \, A, and cosϕ=0.9\cos \phi = 0.9.

Answer: P=324WP = 324 \, W. Using P=VIcosϕ=120×3×0.9=324P = VI\cos\phi = 120 \times 3 \times 0.9 = 324 W.

Flashcard 16: State the expression for power factor in terms of real and apparent power.

Answer: Power Factor=PS\text{Power Factor} = \frac{P}{S}. Power factor equals real power over apparent power.

Flashcard 17: What is the formula for electric power in terms of voltage and resistance?

Answer: P=V2RP = \frac{V^2}{R}. Power equals voltage squared divided by resistance.

Flashcard 18: What is the resistance if P=60WP = 60 \, W and I=2AI = 2 \, A?

Answer: R=15ΩR = 15 \, \text{Ω}. Using P=I2RP = I^2R, so R=PI2=604=15R = \frac{P}{I^2} = \frac{60}{4} = 15 Ω.

Flashcard 19: What is the formula for electric power in terms of current and voltage?

Answer: P=IVP = IV. Power is the product of voltage and current.

Flashcard 20: Calculate the apparent power if V=220VV = 220 \, V and I=5AI = 5 \, A.

Answer: S=1100VAS = 1100 \, VA. Using S=VI=220×5=1100S = VI = 220 \times 5 = 1100 VA.

Flashcard 21: What is the resistance if P=60WP = 60 \, W and I=2AI = 2 \, A?

Answer: R=15ΩR = 15 \, \text{Ω}. Using P=I2RP = I^2R, so R=PI2=604=15R = \frac{P}{I^2} = \frac{60}{4} = 15 Ω.

Flashcard 22: State the formula for electric power in terms of resistance and current.

Answer: P=I2RP = I^2R. Power equals current squared times resistance.

Flashcard 23: Calculate the energy consumed in 2hours2 \, \text{hours} by a 100W100 \, W bulb.

Answer: E=0.2kWhE = 0.2 \, \text{kWh}. Using E=Pt=0.1×2=0.2E = Pt = 0.1 \times 2 = 0.2 kWh.

Flashcard 24: What unit is used to measure apparent power?

Answer: Volt-ampere (VA). VA measures the magnitude of AC power.

Flashcard 25: State the relationship between power, energy, and time.

Answer: P=EtP = \frac{E}{t}. Power is energy divided by time.

Flashcard 26: What does 1kWh1 \, \text{kWh} represent in terms of energy?

Answer: 3.6×106J3.6 \times 10^6 \, \text{J}. One kWh equals 3.6 million joules.

Flashcard 27: Define the term 'active power' in AC circuits.

Answer: Real power. Active power performs actual work in circuits.

Flashcard 28: What is the rate of energy transfer called in an electrical circuit?

Answer: Electric Power. Power is energy transferred per unit time.

Flashcard 29: Find the power factor if the power is 400W400 \, W, I=5AI = 5 \, A, and V=100VV = 100 \, V.

Answer: cosϕ=0.8\cos \phi = 0.8. Power factor = PVI=4005×100=0.8\frac{P}{VI} = \frac{400}{5 \times 100} = 0.8.

Flashcard 30: What is the power loss in a transmission line given I=10AI = 10 \, A and R=2ΩR = 2 \, \text{Ω}?

Answer: Ploss=200WP_{\text{loss}} = 200 \, W. Using Ploss=I2R=102×2=200P_{\text{loss}} = I^2R = 10^2 \times 2 = 200 W.

Flashcard 31: State the formula for electric power in terms of resistance and current.

Answer: P=I2RP = I^2R. Power equals current squared times resistance.

Flashcard 32: Which physical quantity is measured in kilowatt-hours?

Answer: Energy. kWh measures energy consumption over time.

Flashcard 33: Identify the formula for efficiency in terms of power.

Answer: η=PoutPin×100%\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%. Efficiency is output power divided by input power.

Flashcard 34: Calculate the energy consumed in 2hours2 \, \text{hours} by a 100W100 \, W bulb.

Answer: E=0.2kWhE = 0.2 \, \text{kWh}. Using E=Pt=0.1×2=0.2E = Pt = 0.1 \times 2 = 0.2 kWh.

Flashcard 35: Identify the unit of energy used in electric bills.

Answer: Kilowatt-hour (kWh). kWh is the commercial unit for electrical energy.

Flashcard 36: What is the efficiency if Pout=90WP_{\text{out}} = 90 \, W and Pin=100WP_{\text{in}} = 100 \, W?

Answer: η=90%\eta = 90\%. Efficiency = 90100×100%=90%\frac{90}{100} \times 100\% = 90\%.

Flashcard 37: What is the significance of reactive power in AC circuits?

Answer: Non-working power. Reactive power doesn't perform useful work.

Flashcard 38: Calculate the apparent power if V=220VV = 220 \, V and I=5AI = 5 \, A.

Answer: S=1100VAS = 1100 \, VA. Using S=VI=220×5=1100S = VI = 220 \times 5 = 1100 VA.

Flashcard 39: Define apparent power in an AC circuit.

Answer: S=VIS = VI. Apparent power is the total power in AC circuits.

Flashcard 40: How is reactive power denoted in an electrical circuit?

Answer: QQ. QQ represents reactive power in circuits.

Flashcard 41: What is the phase angle for a purely resistive circuit?

Answer: 00^\circ. Resistive circuits have zero phase difference.

Flashcard 42: Which physical quantity is measured in kilowatt-hours?

Answer: Energy. kWh measures energy consumption over time.

Flashcard 43: Calculate the power if V=12VV = 12 \, V and I=0.5AI = 0.5 \, A.

Answer: P=6WP = 6 \, W. Using P=VI=12×0.5=6P = VI = 12 \times 0.5 = 6 W.

Flashcard 44: What is the efficiency if Pout=90WP_{\text{out}} = 90 \, W and Pin=100WP_{\text{in}} = 100 \, W?

Answer: η=90%\eta = 90\%. Efficiency = 90100×100%=90%\frac{90}{100} \times 100\% = 90\%.

Flashcard 45: Calculate the power if I=2AI = 2 \, A and V=5VV = 5 \, V.

Answer: P=10WP = 10 \, W. Using P=IV=2×5=10P = IV = 2 \times 5 = 10 W.

Flashcard 46: Identify the unit of electric power in the International System of Units.

Answer: Watt (W). The SI unit for power is the watt.

Flashcard 47: State the formula for instantaneous power in terms of time.

Answer: p(t)=v(t)i(t)p(t) = v(t) \cdot i(t). Instantaneous power is voltage times current.

Flashcard 48: What is the power loss in a transmission line given I=10AI = 10 \, A and R=2ΩR = 2 \, \text{Ω}?

Answer: Ploss=200WP_{\text{loss}} = 200 \, W. Using Ploss=I2R=102×2=200P_{\text{loss}} = I^2R = 10^2 \times 2 = 200 W.

Flashcard 49: How is power related to current and voltage in an AC circuit?

Answer: P=VIcosϕP = VI \cos \phi. AC power includes the phase angle factor.

Flashcard 50: What does a power factor of 1 signify in an AC circuit?

Answer: Purely resistive load. Unit power factor means no reactive components.

Flashcard 51: State the relationship between power, energy, and time.

Answer: P=EtP = \frac{E}{t}. Power is energy divided by time.

Flashcard 52: Convert 10kW10 \, \text{kW} to watts.

Answer: 10,000W10,000 \, W. 11 kW =1000= 1000 W, so 1010 kW =10,000= 10,000 W.

Flashcard 53: Determine the power consumed by a 4Ω4 \, \text{Ω} resistor with 3A3 \, A current.

Answer: P=36WP = 36 \, W. Using P=I2R=32×4=36P = I^2R = 3^2 \times 4 = 36 W.

Flashcard 54: Find the power factor if apparent power is 1000VA1000 \, VA and real power is 800W800 \, W.

Answer: cosϕ=0.8\cos \phi = 0.8. Power factor = PS=8001000=0.8\frac{P}{S} = \frac{800}{1000} = 0.8.

Flashcard 55: State the formula for instantaneous power in terms of time.

Answer: p(t)=v(t)i(t)p(t) = v(t) \cdot i(t). Instantaneous power is voltage times current.

Flashcard 56: Define the term 'active power' in AC circuits.

Answer: Real power. Active power performs actual work in circuits.

Flashcard 57: What is the formula for electric power in terms of voltage and resistance?

Answer: P=V2RP = \frac{V^2}{R}. Power equals voltage squared divided by resistance.