AP Physics 2 Flashcards: Simple Circuits

Study Simple Circuits 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

Simple Circuits

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QUESTION
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Find the current if V=12VV = 12 \text{V} and R=4ΩR = 4 \text{Ω}.

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ANSWER

I=3AI = 3 \text{A}. Using V=IRV = IR, so I=V/R=12/4I = V/R = 12/4.

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What this deck covers

This deck focuses on Simple Circuits, 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.

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Flashcard 1: Find the current if V=12VV = 12 \text{V} and R=4ΩR = 4 \text{Ω}.

Answer: I=3AI = 3 \text{A}. Using V=IRV = IR, so I=V/R=12/4I = V/R = 12/4.

Flashcard 2: How does internal resistance affect a real battery's voltage?

Answer: Reduces terminal voltage. Internal resistance causes voltage drop.

Flashcard 3: Calculate the total current if I1=2AI_1 = 2 \text{A} and I2=3AI_2 = 3 \text{A} in parallel.

Answer: Itotal=5AI_{total} = 5 \text{A}. Simple addition of parallel currents.

Flashcard 4: Find the equivalent resistance of R1=6ΩR_1 = 6 \text{Ω} and R2=3ΩR_2 = 3 \text{Ω} in parallel.

Answer: Req=2ΩR_{eq} = 2 \text{Ω}. Using parallel formula: 1/Req=1/6+1/31/R_{eq} = 1/6 + 1/3.

Flashcard 5: What is the unit of electrical resistance?

Answer: Ohm (Ω). Named after Georg Simon Ohm.

Flashcard 6: Calculate the resistance if V=24VV = 24 \text{V} and I=3AI = 3 \text{A}.

Answer: R=8ΩR = 8 \text{Ω}. Using R=V/I=24/3R = V/I = 24/3.

Flashcard 7: What happens to power if both voltage and current are doubled?

Answer: Power increases fourfold. Power equals IVIV, so doubling both gives 4P4P.

Flashcard 8: What is the formula for energy consumed by an electrical device?

Answer: E=PtE = Pt. Energy equals power times time.

Flashcard 9: What type of circuit has only one path for current flow?

Answer: Series Circuit. Single continuous current path.

Flashcard 10: State the formula for equivalent resistance in a series circuit.

Answer: Req=R1+R2+...+RnR_{eq} = R_1 + R_2 + \text{...} + R_n. Resistances add directly in series.

Flashcard 11: What happens to the total resistance in a parallel circuit when more resistors are added?

Answer: It decreases. More paths reduce overall resistance.

Flashcard 12: What is the unit of electric current?

Answer: Ampere (A). Named after André-Marie Ampère.

Flashcard 13: What happens to power if both voltage and current are doubled?

Answer: Power increases fourfold. Power equals IVIV, so doubling both gives 4P4P.

Flashcard 14: What happens to the total resistance in a series circuit when more resistors are added?

Answer: It increases. More resistances add to total resistance.

Flashcard 15: Calculate the power if V=10VV = 10 \text{V} and I=2AI = 2 \text{A}.

Answer: P=20WP = 20 \text{W}. Using P=IV=10×2P = IV = 10 \times 2.

Flashcard 16: State Ohm's Law in terms of voltage, current, and resistance.

Answer: V=IRV = IR. Voltage equals current times resistance.

Flashcard 17: Calculate the power dissipated by a 4Ω4 \text{Ω} resistor with 2A2 \text{A} current.

Answer: P=16WP = 16 \text{W}. Using P=I2R=22×4P = I^2R = 2^2 \times 4.

Flashcard 18: What is the effect on resistance if the length of a resistor is doubled?

Answer: Resistance doubles. Resistance is proportional to length.

Flashcard 19: What device is used to measure electric current?

Answer: Ammeter. Connected in series with circuit.

Flashcard 20: What is the unit of electrical power?

Answer: Watt (W). Named after James Watt.

Flashcard 21: What is the direction of conventional current flow?

Answer: From positive to negative. Convention established before electron discovery.

Flashcard 22: What is the voltage across each resistor in a series circuit?

Answer: It varies depending on resistance. Voltage divides proportionally by resistance.

Flashcard 23: Identify the formula for calculating electrical power in a circuit.

Answer: P=IVP = IV. Power equals current times voltage.

Flashcard 24: Find the equivalent resistance of R1=2ΩR_1 = 2 \text{Ω} and R2=3ΩR_2 = 3 \text{Ω} in series.

Answer: Req=5ΩR_{eq} = 5 \text{Ω}. Series resistances add: 2+3=52 + 3 = 5.

Flashcard 25: What is the total current in a parallel circuit?

Answer: Sum of currents through each branch. Kirchhoff's junction rule application.

Flashcard 26: Identify the relationship between power, voltage, and resistance.

Answer: P=V2RP = \frac{V^2}{R}. Derived from P=IVP = IV and I=V/RI = V/R.

Flashcard 27: What happens to the total resistance in a parallel circuit when more resistors are added?

Answer: It decreases. More paths reduce overall resistance.

Flashcard 28: Identify the formula for electric charge.

Answer: Q=ItQ = It. Charge equals current times time.

Flashcard 29: Calculate the total current if I1=2AI_1 = 2 \text{A} and I2=3AI_2 = 3 \text{A} in parallel.

Answer: Itotal=5AI_{total} = 5 \text{A}. Simple addition of parallel currents.

Flashcard 30: What is the unit of electric potential difference?

Answer: Volt (V). Named after Alessandro Volta.

Flashcard 31: What is Kirchhoff's Loop Rule?

Answer: Sum of potential differences in any closed loop is zero. Conservation of energy principle.

Flashcard 32: Calculate the power dissipated by a 4Ω4 \text{Ω} resistor with 2A2 \text{A} current.

Answer: P=16WP = 16 \text{W}. Using P=I2R=22×4P = I^2R = 2^2 \times 4.

Flashcard 33: Find the equivalent resistance of R1=6ΩR_1 = 6 \text{Ω} and R2=3ΩR_2 = 3 \text{Ω} in parallel.

Answer: Req=2ΩR_{eq} = 2 \text{Ω}. Using parallel formula: 1/Req=1/6+1/31/R_{eq} = 1/6 + 1/3.

Flashcard 34: What is the unit of electric current?

Answer: Ampere (A). Named after André-Marie Ampère.

Flashcard 35: What is Kirchhoff's Junction Rule?

Answer: Total current entering a junction equals total current leaving. Conservation of electric charge.

Flashcard 36: What is the current through a 12Ω12 \text{Ω} resistor with 24V24 \text{V} across it?

Answer: I=2AI = 2 \text{A}. Using I=V/R=24/12I = V/R = 24/12.

Flashcard 37: State Ohm's Law in terms of voltage, current, and resistance.

Answer: V=IRV = IR. Voltage equals current times resistance.

Flashcard 38: State the formula for equivalent resistance in a parallel circuit.

Answer: 1Req=1R1+1R2+...+1Rn\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \text{...} + \frac{1}{R_n}. Reciprocal of sum of reciprocals.

Flashcard 39: Identify the relationship between power, voltage, and resistance.

Answer: P=V2RP = \frac{V^2}{R}. Derived from P=IVP = IV and I=V/RI = V/R.

Flashcard 40: What happens to the total resistance in a series circuit when more resistors are added?

Answer: It increases. More resistances add to total resistance.

Flashcard 41: What is the voltage across each resistor in a series circuit?

Answer: It varies depending on resistance. Voltage divides proportionally by resistance.

Flashcard 42: Identify the relationship between power, current, and resistance.

Answer: P=I2RP = I^2R. Derived from P=IVP = IV and V=IRV = IR.

Flashcard 43: What is the effect on current if voltage doubles and resistance remains constant?

Answer: Current doubles. Current is directly proportional to voltage.

Flashcard 44: Identify the formula for calculating electrical power in a circuit.

Answer: P=IVP = IV. Power equals current times voltage.

Flashcard 45: What happens to total voltage in a series circuit with identical voltage sources?

Answer: It adds up. Voltages sum algebraically in series.

Flashcard 46: What is the current through a 12Ω12 \text{Ω} resistor with 24V24 \text{V} across it?

Answer: I=2AI = 2 \text{A}. Using I=V/R=24/12I = V/R = 24/12.

Flashcard 47: What device is used to measure voltage?

Answer: Voltmeter. Connected in parallel with component.

Flashcard 48: State the formula for equivalent resistance in a series circuit.

Answer: Req=R1+R2+...+RnR_{eq} = R_1 + R_2 + \text{...} + R_n. Resistances add directly in series.

Flashcard 49: Calculate the resistance if V=24VV = 24 \text{V} and I=3AI = 3 \text{A}.

Answer: R=8ΩR = 8 \text{Ω}. Using R=V/I=24/3R = V/I = 24/3.

Flashcard 50: Find the equivalent resistance of R1=2ΩR_1 = 2 \text{Ω} and R2=3ΩR_2 = 3 \text{Ω} in series.

Answer: Req=5ΩR_{eq} = 5 \text{Ω}. Series resistances add: 2+3=52 + 3 = 5.

Flashcard 51: What is the effect on current if voltage doubles and resistance remains constant?

Answer: Current doubles. Current is directly proportional to voltage.

Flashcard 52: What is the total current in a parallel circuit?

Answer: Sum of currents through each branch. Kirchhoff's junction rule application.

Flashcard 53: Identify the relationship between power, current, and resistance.

Answer: P=I2RP = I^2R. Derived from P=IVP = IV and V=IRV = IR.

Flashcard 54: What type of circuit allows current to flow through multiple paths?

Answer: Parallel Circuit. Multiple current paths available.

Flashcard 55: What happens to total voltage in a series circuit with identical voltage sources?

Answer: It adds up. Voltages sum algebraically in series.

Flashcard 56: Identify the formula for electric charge.

Answer: Q=ItQ = It. Charge equals current times time.

Flashcard 57: What type of circuit allows current to flow through multiple paths?

Answer: Parallel Circuit. Multiple current paths available.

Flashcard 58: What is the voltage across each resistor in a parallel circuit?

Answer: It is the same across each resistor. All branches share same voltage source.

Flashcard 59: What is the unit of electric potential difference?

Answer: Volt (V). Named after Alessandro Volta.

Flashcard 60: Calculate the energy used by a 100W100 \text{W} bulb in 2 hours.

Answer: E=200WhE = 200 \text{Wh}. Using E=Pt=100×2E = Pt = 100 \times 2.

Flashcard 61: What device is used to measure electric current?

Answer: Ammeter. Connected in series with circuit.

Flashcard 62: Calculate the energy used by a 100W100 \text{W} bulb in 2 hours.

Answer: E=200WhE = 200 \text{Wh}. Using E=Pt=100×2E = Pt = 100 \times 2.

Flashcard 63: How does internal resistance affect a real battery's voltage?

Answer: Reduces terminal voltage. Internal resistance causes voltage drop.

Flashcard 64: What is the effect on resistance if the length of a resistor is doubled?

Answer: Resistance doubles. Resistance is proportional to length.

Flashcard 65: Find the current through a 5Ω5 \text{Ω} resistor with 15V15 \text{V} across it.

Answer: I=3AI = 3 \text{A}. Using Ohm's law: I=V/R=15/5I = V/R = 15/5.

Flashcard 66: What is the voltage across each resistor in a parallel circuit?

Answer: It is the same across each resistor. All branches share same voltage source.

Flashcard 67: Calculate the power if V=10VV = 10 \text{V} and I=2AI = 2 \text{A}.

Answer: P=20WP = 20 \text{W}. Using P=IV=10×2P = IV = 10 \times 2.

Flashcard 68: What device is used to measure voltage?

Answer: Voltmeter. Connected in parallel with component.

Flashcard 69: What is the unit of electrical resistance?

Answer: Ohm (Ω). Named after Georg Simon Ohm.

Flashcard 70: State the formula for equivalent resistance in a parallel circuit.

Answer: 1Req=1R1+1R2+...+1Rn\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \text{...} + \frac{1}{R_n}. Reciprocal of sum of reciprocals.

Flashcard 71: What is the voltage drop across a 10Ω10 \text{Ω} resistor with 0.5A0.5 \text{A} current?

Answer: V=5VV = 5 \text{V}. Using V=IR=0.5×10V = IR = 0.5 \times 10.

Flashcard 72: What type of circuit has only one path for current flow?

Answer: Series Circuit. Single continuous current path.

Flashcard 73: What is the voltage drop across a 10Ω10 \text{Ω} resistor with 0.5A0.5 \text{A} current?

Answer: V=5VV = 5 \text{V}. Using V=IR=0.5×10V = IR = 0.5 \times 10.

Flashcard 74: Find the current through a 5Ω5 \text{Ω} resistor with 15V15 \text{V} across it.

Answer: I=3AI = 3 \text{A}. Using Ohm's law: I=V/R=15/5I = V/R = 15/5.

Flashcard 75: What is the formula for energy consumed by an electrical device?

Answer: E=PtE = Pt. Energy equals power times time.

Flashcard 76: Find the current if V=12VV = 12 \text{V} and R=4ΩR = 4 \text{Ω}.

Answer: I=3AI = 3 \text{A}. Using V=IRV = IR, so I=V/R=12/4I = V/R = 12/4.