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  2. MCAT Chemical and Physical Foundations of Biological Systems
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MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 4c Circuit Elements Ohms Law

Study 4c Circuit Elements Ohms Law in MCAT Chemical and Physical Foundations of Biological Systems with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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This deck focuses on 4c Circuit Elements Ohms Law, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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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.

MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 4c Circuit Elements Ohms Law

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QUESTION

State the formula for equivalent resistance of resistors in parallel.

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ANSWER

1Req=1R1+1R2+⋯\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdotsReq​1​=R1​1​+R2​1​+⋯. In parallel, reciprocals of resistances add due to increased conductance from multiple paths.

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Flashcard 1: State the formula for equivalent resistance of resistors in parallel.

Answer: 1Req=1R1+1R2+⋯\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdotsReq​1​=R1​1​+R2​1​+⋯. In parallel, reciprocals of resistances add due to increased conductance from multiple paths.

Flashcard 2: What is the SI unit of electric current, and what does it represent in terms of charge flow?

Answer: Ampere (A); 1 A=1 C s−11\,\text{A} = 1\,\text{C}\,\text{s}^{-1}1A=1Cs−1. Electric current measures the rate of charge flow, defined as one coulomb per second in SI units.

Flashcard 3: State the formula for electrical energy transferred over time by a circuit element.

Answer: E=PtE = PtE=Pt. Electrical energy is the product of power and time for constant power dissipation.

Flashcard 4: State the power dissipated by a resistor in terms of voltage and resistance.

Answer: P=V2RP = \frac{V^2}{R}P=RV2​. Power in a resistor also derives from Ohm’s law, expressed as voltage squared divided by resistance.

Flashcard 5: State the power dissipated by a resistor in terms of current and resistance.

Answer: P=I2RP = I^2RP=I2R. For resistors, power dissipation derives from combining Ohm’s law with the general power formula, using current squared times resistance.

Flashcard 6: State the formula for electrical power dissipated by a circuit element in terms of III and VVV.

Answer: P=IVP = IVP=IV. Electrical power is the rate of energy transfer, given by the product of current and voltage.

Flashcard 7: Find the power dissipated by a 6 Ω6\,\Omega6Ω resistor with a 12 V12\,\text{V}12V drop across it.

Answer: P=24 WP = 24\,\text{W}P=24W. Power dissipation is voltage squared divided by resistance.

Flashcard 8: What circuit quantity is the same across all branches in a parallel connection?

Answer: Voltage VVV is the same across each parallel branch. In parallel circuits, branches share the same potential difference due to common connection points.

Flashcard 9: State the formula for equivalent resistance of resistors in series.

Answer: Req=R1+R2+⋯R_{\text{eq}} = R_1 + R_2 + \cdotsReq​=R1​+R2​+⋯. In series, resistances add because the same current flows through each, accumulating opposition.

Flashcard 10: What is the definition of an ohmic device in terms of the III-VVV relationship?

Answer: It has constant RRR; I∝VI \propto VI∝V (linear III-VVV curve). An ohmic device obeys Ohm’s law with constant resistance, yielding a linear current-voltage relationship.

Flashcard 11: State the formula for Ohm’s law relating voltage, current, and resistance.

Answer: V=IRV = IRV=IR. Ohm’s law states that potential difference is directly proportional to current, with resistance as the proportionality constant.

Flashcard 12: Find the power dissipated by a 4 Ω4\,\Omega4Ω resistor when the current is 3 A3\,\text{A}3A.

Answer: P=36 WP = 36\,\text{W}P=36W. Power dissipation uses current squared multiplied by resistance.

Flashcard 13: What is the SI unit of electric potential difference (voltage), expressed using base SI units?

Answer: Volt (V); 1 V=1 J C−11\,\text{V} = 1\,\text{J}\,\text{C}^{-1}1V=1JC−1. Voltage represents the potential energy difference per unit charge, equivalent to one joule per coulomb.

Flashcard 14: What is the SI unit of resistance, expressed using volts and amperes?

Answer: Ohm (Ω\OmegaΩ); 1 Ω=1 V A−11\,\Omega = 1\,\text{V}\,\text{A}^{-1}1Ω=1VA−1. Resistance quantifies opposition to current flow, defined as one volt per ampere.

Flashcard 15: State Kirchhoff’s current law (KCL) for a node in a circuit.

Answer: ∑Iin=∑Iout\sum I_{\text{in}} = \sum I_{\text{out}}∑Iin​=∑Iout​. Kirchhoff’s current law enforces charge conservation at a node, balancing incoming and outgoing currents.

Flashcard 16: State Kirchhoff’s voltage law (KVL) for a closed loop in a circuit.

Answer: ∑ΔV=0\sum \Delta V = 0∑ΔV=0 around any closed loop. Kirchhoff’s voltage law upholds energy conservation, summing potential differences to zero in a closed loop.

Flashcard 17: What is the relationship between resistance, resistivity, length, and cross-sectional area?

Answer: R=ρLAR = \rho\frac{L}{A}R=ρAL​. Resistance scales with material resistivity and length while inversely with cross-sectional area.

Flashcard 18: If wire length LLL doubles (same ρ\rhoρ and AAA), how does resistance change?

Answer: RRR doubles: R∝LR \propto LR∝L. Resistance is directly proportional to length, so doubling length doubles resistance.

Flashcard 19: If wire cross-sectional area AAA doubles (same ρ\rhoρ and LLL), how does resistance change?

Answer: RRR halves: R∝1AR \propto \frac{1}{A}R∝A1​. Resistance is inversely proportional to cross-sectional area, so doubling area halves resistance.

Flashcard 20: Find the current when a 12 V12\,\text{V}12V battery is connected to a 3 Ω3\,\Omega3Ω resistor.

Answer: I=4 AI = 4\,\text{A}I=4A. Ohm’s law gives current as voltage divided by resistance.

Flashcard 21: Find the voltage drop across a 5 Ω5\,\Omega5Ω resistor carrying 2 A2\,\text{A}2A of current.

Answer: V=10 VV = 10\,\text{V}V=10V. Ohm’s law calculates voltage as current multiplied by resistance.

Flashcard 22: Find the equivalent resistance of 2 Ω2\,\Omega2Ω and 3 Ω3\,\Omega3Ω connected in series.

Answer: Req=5 ΩR_{\text{eq}} = 5\,\OmegaReq​=5Ω. Equivalent resistance in series is the sum of individual resistances.

Flashcard 23: Find the equivalent resistance of 2 Ω2\,\Omega2Ω and 3 Ω3\,\Omega3Ω connected in parallel.

Answer: Req=65 ΩR_{\text{eq}} = \frac{6}{5}\,\OmegaReq​=56​Ω. Equivalent resistance in parallel is the reciprocal of the sum of reciprocals.

Flashcard 24: Identify the equivalent resistance of three identical resistors RRR connected in parallel.

Answer: Req=R3R_{\text{eq}} = \frac{R}{3}Req​=3R​. For identical parallel resistors, equivalent resistance equals individual resistance divided by the number.

Flashcard 25: What circuit quantity is the same through all elements in a series connection?

Answer: Current III is the same through each series element. In series circuits, conservation of charge ensures identical current through all elements.