AP Physics 2 Quiz: Resistance Resistivity And Ohms Law

What this quiz covers

This quiz focuses on Resistance Resistivity And Ohms Law, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. It increases by a factor of 4 (correct answer)
2. It doubles because the length doubles
3. It decreases because the wire is thinner
4. It is unchanged because resistivity is unchanged

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. When a wire is stretched to twice its length while maintaining constant volume, both its length and cross-sectional area change. Since volume V = L×A must remain constant, if L doubles, then A must halve: A_new = A_original/2. The resistance formula R = ρL/A shows that R_new = ρ(2L)/(A/2) = 4ρL/A = 4R_original. Choice D incorrectly states that resistance only doubles, failing to account for the area change when volume is conserved. When solving stretching problems, always apply the volume constraint to find how both length and area change.

Question 2

1. Increasing the wire's resistivity while keeping $L$ and $A$ constant (correct answer)
2. Increasing the source voltage while keeping the same wire
3. Increasing the wire's cross-sectional area while keeping $\rho$ and $L$ constant
4. Decreasing the wire's length while keeping $\rho$ and $A$ constant

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. Current through a wire is given by I = V/R, where R = ρL/A. To decrease current with a fixed voltage source, we must increase the wire's resistance. Increasing cross-sectional area (choice A) decreases R, increasing current; decreasing length (choice B) decreases R, increasing current; increasing resistivity (choice C) increases R, decreasing current; increasing source voltage (choice D) increases V, increasing current. Choice D reflects the misconception that changing voltage changes resistance—students confuse the wire's fixed resistance property with the variable current that results from different voltages. When analyzing current changes, always determine first whether resistance increases or decreases, then apply Ohm's law.

Question 3

1. $4.2\times10^{-2}\ \Omega$ (correct answer)
2. $2.1\times10^{-5}\ \Omega$
3. $4.2\times10^{-5}\ \Omega$
4. $2.1\times10^{-2}\ \Omega$

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. Resistance is calculated using R = ρL/A, where ρ is the material's resistivity, L is length, and A is cross-sectional area. Substituting the given values: R = (2.8×10⁻⁸ Ω·m)(3.0 m)/(2.0×10⁻⁶ m²) = (8.4×10⁻⁸)/(2.0×10⁻⁶) = 4.2×10⁻² Ω. Choice C (4.2×10⁻⁵ Ω) represents a calculation error in handling scientific notation, likely from incorrectly dividing the exponents. When calculating resistance, carefully track units and exponents: the result should have units of ohms with reasonable magnitude for common wire dimensions.

Question 4

1. one-fourth as large, because the wire is longer
2. four times as large, because the area decreases (correct answer)
3. one-half as large, because the resistivity decreases
4. twice as large, because the wire is longer

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. When a wire is stretched to twice its length while maintaining constant volume, both its length and cross-sectional area change. Since volume V = LA remains constant, if L doubles, then A must halve. The resistance formula R = ρL/A shows that doubling L increases R by a factor of 2, while halving A increases R by another factor of 2, giving a total increase of 4. Choice A incorrectly assumes only length matters and confuses the direction of the effect. To solve stretching problems, always use volume conservation to find how both length and area change together.

Question 5

1. Doubling the length while keeping the same material and area (correct answer)
2. Doubling the applied voltage while keeping the same wire
3. Replacing the wire with the same length and area but lower resistivity
4. Doubling the cross-sectional area while keeping the same material and length

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. Resistance is given by R = ρL/A, where ρ is the material's resistivity, L is length, and A is cross-sectional area. To increase resistance, we need to either increase the numerator (ρ or L) or decrease the denominator (A). Doubling the length (choice A) directly doubles the resistance, while doubling the area (choice C) would halve the resistance. Choice B incorrectly assumes that changing voltage changes the wire's resistance—voltage affects current through Ohm's law (V = IR) but doesn't change the wire's inherent resistance. When analyzing resistance changes, always check how each geometric or material parameter affects the R = ρL/A formula.

Question 6

1. one-third as large
2. three times as large (correct answer)
3. unchanged because resistivity is unchanged
4. three times as small because thicker wires have more resistance

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. The resistance of a wire is R = ρL/A, where A is the cross-sectional area. When area is tripled while length and resistivity remain constant, the new resistance becomes R_new = ρL/(3A) = R_original/3. According to Ohm's law (V = IR), if voltage V remains constant and resistance is reduced to one-third, the current must triple: I_new = V/(R/3) = 3V/R = 3I_original. Choice D incorrectly claims that thicker wires have more resistance, which contradicts the inverse relationship between resistance and area. Remember that resistance decreases as cross-sectional area increases: wider paths allow easier current flow.

Question 7

1. quadrupled because $I\propto V^2$
2. unchanged because resistance depends on voltage
3. doubled (correct answer)
4. halved

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. For an ohmic resistor, resistance R remains constant regardless of the applied voltage or current. Using Ohm's law (V = IR), the initial current is I₁ = 3.0V/150Ω = 0.02A. When voltage doubles to 6.0V, the new current is I₂ = 6.0V/150Ω = 0.04A, which is exactly double the original current. Choice D incorrectly suggests a quadratic relationship (I ∝ V²), confusing Ohm's law with power relationships. For ohmic materials, always apply the linear relationship V = IR, where doubling voltage doubles current when resistance is constant.

Question 8

1. smaller by a factor of about 6, because Y has higher resistivity (correct answer)
2. the same, because resistivity does not affect resistance
3. larger by a factor of about 6, because Y has higher resistivity
4. the same, because current depends only on applied voltage

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. With identical length and area, the resistance of each wire is R = ρL/A, so the ratio of resistances is R_Y/R_X = ρ_Y/ρ_X = (1.0×10⁻⁷)/(1.7×10⁻⁸) ≈ 6. Since current I = V/R with the same voltage applied, the current ratio is I_Y/I_X = R_X/R_Y = 1/6. Wire Y has about 6 times higher resistance, so it carries about 1/6 the current of wire X. Choice B incorrectly inverts the relationship between resistivity and current—higher resistivity means higher resistance and thus lower current. When comparing materials, remember that current is inversely proportional to resistance and resistivity.

Question 9

1. one-half as large, because the diameter is smaller
2. one-fourth as large, because the cross-sectional area is smaller (correct answer)
3. four times as large, because the cross-sectional area is smaller
4. unchanged, because resistivity depends only on the material

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. When diameter is halved, the cross-sectional area A = πr² = π(d/2)² decreases by a factor of 4. Since resistance R = ρL/A and only A changes, the resistance increases by a factor of 4. With fixed voltage and I = V/R, the current decreases by a factor of 4, making it one-fourth as large. Choice A incorrectly uses diameter instead of area in the resistance formula—resistance depends on cross-sectional area, which varies with the square of diameter. Always convert diameter changes to area changes when analyzing resistance.

Question 10

1. $0.50\ \text{A}$
2. $2.0\ \text{A}$ (correct answer)
3. $8.0\ \text{A}$
4. $4.0\ \text{A}$

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. For an ohmic resistor, the resistance R remains constant regardless of applied voltage, and current follows Ohm's law: I = V/R. With R = 6.0 Ω and initial voltage 3.0 V, the initial current is I₁ = 3.0/6.0 = 0.50 A. When voltage increases to 12 V, the new current is I₂ = 12/6.0 = 2.0 A. Choice C (4.0 A) represents the misconception that current increases with the square of voltage, confusing power relationships with Ohm's law. Remember that for ohmic materials, current is directly proportional to voltage when resistance stays constant.

Question 11

1. Replace it with the same geometry but higher resistivity
2. Halve the applied potential difference across the conductor
3. Double the length while keeping area the same
4. Double the cross-sectional area while keeping length the same (correct answer)

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. The resistance of a conductor is R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area. To decrease resistance, we can decrease L, increase A, or use a material with lower ρ. Doubling the cross-sectional area (choice C) halves the resistance since R is inversely proportional to A. Choice A incorrectly focuses on changing the applied voltage, which affects current but not resistance—resistance is an intrinsic property of the conductor determined by its material and geometry. To maximize resistance changes, focus on geometric factors (L and A) rather than circuit conditions (V).

Question 12

1. the same because length and area are the same
2. half as large
3. twice as large (correct answer)
4. four times as large because resistivity is squared in $R$

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. Resistance is calculated as R = ρL/A, where ρ is the material's resistivity. For wires with identical length L and area A, resistance is directly proportional to resistivity. Since wire Y has resistivity ρ_Y = 2ρ_X, its resistance is R_Y = (2ρ_X)L/A = 2(ρ_X L/A) = 2R_X. Choice D incorrectly claims that resistivity is squared in the resistance formula, which would give R ∝ ρ²—this is a mathematical misconception. When comparing wires of identical geometry, resistance ratios equal resistivity ratios: R₂/R₁ = ρ₂/ρ₁.

Question 13

1. Decrease the cross-sectional area to $5.0\times10^{-7}\,\text{m}^2$ (correct answer)
2. Decrease the applied potential difference across the wire
3. Increase the cross-sectional area to $2.0\times10^{-6}\,\text{m}^2$
4. Replace copper with a material of the same resistance but different resistivity

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. Resistance is given by R = ρL/A, where ρ is the material's resistivity, L is length, and A is cross-sectional area. To increase resistance, we need to either increase L, decrease A, or use a material with higher ρ. Decreasing the cross-sectional area from 1.0×10⁻⁶ m² to 5.0×10⁻⁷ m² (halving it) will double the resistance since R is inversely proportional to A. Choice B incorrectly suggests that changing the applied voltage affects resistance—this is a common misconception, as resistance is a property of the conductor itself, not dependent on the applied voltage. When analyzing resistance problems, always separate the conductor's properties (R = ρL/A) from the circuit conditions (V = IR).

Question 14

1. unchanged because resistivity is unchanged
2. twice as large
3. one-fourth as large (correct answer)
4. four times as large

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. For a cylindrical wire, the cross-sectional area is A = πr², where r is the radius. When radius is doubled, the new area becomes A_new = π(2r)² = 4πr² = 4A_original. Since resistance R = ρL/A and area appears in the denominator, quadrupling the area reduces resistance to one-fourth: R_new = ρL/(4A) = R_original/4. Choice D incorrectly assumes that resistance is unchanged when only resistivity is unchanged, missing the crucial geometric dependence. When dealing with cylindrical conductors, remember that area scales with the square of the radius, leading to R ∝ 1/r².

Question 15

1. four times as large because $R\propto L^2$
2. twice as large (correct answer)
3. half as large because the length is smaller
4. unchanged because resistivity is unchanged

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. When a wire is cut to half its original length while maintaining the same cross-sectional area, its resistance becomes R_new = ρ(L/2)/A = R_original/2. According to Ohm's law (V = IR), if the same voltage is applied to a wire with half the resistance, the current doubles: I_new = V/(R/2) = 2V/R = 2I_original. Choice C incorrectly claims that current is unchanged because resistivity is unchanged, confusing the material property (ρ) with the geometric effect on resistance. Remember that changing a wire's length directly affects its resistance, which in turn affects current flow at constant voltage.

Question 16

1. half as large (correct answer)
2. one-fourth as large
3. unchanged because resistivity is unchanged
4. twice as large because the voltage is the same

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. The resistance of a uniform wire is R = ρL/A, where ρ is resistivity (a material property), L is length, and A is cross-sectional area. When length is doubled while A remains constant, the resistance doubles: R_new = ρ(2L)/A = 2R_original. By Ohm's law (V = IR), if voltage V stays constant and resistance doubles, the current must be halved: I_new = V/(2R) = I_original/2. Choice B incorrectly assumes that resistance doesn't change because resistivity is unchanged, confusing the material property (ρ) with the geometric property (R). To avoid this error, remember that resistance depends on both material properties and geometry: R = ρL/A.

Question 17

1. one-third as large, because the length is shorter
2. three times as large, because the resistance is smaller (correct answer)
3. unchanged, because resistivity is the same material property
4. nine times as large, because resistance scales with $1/L^2$

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. When a wire is cut into three equal pieces, each piece has one-third the original length but the same cross-sectional area and resistivity. Using R = ρL/A, each piece has resistance R_piece = ρ(L/3)/A = R/3 = 1.0 Ω. With the same voltage applied and I = V/R, the current through one piece is three times larger than through the original wire. Choice D incorrectly suggests resistance scales with 1/L²—resistance is directly proportional to length, not inversely proportional to length squared. To analyze cut wires, apply the resistance formula to the new dimensions of each piece.

Question 18

1. halved
2. quadrupled
3. unchanged (correct answer)
4. doubled

Explanation: This question tests understanding of resistance, resistivity, and Ohm's law. Resistance is given by R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area. When both length and area are doubled, the new resistance becomes R_new = ρ(2L)/(2A) = 2ρL/2A = ρL/A = R_original. The factor of 2 in the numerator (from doubled length) cancels with the factor of 2 in the denominator (from doubled area), leaving resistance unchanged. This demonstrates that resistance depends on the ratio L/A, not on L and A individually. When multiple geometric parameters change, calculate the new resistance step by step using R = ρL/A.