Resistance, Resistivity, and Ohm's Law
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AP Physics 2 › Resistance, Resistivity, and Ohm's Law
A cylindrical conductor has resistivity $\rho=2.0\times10^{-7},\Omega\cdot\text{m}$ (a material property). Its length is $1.0,\text{m}$ and area is $2.0\times10^{-6},\text{m}^2$. Which change decreases its resistance the most?
Double the length while keeping area the same
Halve the applied potential difference across the conductor
Double the cross-sectional area while keeping length the same
Replace it with the same geometry but higher resistivity
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).
Two wires have the same length and cross-sectional area. Wire X has resistivity $\rho_X$ and wire Y has resistivity $\rho_Y=2\rho_X$. Resistivity is a material property; resistance depends on $\rho L/A$. Compared to X, Y’s resistance is
the same because length and area are the same
four times as large because resistivity is squared in $R$
half as large
twice as large
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.
A copper wire has length $2.0,\text{m}$ and cross-sectional area $1.0\times10^{-6},\text{m}^2$. Copper’s resistivity $\rho$ is constant. Which change increases the wire’s resistance? (Resistance depends on geometry and $\rho$, not applied voltage.)
Decrease the applied potential difference across the wire
Increase the cross-sectional area to $2.0\times10^{-6},\text{m}^2$
Replace copper with a material of the same resistance but different resistivity
Decrease the cross-sectional area to $5.0\times10^{-7},\text{m}^2$
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).
A carbon resistor is ohmic with $R=150,\Omega$. Resistance is a circuit property; resistivity is a material property. When the potential difference across it increases from $3.0,\text{V}$ to $6.0,\text{V}$ at constant temperature, the current is
unchanged because resistance depends on voltage
doubled
halved
quadrupled because $I\propto V^2$
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.
An ohmic wire has resistance $R=\rho L/A$. Its resistivity $\rho$ stays constant. If both the length and cross-sectional area are doubled, compared to before, the resistance is
quadrupled
unchanged
doubled
halved
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.
A metal wire has resistance $R=3.0,\Omega$ at room temperature. Resistivity $\rho$ is a property of the metal; resistance depends on $\rho$, length, and area. The wire is uniformly stretched to twice its length while its volume stays constant. Which statement is correct about its resistance?
It is unchanged because resistivity is unchanged
It increases by a factor of 4
It doubles because the length doubles
It decreases because the wire is thinner
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.
A cylindrical wire has resistance $R=\rho L/A$. Resistance depends on geometry; resistivity $\rho$ depends on material. If the wire’s radius is doubled while $L$ and $\rho$ stay constant, the resistance becomes
four times as large
one-fourth as large
twice as large
unchanged because resistivity is unchanged
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.
A uniform wire has resistivity $\rho$ (material property), length $L$, and area $A$. It is cut to half its original length, keeping the same cross-sectional area. With the same applied voltage, compared to before, the current is
half as large because the length is smaller
unchanged because resistivity is unchanged
twice as large
four times as large because $R\propto L^2$
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.
A uniform wire has resistance $R$ and obeys Ohm’s law. Resistivity $\rho$ is fixed for the material. If the wire’s cross-sectional area is tripled while the same voltage is applied, compared to before, the current is
unchanged because resistivity is unchanged
three times as small because thicker wires have more resistance
one-third as large
three times as large
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.
A uniform nichrome wire obeys $R=\rho L/A$. Its resistivity $\rho$ is a material property and stays constant. If the wire’s length is doubled while $A$ is unchanged, compared to before, the current at the same voltage is
unchanged because resistivity is unchanged
half as large
twice as large because the voltage is the same
one-fourth as large
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.