Resistance, Resistivity, and Ohm's Law
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AP Physics 2 › Resistance, Resistivity, and Ohm's Law
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 doubles because the length doubles
It increases by a factor of 4
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 metal wire obeys Ohm’s law. Its resistivity is $\rho$, and its resistance is $R=\rho L/A$ for length $L$ and area $A$. With a fixed $9.0\ \text{V}$ source across the wire, which change decreases the current through the wire?
Decreasing the wire’s length while keeping $\rho$ and $A$ constant
Increasing the wire’s cross-sectional area while keeping $\rho$ and $L$ constant
Increasing the source voltage while keeping the same wire
Increasing the wire’s resistivity while keeping $L$ 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.
A uniform aluminum wire ($\rho=2.8\times10^{-8}\ \Omega\cdot\text{m}$) has length $L=3.0\ \text{m}$ and area $A=2.0\times10^{-6}\ \text{m}^2$. Resistivity is a material property; resistance is $R=\rho L/A$. Approximately what is the wire’s resistance?
$2.1\times10^{-2}\ \Omega$
$4.2\times10^{-2}\ \Omega$
$2.1\times10^{-5}\ \Omega$
$4.2\times10^{-5}\ \Omega$
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.
A student measures a uniform wire’s resistance as $R=4.0\ \Omega$ at room temperature. The wire’s resistivity $\rho$ is a property of the material, while $R$ depends on $L$ and $A$ via $R=\rho L/A$. The wire is stretched to twice its original length while its volume stays constant. Compared to before, the new resistance is
one-half as large, because the resistivity decreases
four times as large, because the area decreases
twice as large, because the wire is longer
one-fourth 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.
A uniform copper wire has resistivity $\rho=1.7\times10^{-8}\ \Omega\cdot\text{m}$, length $L=2.0\ \text{m}$, and cross-sectional area $A=1.0\times10^{-6}\ \text{m}^2$. (Resistivity is a material property; resistance depends on geometry.) Which change increases the wire’s resistance?
Doubling the applied voltage while keeping the same wire
Doubling the cross-sectional area while keeping the same material and length
Doubling the length while keeping the same material and area
Replacing the wire with the same length and area but lower 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 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.
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
doubled
unchanged
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 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
three times as small because thicker wires have more resistance
three times as large
unchanged because resistivity is unchanged
one-third 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 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
doubled
quadrupled because $I\propto V^2$
unchanged because resistance depends on voltage
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.
Two uniform wires have the same length and cross-sectional area. Wire X is copper with $\rho_X=1.7\times10^{-8}\ \Omega\cdot\text{m}$; wire Y is iron with $\rho_Y=1.0\times10^{-7}\ \Omega\cdot\text{m}$. (Resistivity is intrinsic; resistance is $R=\rho L/A$.) If the same voltage is applied across each wire separately, compared to X the current in Y is
the same, because resistivity does not affect resistance
smaller by a factor of about 6, because Y has higher resistivity
the same, because current depends only on applied voltage
larger by a factor of about 6, because Y has higher resistivity
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.
A uniform wire of resistivity $\rho$ and length $L$ carries current when connected to a fixed voltage source. (Resistivity is intrinsic; resistance is $R=\rho L/A$.) The wire is replaced by one made of the same material and same length but with half the original diameter. Compared to before, the current is
four times as large, because the cross-sectional area is smaller
unchanged, because resistivity depends only on the material
one-half as large, because the diameter is smaller
one-fourth as large, because the cross-sectional area is smaller
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.