11.3 Resistance, Resistivity, and Ohm’s Law

1. AWith the same potential difference across each, X carries the smaller current. Correct
   Resistance measures how strongly an object opposes the movement of charge: R = ΔV/I is the potential difference needed per unit current. With the same ΔV across each, I = ΔV/R is smaller for the larger resistance, X.
2. BConnected in turn across one ideal battery, X and Y carry equal currents.
   A student who thinks a battery supplies a fixed current picks this. An ideal battery fixes the potential difference; with the same ΔV across each, the current I = ΔV/R is smaller in X.
3. CX uses up more of the current entering it, so less current leaves X than leaves Y.
   A student who thinks a resistor uses up current picks this. Charge is conserved: the current leaving each resistor equals the current entering it. A larger resistance means a smaller current for a given ΔV, not a loss of current inside.
4. DWhatever the potential differences across them, X has less current than Y.
   A student who thinks resistance alone fixes the current picks this. The current depends on ΔV as well: with a large enough potential difference across X, I = ΔV/R in X can exceed the current in Y.

CED 11.3.A.1 · Read this in Fix

1. AThe resistance increases, and the current decreases.
   A student who thinks a hot metal conducts better, so that cooling makes it conduct worse, picks this. For a metal the resistivity typically rises with temperature; cooling lowers it.
2. BThe resistance and the current both remain the same.
   A student who thinks a wire's resistance is fixed by its length and area picks this. The resistivity itself depends on temperature, so the resistance changes even though the dimensions do not.
3. CThe resistance decreases; the current does not vary.
   A student who thinks a battery supplies a fixed current picks this. The ideal battery fixes the potential difference; with a smaller resistance, I = ΔV/R is larger.
4. DThe resistance decreases, and the current increases. Correct
   The resistivity of a conductor such as copper typically increases with temperature, so cooling the wire lowers its resistivity and, with ℓ and A unchanged, its resistance R = ρℓ/A. The battery keeps ΔV fixed, so I = ΔV/R increases.

CED 11.3.A.2.ii · Read this in Fix

1. AI and 3ε
   A student who thinks the battery supplies a fixed current picks this, keeping I and finding ΔV = I × 3R = 3ε. The ideal battery fixes ε, so the current falls to I/3.
2. BI/3 and ε Correct
   The ideal battery keeps the potential difference across whatever is connected equal to ε. By Ohm’s law, I_new = ε/(3R) = I/3, and ΔV = I_new × 3R = ε.
3. CI/3 and ε/3
   A student who thinks the potential difference falls in step with the current picks this. With three times the resistance and one-third the current, ΔV = (I/3)(3R) = ε is unchanged.
4. DI/3 and 3ε
   A student who thinks the potential difference across a resistor is proportional to its resistance in any circuit picks this. A single resistor across an ideal battery has ΔV = ε whatever its resistance, and 3ε would exceed the battery's emf.

Working ΔV = ε (ideal battery). I_new = ε/(3R) = I/3. Check: I_new × 3R = ε.

CED 11.3.B.1 · Read this in Fix

1. AρL/(81πr²)
   A student who takes the cross-section of a hollow cylinder as a full circle of the outer radius, π(9r)² = 81πr², picks this. The core carries no current; the conducting area is the ring, 81πr² − 25πr² = 56πr².
2. BρL/(16πr²)
   A student who uses the wall thickness, 9r − 5r = 4r, as the radius in A = πr² picks this: π(4r)² = 16πr². That is the area of a circle of radius 4r, not the area of the ring.
3. CρL/(56πr²) Correct
   The current passes along the length L through the wall only, so the cross-sectional area is that of the ring: A = π(9r)² − π(5r)² = 56πr². Then R = ρL/A = ρL/(56πr²).
4. DρL/(72πr²)
   A student who finds the ring's area as its outer circumference times its thickness, 2π(9r)(4r) = 72πr², picks this. That shortcut only approximates a thin ring; the exact area is 56πr², the thickness times the mean circumference 2π(7r).

Working A = π(9r)² − π(5r)² = (81 − 25)πr² = 56πr²; R = ρL/A = ρL/(56πr²). All options have units Ω·m·m/m² = Ω. Distractors: full disk 81πr²; circle of radius 4r, 16πr²; outer circumference × thickness 2π(9r)(4r) = 72πr².

CED 11.3.A.2 · Read this in Fix

1. A0.50R₀
   A student who thinks the area is proportional to the diameter picks this, doubling A instead of quadrupling it. A = πd²/4, so twice the diameter gives four times the area.
2. B1.00R₀
   A student who thinks a wire's resistance depends on its material and length but not its thickness picks this. R = ρℓ/A: a larger cross-sectional area gives a smaller resistance.
3. C4.00R₀
   A student who thinks a thicker wire has more resistance, because there is more material in the way, picks this: four times the area taken as four times the resistance. Resistance is inversely proportional to area; the thicker wire has one-fourth the resistance.
4. D0.25R₀ Correct
   R = ρℓ/A with ρ and ℓ unchanged. The cross-sectional area is proportional to the square of the diameter, A = πd²/4, so doubling d makes A four times as large and R one-fourth as large: 0.25R₀.

Working R ∝ 1/A ∝ 1/d². d → 2d: A → 4A, R → R₀/4 = 0.25R₀.

CED 11.3.A.2 · Read this in Fix

1. A2.2 Ω
   A student who thinks only the length changes when a wire is stretched picks this: twice the length, twice the resistance. The volume is unchanged, so the area also halves, which doubles the resistance again.
2. B4.4 Ω Correct
   The length doubles. Because the volume Aℓ is unchanged, the cross-sectional area halves. R = ρℓ/A therefore increases by a factor of 2 × 2 = 4: 4 × 1.1 Ω = 4.4 Ω.
3. C1.1 Ω
   A student who thinks the resistance stays the same because the wire contains the same amount of the same metal picks this. Resistance depends on shape: a longer, thinner wire of the same volume has a larger resistance.
4. D8.8 Ω
   A student who thinks the diameter halves when the length doubles picks this: the area then falls to one-fourth and R = 2 × 4 × 1.1 Ω = 8.8 Ω. That would reduce the volume to half; with constant volume the area halves, not the diameter.

Working Constant volume: A₂ = A₁ℓ₁/ℓ₂ = A₁/2. R₂ = ρℓ₂/A₂ = R₁ × (ℓ₂/ℓ₁)² = 1.1 Ω × 2² = 4.4 Ω.

CED 11.3.A.2 · Read this in Fix

1. A2.00 Correct
   Resistivity is a property of the metal, found from ρ = RA/ℓ with A = πd²/4. For P: ρ_P = R × π(2d)²/4 ÷ L = πRd²/L. For Q: ρ_Q = 4R × πd²/4 ÷ 2L = πRd²/(2L). So ρ_P/ρ_Q = 2.00, although P has the smaller resistance.
2. B0.25
   A student who treats resistivity and resistance as the same property compares the resistances, R and 4R, and picks this. Resistance also depends on each wire's length and area; removing them with ρ = RA/ℓ shows that P's metal has twice the resistivity.
3. C1.00
   A student who takes the area as proportional to the diameter picks this: ρ ∝ Rd/ℓ gives R(2d)/L for P and 4R(d)/(2L) for Q, which are equal. The area is proportional to d², so P's area is four times Q's, not twice.
4. D0.50
   A student who ignores the thickness of the wires picks this: ρ ∝ R/ℓ gives R/L for P and 4R/(2L) for Q. The resistance also depends on the cross-sectional area, and P's area is four times Q's.

Working ρ = RA/ℓ = Rπd²/(4ℓ). P: R·π(2d)²/(4L) = πRd²/L. Q: 4R·πd²/(4·2L) = πRd²/(2L). Ratio = 2.00.

CED 11.3.A.2.i · Read this in Fix

1. A2.00ρ₀ℓ/A
   A student who uses the resistivity at the far end, ρ(ℓ) = 2ρ₀, as if it applied to the whole rod picks this. The resistivity is 2ρ₀ only at x = ℓ; the integral adds the smaller values along the rest of the rod.
2. B1.33ρ₀ℓ/A Correct
   Each slice of length dx contributes ρ(x)dx/A, so R = (1/A)∫₀^ℓ ρ₀(1 + x²/ℓ²)dx = (ρ₀/A)(ℓ + ℓ/3) = (4/3)ρ₀ℓ/A ≈ 1.33ρ₀ℓ/A.
3. C1.50ρ₀ℓ/A
   A student who takes the average resistivity as the mean of the end values, (ρ₀ + 2ρ₀)/2, picks this. That shortcut works only for a linear variation; here ρ rises slowly at first, so its average over the length is (4/3)ρ₀.
4. D1.25ρ₀ℓ/A
   A student who takes the resistivity at the midpoint, ρ(ℓ/2) = 1.25ρ₀, as the average picks this. The midpoint value equals the average only for special variations such as a linear one; the integral gives (4/3)ρ₀.

Working R = (1/A)∫₀^ℓ ρ₀(1 + x²/ℓ²)dx = (ρ₀/A)[x + x³/(3ℓ²)]₀^ℓ = (4/3)ρ₀ℓ/A = 1.33ρ₀ℓ/A. (End value 2ρ₀; mean of ends 1.5ρ₀; midpoint 1.25ρ₀.)

CED 11.3.A.2.iii · Read this in Fix

1. A3.0 Ω
   A student who uses the resistivity at the far end, 6.0 × 10⁻⁶ Ω·m, for the whole rod picks this: (6.0 × 10⁻⁶)(0.50)/(1.0 × 10⁻⁶) = 3.0 Ω. The resistivity is smaller over most of the rod; the area under the graph is needed.
2. B4.0 Ω
   A student who adds the resistivities at the two ends and applies the sum over the whole length picks this: (2.0 + 6.0) × 10⁻⁶ × 0.50/(1.0 × 10⁻⁶) = 4.0 Ω. Each value applies only near its own position; the integral averages them.
3. C2.0 Ω Correct
   R = ∫ρ dx/A. The integral is the area under the graph from x = 0 to 0.50 m: ½(2.0 + 6.0) × 10⁻⁶ Ω·m × 0.50 m = 2.0 × 10⁻⁶ Ω·m². Dividing by A = 1.0 × 10⁻⁶ m² gives R = 2.0 Ω.
4. D8.0 Ω
   A student who reads the slope of the graph in place of the area picks this: slope = 4.0 × 10⁻⁶ Ω·m/0.50 m = 8.0 × 10⁻⁶ Ω, divided by A gives 8.0. R needs ∫ρ dx, the area under the graph, not the slope.

Working ∫ρ dx = area of trapezoid = ½(2.0 + 6.0)×10⁻⁶ Ω·m × 0.50 m = 2.0×10⁻⁶ Ω·m². R = 2.0×10⁻⁶/1.0×10⁻⁶ = 2.0 Ω.

CED 11.3.A.2.iii · Read this in Fix

1. A3.5ρL/A Correct
   R = ∫ρ dℓ/A. The resistivity is ρ over a length 3L and ρ/2 over a length L, so ∫ρ dℓ = 3ρL + 0.5ρL = 3.5ρL and R = 3.5ρL/A.
2. B6.0ρL/A
   A student who adds the two resistivities, ρ + ρ/2 = 1.5ρ, and applies the sum over the whole length 4L picks this. Each resistivity applies only over its own length: ρ over 3L and ρ/2 over L.
3. C3.0ρL/A
   A student who averages the two resistivities without weighting them by length, (ρ + ρ/2)/2 = 0.75ρ, and uses the whole length 4L picks this. The resistivity ρ applies over three times the length that ρ/2 does, so the length-weighted average is 0.875ρ.
4. D4.0ρL/A
   A student who uses one resistivity for the whole rod, here ρ of the longer section, picks this: ρ × 4L. The length L of resistivity ρ/2 contributes only 0.5ρL/A.

Working ∫ρ dℓ = ρ(3L) + (ρ/2)(L) = 3.5ρL; R = 3.5ρL/A. (Sum over total length (1.5ρ)(4L) = 6.0ρL; unweighted average 0.75ρ × 4L = 3.0ρL; single value ρ × 4L = 4.0ρL.)

CED 11.3.A.2.iii · Read this in Fix

1. A3.0 Ω
   A student who finds the current as q/t at one point picks this: 4.0 C/2.0 s = 2.0 A and R = 6.0 V/2.0 A = 3.0 Ω. The line does not pass through the origin, so q/t is not the current; the slope is.
2. B1.5 Ω
   A student who uses the charge in place of the current picks this: R = 6.0 V/4.0 C. Current is the rate at which charge passes, the slope of the graph, 1.5 A.
3. C1.2 Ω
   A student who takes the area under the graph as the current picks this: area = ½(1.0 + 4.0) × 2.0 = 5.0, and 6.0/5.0 = 1.2. The current is the slope of a charge–time graph, not the area under it.
4. D4.0 Ω Correct
   The current is the slope of the charge–time graph: I = Δq/Δt = (4.0 C − 1.0 C)/(2.0 s) = 1.5 A. Ohm’s law gives R = ΔV/I = 6.0 V/1.5 A = 4.0 Ω.

Working I = dq/dt = (4.0 − 1.0) C/2.0 s = 1.5 A; R = 6.0 V/1.5 A = 4.0 Ω.

CED 11.3.B.1 · Read this in Fix

1. AX and Z, because the graphs of both of these elements are straight lines
   A student who thinks any straight-line graph shows ohmic behavior picks this. Z's line meets the axis at ΔV = 1 V, so ΔV/I is different at every point along it; its resistance is not constant.
2. BX and Y, because the graphs of both elements pass through the origin
   A student who thinks passing through the origin is enough picks this. A passive element such as Y has zero current at zero potential difference; Y's graph curves, so its ΔV/I changes with the current.
3. CX only, because only its current is proportional to the potential difference Correct
   An ohmic element has the same resistance ΔV/I at all currents, so its I–ΔV graph is a straight line through the origin. Only X's graph is. Y's graph curves, and Z's line does not pass through the origin, so ΔV/I changes along each.
4. DX, Y and Z, because R = ΔV/I can be found at every point of each graph
   A student who thinks every element with a resistance ΔV/I obeys Ohm’s law picks this. ΔV/I can be calculated for any element; an element is ohmic only if that value is the same for all currents.

CED 11.3.B.1.i · Read this in Fix

1. AIt is not made of a conductor, since the resistivity of a conductor increases as its temperature rises.
   A student who thinks every conductor's resistivity must increase with temperature picks this. The increase is typical, not universal: an ohmic material's resistivity is constant regardless of temperature, and the sample does conduct.
2. BIts resistivity rises with temperature, but its resistance is set by its length and area alone.
   A student who thinks resistance is fixed by the dimensions alone picks this. R = ρℓ/A with ℓ and A unchanged, so if ρ rose, R would rise too; a constant R means a constant ρ.
3. CThe current in it would be the same whatever potential difference were applied to it.
   A student who confuses a constant resistance with a constant current picks this. A constant resistance means the current is proportional to the potential difference, I = ΔV/R, not independent of it.
4. DOver this range its resistivity is constant regardless of temperature, as for an ohmic material. Correct
   With ℓ and A unchanged, a constant R = ρℓ/A means a constant resistivity over the range tested. A resistivity that is constant regardless of temperature is what the ohmic model describes, so the evidence is consistent with an ohmic material; it would not be for a typical metal, whose resistivity rises with temperature.

CED 11.3.B.1.ii · Read this in Fix

1. ALess current leaves it than enters it, because some of the charge is converted into thermal energy.
   A student who thinks a resistor uses up current picks this. Charge is not converted into energy; the current leaving equals the current entering, and what is converted is electrical energy.
2. BThe charges leave the resistor moving more slowly than they entered, having lost kinetic energy in it.
   A student who pictures the energy converted in the resistor as kinetic energy of the charges picks this. The current, and so the drift speed in wires of equal cross-section, is the same on both sides; the energy comes from the charges' electric potential energy.
3. CThe current leaving equals the current entering, and electrical energy becomes thermal energy. Correct
   Charge is conserved, so in a steady state the current is the same on both sides of the resistor. Electrical energy is converted to thermal energy in the resistor, which may raise the temperature of the resistor and its surroundings.
4. DElectrical energy becomes thermal energy that stays inside the resistor, not in its surroundings.
   A student who thinks the thermal energy stays where it is produced picks this. Thermal energy in a resistor can raise the temperature of both the resistor and its surroundings.

CED 11.3.B.1.iii · Read this in Fix

1. A6.5 × 10⁻⁹ Ω·m
   A student who takes the slope of the I–ΔV graph, 0.050 A/V, as the resistance picks this: 0.050 × 1.96 × 10⁻⁷/1.5. The slope of an I–ΔV graph is 1/R, so R = 20 Ω.
2. B2.6 × 10⁻⁶ Ω·m Correct
   The graph is a straight line through the origin, so R is the reciprocal of its slope: R = 5.0 V/0.25 A = 20 Ω. The cross-sectional area is A = π(0.25 × 10⁻³ m)² = 1.96 × 10⁻⁷ m², so ρ = RA/ℓ = (20 Ω)(1.96 × 10⁻⁷ m²)/(1.5 m) = 2.6 × 10⁻⁶ Ω·m.
3. C1.0 × 10⁻⁵ Ω·m
   A student who uses the diameter, 0.50 mm, as the radius in A = πr² picks this, making the area four times too large. The radius is 0.25 mm.
4. D3.1 × 10⁻² Ω·m
   A student who uses the wire's curved surface area, πdℓ, in place of its cross-sectional area picks this: ρ = R(πdℓ)/ℓ = πdR. The area in R = ρℓ/A is the area of a slice across the wire, πd²/4.

Working Slope = 0.25 A/5.0 V = 0.050 A/V; R = 1/slope = 20 Ω. A = π(0.25×10⁻³)² = 1.963×10⁻⁷ m². ρ = RA/ℓ = 20 × 1.963×10⁻⁷/1.5 = 2.62×10⁻⁶ Ω·m ≈ 2.6×10⁻⁶ Ω·m.

CED 11.3.B.1.iv · Read this in Fix

1. A2ρN/a
   A student who takes A to be the area enclosed by each turn, πa², picks this: ρ(2πNa)/(πa²) = 2ρN/a. The current passes through the wire's own cross section, πd²/4; the space inside the turns carries no current.
2. B8ρNa/d² Correct
   The current follows the wire, so ℓ is the length of the wire: N turns of circumference 2πa, ℓ = 2πNa. A is the wire's cross-sectional area, π(d/2)² = πd²/4. Then R = ρℓ/A = ρ(2πNa)/(πd²/4) = 8ρNa/d².
3. C4ρh/(πd²)
   A student who takes ℓ to be the coil's length h along its axis picks this. The charges travel along the wire itself, around every turn, a distance 2πNa, which is much longer than h.
4. Dρ/(πd)
   A student who uses the wire's curved outer surface area, πd(2πNa), as A picks this; the length then cancels, leaving ρ/(πd), which does not depend on N. The area in R = ρℓ/A is the cross section through which the current passes, πd²/4.

Working Length of wire ℓ = N(2πa) = 2πNa; cross-sectional area A = π(d/2)² = πd²/4. R = ρℓ/A = ρ(2πNa)/(πd²/4) = 8ρNa/d² (coefficient 8 in the key). Units: Ω·m·m/m² = Ω. Distractors (sympy-checked, all in Ω, each different from the key at test values): A = πa² gives 2ρN/a; ℓ = h gives 4ρh/(πd²); A = curved surface πd(2πNa) gives ρ/(πd).

CED 11.3.A.2 · Read this in Fix

1. A4.0ε/ℓ
   A student who finds the resistance from the resistivity at x = 0, R = ρ₀ℓ/A, as if it applied to the whole rod, picks this. That value of R is half the true resistance, so the current, and with it the field, comes out twice as large.
2. B1.6ε/ℓ
   A student who takes the average resistivity as the mean of the end values, (ρ₀ + 4ρ₀)/2 = 2.5ρ₀, picks this. The resistivity rises slowly near x = 0, so its average over the length is 2ρ₀, not 2.5ρ₀, and the current is larger than this student finds.
3. C2.0ε/ℓ Correct
   The resistance is R = (1/A)∫₀^ℓ ρ₀(1 + 3x²/ℓ²)dx = 2ρ₀ℓ/A, so I = ε/R = εA/(2ρ₀ℓ). The current density J = I/A is the same at every x, and at x = ℓ the resistivity is 4ρ₀, so E = ρJ = 4ρ₀ε/(2ρ₀ℓ) = 2.0ε/ℓ.
4. D1.0ε/ℓ
   A student who assumes the field is ε/ℓ everywhere in the rod, as in a uniform wire, picks this. The current density is the same at every x, so E = ρJ is largest where the resistivity is largest; ε/ℓ is only the average field along the rod.

Working R = (1/A)∫₀^ℓ ρ₀(1 + 3x²/ℓ²)dx = (ρ₀/A)(ℓ + ℓ) = 2ρ₀ℓ/A. I = ε/R = εA/(2ρ₀ℓ); J = I/A = ε/(2ρ₀ℓ), the same at every x. At x = ℓ, ρ = 4ρ₀, so E = ρJ = 4ρ₀ε/(2ρ₀ℓ) = 2.0ε/ℓ. Check: ∫₀^ℓ ρ(x)J dx = ε. Units: V/m. Distractors (sympy-checked): R from ρ(0) = ρ₀ gives 4.0ε/ℓ; average resistivity (ρ₀ + 4ρ₀)/2 = 2.5ρ₀ gives 4/2.5 = 1.6ε/ℓ; a uniform field gives ε/ℓ.

CED 11.3.B.1 · Read this in Fix