11.3 Resistance, Resistivity, and Ohm’s Law

1. AP uses up a larger share of the current that flows into it than Q does.
   A student who thinks a resistor uses up current picks this. The current leaving any resistor equals the current entering it; a larger resistance limits the current for a given ΔV, it does not consume any of it.
2. BWith equal ΔV across each, less charge passes through P per second. Correct
   Resistance measures how strongly an object opposes the movement of charge. With the same potential difference across each resistor, the one that opposes the movement more, P, has less charge passing through it each second: a smaller current, I = ΔV/R.
3. CP is made of a material that conducts charge less well than Q's material.
   A student who thinks resistance depends only on the material picks this. P and Q could be made of the same material: P could simply be longer or thinner. Resistance belongs to the object, R = ρℓ/A.
4. DP has the larger potential difference across it in every circuit.
   A student who reads R = ΔV/I as R ∝ ΔV picks this. The potential difference across a resistor depends on the circuit it is in; a larger resistance does not mean a larger ΔV in every circuit, for example when P and Q are connected to different batteries.

CED 11.3.A.1 · Read this in Fix

1. AI = πd²ΔV/(4ρℓ) Correct
   The cross-sectional area is A = π(d/2)² = πd²/4, so R = ρℓ/A = 4ρℓ/(πd²). Ohm's law then gives I = ΔV/R = πd²ΔV/(4ρℓ).
2. BI = 4ρℓΔV/(πd²)
   A student who multiplies ΔV by R instead of dividing picks this. Ohm's law is I = ΔV/R: a larger resistance gives a smaller current, so R belongs in the denominator. (Units check: V·Ω is not A.)
3. CI = 4ΔV/(πd²ρℓ)
   A student who thinks a thicker wire has more resistance, R = ρℓA, picks this. Resistance is inversely proportional to cross-sectional area, R = ρℓ/A, so the area belongs in the numerator of I.
4. DI = πd²ΔV/(2ρℓ)
   A student who squares only the numerator of d/2, writing A = πd²/2, picks this. (d/2)² = d²/4, so A = πd²/4 and the current is half this value.

Working A = π(d/2)² = πd²/4. R = ρℓ/A = 4ρℓ/(πd²). I = ΔV/R = πd²ΔV/(4ρℓ). (Errors: I = ΔV·R gives 4ρℓΔV/(πd²); R = ρℓA gives I = ΔV/(ρℓ·πd²/4) = 4ΔV/(πd²ρℓ); A = πd²/2 gives πd²ΔV/(2ρℓ).)

CED 11.3.A.2 · Read this in Fix

1. AIts resistivity is small, as it is for a good conductor.
   A student who links 'ohmic' with good conductors picks this. Being ohmic concerns whether the resistivity stays constant, not how small it is; tungsten is a good conductor, but a tungsten filament is not ohmic over its working range.
2. BIts resistance increases in proportion to the applied ΔV.
   A student who reads R = ΔV/I as R ∝ ΔV picks this. An ohmic material has the opposite property: its resistance stays constant when ΔV changes.
3. CIts dimensions stay fixed as its temperature changes.
   A student who thinks temperature affects resistance only through the size of a resistor picks this. Temperature changes resistance mainly by changing the resistivity; the ohmic model requires the resistivity to stay constant.
4. DIts resistivity stays the same as its temperature changes. Correct
   An ohmic material has a constant resistance for all currents, and its resistivity is constant regardless of temperature. So even as a resistor made from it warms up, its resistance does not change.

CED 11.3.B.1.ii · Read this in Fix

1. AR_Y/R_X = 1.5
   A student who takes the area to be proportional to the diameter picks this: 3/2 = 1.5. Area is proportional to d², so doubling the diameter makes the area four times as large.
2. BR_Y/R_X = 12
   A student who thinks a thicker wire has more resistance picks this, multiplying by the area ratio instead of dividing: 3 × 4 = 12. A larger cross-sectional area lowers the resistance.
3. CR_Y/R_X = 0.75 Correct
   Resistance is proportional to length and inversely proportional to cross-sectional area, and the area is proportional to the square of the diameter. Tripling ℓ multiplies R by 3; doubling d multiplies A by 4 and so divides R by 4: R_Y/R_X = 3/4 = 0.75.
4. DR_Y/R_X = 1
   A student who thinks resistance depends only on the material picks this. The material fixes the resistivity; the resistance also depends on the wire's length and cross-sectional area.

Working R = ρℓ/A with A = πd²/4, so R ∝ ℓ/d². R_Y/R_X = (ℓ_Y/ℓ_X)(d_X/d_Y)² = 3 × (1/2)² = 3/4 = 0.75. (Errors: A ∝ d gives 3/2 = 1.5; R ∝ ℓA gives 3 × 4 = 12; R set by material alone gives 1.)

CED 11.3.A.2 · Read this in Fix

1. AIts resistance and its resistivity are both halved.
   A student who thinks resistivity depends on the size of the sample picks this. Resistivity is a property of copper itself; cutting the wire changes the object's resistance, not the material's resistivity.
2. BIts resistance and its resistivity are both the same.
   A student who thinks resistance is a property of the material alone picks this. The resistivity is unchanged, but each piece is half as long, so its resistance is halved.
3. CIts resistance is the same, but its resistivity is halved.
   A student who has exchanged the meanings of resistance and resistivity picks this. Resistivity belongs to the material and stays the same; resistance belongs to the object and halves with its length.
4. DIts resistance is halved; its resistivity is unchanged. Correct
   Resistance belongs to the object: each piece has half the length with the same material and cross section, so R = ρℓ/A is halved. Resistivity belongs to the material, set by its atomic structure, so each piece has the same resistivity as the whole wire.

Working R = ρℓ/A: ρ and A unchanged, ℓ halved, so R is halved; ρ is a material property and is unchanged.

CED 11.3.A.2.i · Read this in Fix

1. A1.3 × 10⁻⁶ Ω·m
   A student who writes the area as πd²/2, squaring only the numerator of d/2, picks this: A = π(0.40 × 10⁻³ m)²/2 = 2.5 × 10⁻⁷ m². (d/2)² = d²/4, so the area is half that.
2. B6.3 × 10⁻⁷ Ω·m Correct
   For a uniform wire R = (ρ/A)ℓ, so the slope of the R–ℓ graph is ρ/A: 7.5 Ω ÷ 1.5 m = 5.0 Ω/m. The area is A = π(0.20 × 10⁻³ m)² = 1.26 × 10⁻⁷ m², so ρ = (5.0 Ω/m)(1.26 × 10⁻⁷ m²) = 6.3 × 10⁻⁷ Ω·m.
3. C7.1 × 10⁻⁷ Ω·m
   A student who uses the area under the line instead of its slope picks this: ½(1.5 m)(7.5 Ω) = 5.6 Ω·m, times A. The equation R = (ρ/A)ℓ makes ρ/A the slope of the graph; the area under an R–ℓ line has no physical meaning.
4. D3.1 × 10⁻³ Ω·m
   A student who forgets to square the radius picks this: A = π(0.20 × 10⁻³ m) = 6.3 × 10⁻⁴, which is not even an area in m². The area is A = πr² = 1.26 × 10⁻⁷ m².

Working Slope = ΔR/Δℓ = 7.5 Ω/1.5 m = 5.0 Ω/m = ρ/A. A = π(d/2)² = π(0.20 × 10⁻³ m)² = 1.26 × 10⁻⁷ m². ρ = (5.0 Ω/m)(1.26 × 10⁻⁷ m²) = 6.3 × 10⁻⁷ Ω·m.

CED 11.3.A.2.i · Read this in Fix

1. AThe hot filament expands, and its greater length gives it a larger resistance.
   A student who thinks temperature changes resistance only through the size of the wire picks this. Thermal expansion changes the filament's length (and its cross-sectional area) by only about 1%, far too little to explain a tenfold increase; the change is in the resistivity.
2. BAt the higher potential difference, more of the current is used up in the filament.
   A student who thinks current is used up in an element picks this. The current leaving the filament equals the current entering it; the larger resistance comes from the hotter filament's greater resistivity.
3. CThe filament is hotter, and tungsten's resistivity rises with temperature. Correct
   A conductor's resistivity usually rises as its temperature rises. At 6.0 V the filament is far hotter, so its resistivity, and with it R = ρℓ/A, is much larger: ten times larger here.
4. DSince R = ΔV/I, the resistance grows in proportion to the potential difference.
   A student who reads R = ΔV/I as saying R is proportional to ΔV picks this. ΔV increased 60 times but R only 10 times, so R is not proportional to ΔV; for an ohmic element R would not change at all. The increase is caused by the rise in temperature.

Working R = ΔV/I: 0.10 V/0.050 A = 2.0 Ω; 6.0 V/0.30 A = 20 Ω. ΔV rises by a factor of 60 while R rises by a factor of 10, so R is not proportional to ΔV. The filament temperature rises greatly, and the resistivity of a conductor typically increases with temperature; expansion changes ℓ and A by only about 1%.

CED 11.3.A.2.ii · Read this in Fix

1. AIt is smaller at Q, because the ratio I/ΔV is smaller at Q than it is at P.
   A student who takes I/ΔV, the slope of a line from the origin, as the resistance picks this. I/ΔV is 1/R: it is smaller at Q, so R = ΔV/I is larger at Q.
2. BIt is the same at both points, since the current rises whenever ΔV rises.
   A student who thinks a current that always rises with ΔV means an ohmic element picks this. The graph is curved: ΔV/I is 4.0 Ω at P and 7.5 Ω at Q, so the resistance is not constant.
3. CIt cannot be found at either point, because the bulb is not ohmic.
   A student who thinks R = ΔV/I applies only to ohmic elements picks this. The resistance of any element at a point is ΔV/I there; for this nonohmic bulb it is 4.0 Ω at P and 7.5 Ω at Q.
4. DIt is greater at Q, because the ratio ΔV/I is greater at Q than at P. Correct
   At any point, R = ΔV/I. At P, R = 1.0 V ÷ 0.250 A = 4.0 Ω; at Q, R = 3.0 V ÷ 0.400 A = 7.5 Ω. The resistance rises as the filament heats, which is why the graph curves over.

Working R = ΔV/I at each point. P: 1.0 V/0.250 A = 4.0 Ω. Q: 3.0 V/0.400 A = 7.5 Ω. R_Q > R_P. (I/ΔV: 0.25 Ω⁻¹ at P, 0.13 Ω⁻¹ at Q.)

CED 11.3.B.1 · Read this in Fix

1. AThe ratio ΔV/I has the same value for every current measured. Correct
   An ohmic material has a constant resistance for all currents. The ratio ΔV/I is the resistance, so if it is the same at every current, the resistor is ohmic; its I–ΔV graph is then a straight line through the origin.
2. BThe current increases each time the potential difference increases.
   A student who thinks 'ohmic' just means 'more ΔV gives more current' picks this. A filament bulb's current also rises with ΔV, yet its resistance changes; an ohmic element needs ΔV/I to stay constant.
3. CA smooth curve drawn through the data points passes through the origin.
   A student who thinks passing through the origin is enough picks this. A resistor carries no current when ΔV = 0 whether or not it is ohmic; the graph must also be a straight line.
4. DThe equation ΔV = IR gives a value of R for each current measured.
   A student who confuses the definition of resistance with Ohm's law picks this. R = ΔV/I can be calculated for any element at any current; the element is ohmic only if that value is the same at every current.

Working Ohmic ⇔ R constant for all currents ⇔ ΔV/I the same for every measurement (a straight-line I–ΔV graph through the origin).

CED 11.3.B.1.i · Read this in Fix

1. APart of the charge that flows into the resistor is used up, becoming thermal energy.
   A student who thinks charge is used up in a resistor picks this. Charge is conserved: the current leaving the resistor equals the current entering it. Energy, not charge, is converted.
2. BHeat that was stored inside the battery flows along the wires with the current.
   A student who pictures heat as a substance that flows with the current picks this. The battery stores chemical energy, not thermal energy; the thermal energy appears in the resistor, where electrical energy is converted.
3. CEnergy the charges carry from the battery becomes thermal energy in the resistor. Correct
   The battery transfers energy to the moving charges, and in the resistor that electrical energy is converted to thermal energy. The resistor's temperature rises, and energy then passes to the cooler water, raising its temperature too. The current entering and leaving the resistor is the same.
4. DThe charges slow down in the resistor, and their lost kinetic energy heats it.
   A student who thinks charges slow down as they give up energy in a resistor picks this. With a steady current, the charges leave at the same average speed as they had in the identical wire before the resistor; the energy they transfer is electric potential energy.

CED 11.3.B.1.iii · Read this in Fix

1. A7.0 Ω
   A student who calculates the slope from the first and last data points picks this: (4.0 V − 0.5 V)/(0.55 A − 0.05 A) = 7.0 Ω. Both points lie off the line; the slope must be taken from points on the best-fit line.
2. B8.0 Ω Correct
   The best-fit line passes through the origin and (4.0 V, 0.50 A), so its slope is 0.50 A ÷ 4.0 V = 0.125 A/V. For an ohmic element I = (1/R)ΔV, so the slope is 1/R: R = 4.0 V ÷ 0.50 A = 8.0 Ω. The line uses all six measurements and averages out their scatter.
3. C7.3 Ω
   A student who trusts only the largest measurement picks this: 4.0 V ÷ 0.55 A = 7.3 Ω. That point lies above the line; the line through all six points gives the best-supported value, 8.0 Ω.
4. D1.0 Ω
   A student who uses the area under the line picks this: ½(4.0 V)(0.50 A) = 1.0. The resistance comes from the slope of the I–ΔV line, not from the area under it.

Working Best-fit line: through (0, 0) and (4.0 V, 0.50 A). Slope = 0.50 A/4.0 V = 0.125 A/V = 1/R. R = 4.0 V/0.50 A = 8.0 Ω. (Least-squares fit through the origin of the six points gives 7.9 Ω, consistent with the drawn line.)

CED 11.3.B.1.iv · Read this in Fix

1. Aρ = πd²/(4kℓ) Correct
   With current on the vertical axis, the slope is k = I/ΔV = 1/R, so R = 1/k. The area is A = π(d/2)² = πd²/4, and rearranging R = ρℓ/A gives ρ = RA/ℓ = πd²/(4kℓ).
2. Bρ = πd²k/(4ℓ)
   A student who takes the slope of the I–ΔV graph as the resistance picks this: R = k, so ρ = kA/ℓ. With I on the vertical axis the slope is I/ΔV = 1/R, so the resistance is the reciprocal of the slope, 1/k.
3. Cρ = 4/(πd²kℓ)
   A student who thinks a thicker wire has more resistance puts the area in the numerator, R = ρℓA, and picks this: ρ = R/(ℓA) = 4/(πd²kℓ). A larger area gives charge more room to move, so R = ρℓ/A and ρ = RA/ℓ.
4. Dρ = πd²/(2kℓ)
   A student who squares d/2 as d²/2 takes the area as πd²/2 and picks this. Squaring d/2 squares both parts: A = πd²/4, which gives ρ = πd²/(4kℓ).

Working The slope of the I–ΔV graph is k = ΔI/Δ(ΔV) = 1/R, so R = 1/k. The cross-sectional area is A = π(d/2)² = πd²/4. From R = ρℓ/A, ρ = RA/ℓ = (1/k)(πd²/4)/ℓ = πd²/(4kℓ). (Errors: taking the slope as R gives ρ = kA/ℓ = πd²k/(4ℓ); putting A in the numerator of R, R = ρℓA, gives ρ = R/(ℓA) = 4/(πd²kℓ); A = πd²/2 gives πd²/(2kℓ).)

CED 11.3.B.1.iv · Read this in Fix

1. A1.3 m
   A student who uses I = ΔV·R picks this: R = I/ΔV = 0.80 Ω, which gives 1.3 m. The units show the slip: A/V is not an ohm. Ohm's law gives R = ΔV/I = 1.25 Ω and a length of 2.0 m.
2. B4.0 m
   A student who squares d/2 as d²/2 picks this: the area comes out as πd²/2, twice the true value, and so does the length. (d/2)² = d²/4, so A = 1.77 × 10⁻⁶ m² and ℓ = 2.0 m.
3. C8.0 m
   A student who puts the diameter in place of the radius in A = πr² picks this: the area comes out as πd², four times the true value, and so does the length. The radius is d/2 = 0.75 mm, so A = 1.77 × 10⁻⁶ m² and ℓ = 2.0 m.
4. D2.0 m Correct
   The resistance needed is R = ΔV/I = 3.0 V/2.4 A = 1.25 Ω. The cross-sectional area is A = π(d/2)² = π(0.75 × 10⁻³ m)² = 1.77 × 10⁻⁶ m². From R = ρℓ/A, ℓ = RA/ρ = (1.25 Ω)(1.77 × 10⁻⁶ m²)/(1.1 × 10⁻⁶ Ω·m) = 2.0 m.

Working R = ΔV/I = 3.0 V/2.4 A = 1.25 Ω. A = π(d/2)² = π(0.75 × 10⁻³ m)² = 1.77 × 10⁻⁶ m². R = ρℓ/A, so ℓ = RA/ρ = (1.25 Ω)(1.77 × 10⁻⁶ m²)/(1.1 × 10⁻⁶ Ω·m) = 2.0 m. Errors: I = ΔV·R, so R = I/ΔV = 0.80 Ω (m09): 1.3 m; A = πd²/2 (m04): 4.0 m; A = πd² (new): 8.0 m.

CED 11.3.A.2 · Read this in Fix