11.4 Electric Power

1. AI ×1, P ×½
   A student who thinks the battery supplies a fixed current picks this: with I unchanged, P = I²R halves. The battery fixes ΔV, so the current doubles.
2. BI ×2, P ×4
   A student who takes the power as proportional to I², leaving out the change in R, picks this. P = I²R = (2I)²(R/2): the doubled current and the halved resistance give a factor of 2.
3. CI ×2, P ×2 Correct
   The ideal battery keeps ΔV the same. I = ΔV/R, so halving R doubles I. P = ΔV²/R, so halving R doubles P; equivalently, P = I²R = (2I)²(R/2) = 2I²R.
4. DI ×1, P ×1
   A student who thinks a battery delivers energy at a fixed rate, and a fixed current, whatever is connected picks this. With ΔV fixed, the smaller resistance draws twice the current, and P = IΔV doubles.

Working ΔV fixed. I' = ΔV/(R/2) = 2I. P' = ΔV²/(R/2) = 2P (check: (2I)²(R/2) = 2I²R).

CED 11.4.A.1 · Read this in Fix

1. AX > Y > Z
   A student who ranks brightness by current alone picks this: 0.30 A > 0.20 A > 0.10 A. X has the largest current, but its potential difference is only half that of Y, so its power is smaller.
2. BY = Z > X
   A student who ranks brightness by potential difference alone picks this: Y and Z each have 6.0 V. Their currents differ, so their powers, 1.2 W and 0.60 W, differ.
3. CZ > Y > X
   A student who thinks the larger resistance gives the greater power picks this: R = ΔV/I is 60 Ω for Z, 30 Ω for Y and 10 Ω for X. Power depends on the current and potential difference together: Z's power, 0.60 W, is the smallest.
4. DY > X > Z Correct
   Brightness increases with power, P = IΔV. X: (0.30 A)(3.0 V) = 0.90 W. Y: (0.20 A)(6.0 V) = 1.2 W. Z: (0.10 A)(6.0 V) = 0.60 W. So Y is brightest, then X, then Z.

Working P = IΔV: X 0.30 × 3.0 = 0.90 W; Y 0.20 × 6.0 = 1.2 W; Z 0.10 × 6.0 = 0.60 W. Y > X > Z. (Consistent circuit: 0.30 A = 0.20 A + 0.10 A; battery 9.0 V = 3.0 V + 6.0 V.)

CED 11.4.A.2 · Read this in Fix

1. AIΔV
   A student who thinks a motor converts all the electrical energy it receives into mechanical energy picks this. The current in the coil's resistance dissipates energy at the rate I²R, which must be subtracted.
2. BIΔV − I²R Correct
   Energy is transferred to the motor at the rate IΔV. The current I in the coil's resistance dissipates energy at the rate I²R. The rest becomes mechanical energy: P_mech = IΔV − I²R.
3. CIΔV − ΔV²/R
   A student who finds the coil's dissipation from ΔV²/R using the potential difference across the whole motor picks this. Only part of ΔV is across the coil's resistance; its dissipation is I²R, found from the current.
4. DΔV²/R
   A student who treats the motor as if it were simply a resistor R across ΔV picks this. A running motor is not just its coil's resistance: ΔV²/R would exceed the actual input IΔV, since ΔV > IR. The mechanical power is IΔV − I²R.

Working Input P = IΔV; dissipation in coil = I²R; P_mech = IΔV − I²R. All options have units of watts. (ΔV²/R ≥ IΔV for a running motor, since ΔV > IR, so IΔV − ΔV²/R would be zero or negative.)

CED 11.4.A.1 · Read this in Fix

1. A0.63I₀²RT
   A student who thinks the power falls in proportion to the current picks this: ∫₀^T I₀²R e^−t/T dt = (1 − e⁻¹)I₀²RT. Since P = I²R, the power decays as e^−2t/T, twice as fast.
2. B0.40I₀²RT
   A student who squares the average current picks this: the average current over the interval is I₀(1 − e⁻¹), and (1 − e⁻¹)² ≈ 0.40. The energy needs the integral of I², which is not the square of the average of I.
3. C1.00I₀²RT
   A student who uses the initial power, I₀²R, for the whole interval picks this. The power falls throughout the interval, so the energy is the integral of P, which is less than I₀²RT.
4. D0.43I₀²RT Correct
   P = I²R = I₀²R e^−2t/T. E = ∫₀^T I₀²R e^−2t/T dt = I₀²R(T/2)(1 − e⁻²) ≈ 0.43I₀²RT. The power decays twice as fast as the current because it depends on I².

Working E = ∫₀^T I₀²R e^−2t/T dt = I₀²R · (T/2)(1 − e⁻²) = 0.432 I₀²RT ≈ 0.43I₀²RT.

CED 11.4.A.1 · Read this in Fix

1. A20 J Correct
   Over this interval the current rises linearly, I = (2.0 A/3.0 s)t, so P = I²R rises as t². E = ∫₀^3.0 (5.0 Ω)(2.0/3.0)²t² dt = (5.0)(4/9)(3.0)³/3 = 20 J.
2. B30 J
   A student who thinks the power rises in proportion to the current, and so linearly with time, takes the average power as half the final power: ½ × (2.0 A)²(5.0 Ω) × 3.0 s = 30 J. Since P = I²R, the power rises as t², and its average is one-third of the final value.
3. C15 J
   A student who uses the average current, 1.0 A, in P = I²R picks this: (1.0 A)²(5.0 Ω)(3.0 s) = 15 J. The average of I² is larger than the square of the average current.
4. D60 J
   A student who uses the final power, (2.0 A)²(5.0 Ω) = 20 W, for the whole 3.0 s picks this. The power is smaller at every earlier time; the energy is the integral of P = I²R.

Working I = (2/3)t A for 0 ≤ t ≤ 3.0 s. E = ∫I²R dt = 5.0 × (4/9) × (3.0³/3) = 5.0 × (4/9) × 9 = 20 J. (Average power = (1/3) × 20 W.)

CED 11.4.A.1 · Read this in Fix

1. A0.5
   A student who reads the steeper line, X, as the larger resistance picks this, taking the slopes 0.10 A/V and 0.050 A/V as the resistances. The slope of an I–ΔV graph is 1/R, so X has the smaller resistance, 10 Ω.
2. B1.0
   A student who thinks equal currents mean equal power picks this. Power depends on the potential difference too: with the same current, the resistor with the larger ΔV, Y, dissipates more.
3. C2.0 Correct
   At 0.20 A, the graph gives ΔV = 2.0 V for X and 4.0 V for Y. P = IΔV gives 0.40 W for X and 0.80 W for Y, so P_Y/P_X = 2.0. (Equivalently, R_X = 10 Ω, R_Y = 20 Ω, and P = I²R with the same current.)
4. D4.0
   A student who takes the power as proportional to ΔV², leaving out the different resistances, picks this: (4.0 V/2.0 V)² = 4. With the same current, P = IΔV, so the ratio is 4.0/2.0 = 2.0.

Working From graph at I = 0.20 A: ΔV_X = 2.0 V, ΔV_Y = 4.0 V. P_X = 0.40 W, P_Y = 0.80 W; ratio 2.0.

CED 11.4.A.1 · Read this in Fix

1. ABoth hold; which one shows the effect of R depends on whether I or ΔV is fixed. Correct
   For a resistor, P = I²R = ΔV²/R always. If two resistors carry the same current, I²R shows the larger R dissipates more; if they have the same ΔV, ΔV²/R shows the smaller R dissipates more. The two forms describe different comparisons.
2. BP = I²R is the right one: a larger resistance converts energy at a greater rate.
   A student who thinks the larger resistance always dissipates more picks this. That holds only when the currents are equal; across the same potential difference, the smaller resistance dissipates more.
3. CP = ΔV²/R is the right one: a larger resistance converts energy at a smaller rate.
   A student who thinks the smaller resistance always dissipates more picks this. That holds only when the potential differences are equal; with the same current, the larger resistance dissipates more.
4. DNeither: a battery sets the power, whatever resistance is connected.
   A student who thinks a battery transfers energy at a fixed rate picks this. P = IΔV depends on the current the resistance draws, so the power does depend on R.

CED 11.4.A.1 · Read this in Fix

1. AA smaller resistance dissipates energy at a greater rate in any circuit, so X has the larger power.
   A student who thinks the smaller resistance dissipates energy at the greater rate whatever the circuit picks this. The conclusion is right here, but the general premise is false: with equal currents, the larger resistance dissipates energy at the greater rate (P = I²R). X is brighter because both bulbs have the same ΔV, so X, with the larger current, has the larger IΔV.
2. BBoth have the same ΔV, so X, with the smaller resistance, has the larger current and the larger IΔV. Correct
   Each bulb has the battery's potential difference across it. With the same ΔV, X's current, ΔV/R, is twice Y's, so X's power IΔV is twice Y's, and brightness increases with power.
3. CCurrent takes the path of least resistance, so more of the battery current goes to X than to Y.
   A student who applies 'current takes the path of least resistance' to separate circuits picks this. The bulbs are not paths between the same two points; each has its own battery and its own current, ΔV/R.
4. DLess energy is lost overcoming X's smaller resistance, so more is left over to produce light.
   A student who pictures resistance as wasting energy that would otherwise become light picks this. The energy is converted in the bulb's resistance; X is brighter because energy is transferred to it at a greater rate, IΔV.

CED 11.4.A.2 · Read this in Fix

1. A0.14ε²A/(ρℓ) Correct
   The two segments carry the same current, I = ε/(R₁ + R₂) with R₁ = ρℓ/A and R₂ = 5ρℓ/A, so I = εA/(6ρℓ). The segment of resistivity 5ρ dissipates energy at the rate P = I²R₂ = 5ε²A/(36ρℓ) ≈ 0.14ε²A/(ρℓ).
2. B0.20ε²A/(ρℓ)
   A student who uses P = ΔV²/R with the emf ε, the potential difference across the whole rod, and the segment's resistance 5ρℓ/A picks this. The potential difference across the segment is IR₂ = 5ε/6, not ε.
3. C0.08ε²A/(ρℓ)
   A student who thinks that parts carrying the same current dissipate energy at the same rate picks this, giving each segment half of the total rate ε²/(R₁ + R₂). With the same current, P = I²R, so the segment with five times the resistance dissipates energy at five times the rate of the other.
4. D0.05ε²A/(ρℓ)
   A student who splits the emf equally between the two equal-length segments, ε/2 each, picks this: (ε/2)²/(5ρℓ/A). The potential difference across each segment is IR for that segment, so the segment of resistivity 5ρ has 5ε/6 across it.

Working R₁ = ρℓ/A and R₂ = 5ρℓ/A; the same current passes through both segments, I = ε/(R₁ + R₂) = εA/(6ρℓ). P₂ = I²R₂ = [ε²A²/(36ρ²ℓ²)](5ρℓ/A) = 5ε²A/(36ρℓ) ≈ 0.14ε²A/(ρℓ). Check: ΔV₂ = IR₂ = 5ε/6 and P₂ = IΔV₂ give the same. Units: V²·m²/(Ω·m·m) = W. Distractors (sympy-checked): ε²/R₂ = 0.20ε²A/(ρℓ); half of ε²/(R₁ + R₂) = ε²A/(12ρℓ) ≈ 0.08ε²A/(ρℓ); (ε/2)²/R₂ = 0.05ε²A/(ρℓ).

CED 11.4.A.1 · Read this in Fix

1. A4.0 Ω
   A student who treats the running motor as a resistor equal to its coil's resistance picks this, from R = ΔV/I = (12 V)/(3.0 A). If the coil had 4.0 Ω, the coil alone would dissipate all 36 W and none would be left to lift the load; the coil's potential difference IR is only part of the 12 V.
2. B7.2 Ω
   A student who finds the coil's dissipation from ΔV²/R, using the 12 V across the whole motor, picks this: (12 V)²/R = 20 W. Only part of the 12 V is across the coil's resistance, so its dissipation must be found from the current, I²R.
3. C2.2 Ω Correct
   Energy reaches the motor at the rate IΔV = 36 W. The load rises at constant speed, so the motor does work at the rate mgv = (2.0)(10)(0.80) = 16 W. The rest, 20 W, is dissipated in the coil at the rate I²R, so R = 20 W/(3.0 A)² = 2.2 Ω.
4. D1.8 Ω
   A student who takes the motor's mechanical output to be I²R picks this: (3.0 A)²R = 16 W. I²R is the rate at which energy is dissipated in the coil, not the rate of mechanical work; the energy balance gives I²R = 36 W − 16 W.

Working Input rate IΔV = (3.0 A)(12 V) = 36 W. Mechanical rate: the load rises at constant speed, so the motor's force equals mg and P_mech = mgv = (2.0)(10)(0.80) = 16 W. Energy balance: IΔV = P_mech + I²R ⇒ I²R = 36 − 16 = 20 W ⇒ R = 20/(3.0)² = 2.22 Ω ≈ 2.2 Ω. (The running motor is not an ohmic resistor: ΔV > IR = 6.7 V.) Distractors: ΔV/I = 4.0 Ω (motor treated as a resistor); dissipation set equal to ΔV²/R: R = (12)²/20 = 7.2 Ω; mechanical rate set equal to I²R: R = 16/9.0 = 1.8 Ω.

CED 11.4.A.1 · Read this in Fix