AP Unit 11 Solutions

The graph has horizontal steps at 0.20 A from 0 to 3 s, 0.50 A from 3 to 5 s, and 0 A from 5 to 9 s. Area is (0.20)(3.0) + (0.50)(2.0) = 1.60 C.

N = 1.60/(1.60 × 10⁻¹⁹) = 1.0 × 10¹⁹ electrons.

Electrons drift right to left, opposite conventional current. At a bend current follows the path without a vector-component reduction.

During zero current the average directed flow is zero, though random microscopic carrier speeds need not vanish.

IA = 0.90/0.30 = 3.0 A and IB = 1.20/0.80 = 1.5 A. In 2.0 s, A transfers 6.0 C and B 3.0 C.

Emf is energy supplied per charge, J/C = V. A maintained potential difference supports the field that drives the circuit’s net carrier motion.

The same charge rate crosses both legs of an unbranched steady path. Current magnitudes are equal even though the legs point in different spatial directions.

In each leg electron drift is opposite its local conventional-current arrow.

Draw the battery across A–B, a resistor across A–B, and switch–bulb in a separate A–B path. Put the voltmeter across only the bulb. Label the battery positive end A and use arrows from A toward B in the external conducting branches.

The loops are battery–resistor and battery–switch–bulb. The battery belongs to both. The adjustable resistor is the usual resistor symbol with a diagonal crossing arrow.

The bulb branch becomes open and has no steady conducting current; the independent resistor branch retains a closed battery path.

The open switch can have the battery’s voltage across it, even with no branch current. An open gap is not an equal-potential connection.

Draw the battery, switch and 2 Ω feed in series up to node B. Between B and return node D, put the 6 Ω load in parallel with an ideal wire.

The ideal wire makes B and D the same potential, so ΔVload = 0 and Iload = 0/6 = 0 A. The feed takes the full 12 V, so Isource = 12/2 = 6 A, all through the shunt.

Before the shunt, source and load current were 12/(2 + 6) = 1.5 A and load voltage was 9 V. Source current increases while load current falls to zero.

The ideal source requires 12 V between its terminals while the ideal wire requires 0 V between the same terminals. Equivalently a loop has +12 V and no compensating drop. No finite steady solution satisfies these constraints. Restore battery internal resistance and wire/contact resistance; heating can change them.

Bending a line with unchanged endpoints preserves topology; adding the shunt creates a new connection and changes the circuit.

RA = ρL/(πa²) and RB = ρ(3L)/(π(2a)²), so RB/RA = 3/4.

a = 1.0 × 10⁻⁴ m gives RA = 0.637 Ω and RB = 0.477 Ω. R describes the object including geometry; ρ describes the material’s opposition to current through its atomic structure.

Real metallic ρ typically rises with temperature, so RA = ρL/A rises too at unchanged geometry.

The ideal ohmic model holds R and ρ fixed, including the framework’s temperature-independent idealization. Real ohmic behavior at a fixed temperature does not establish that ρ stays unchanged across temperatures.

The first data lie on a straight line through zero with slope 0.040 A/V, so R = 1/slope = 25 Ω. On a ΔV-versus-I plot its slope is 25 V/A.

For the bulb the ratios V/I are 10, 12.5, and 15 Ω, so its data do not support one constant resistance. Its I–V curve bends as the resistance changes.

Measure paired I and V with ammeter in series and voltmeter across the bulb, recording temperature if a suitable sensor is available. Repeat upward/downward voltage sweeps with short readings and controlled cooldowns; keep the same bulb and meter arrangement.

Electrical energy is converted to thermal energy in the filament and surroundings. Real temperature change can alter ρ, unlike the fixed-ρ ideal model. Look for reproducible differences related to temperature or cooldown, while acknowledging other nonohmic mechanisms are possible.

Substitute ΔV = IR to get P = I²R; substitute I = ΔV/R to get P = (ΔV)²/R.

I = 6.0/30 = 0.20 A, P = 36/30 = 1.2 W, and energy = (1.2)(120) = 144 J.

Doubling V gives four times the power, 4.8 W. Doubling R at fixed 6 V gives half the power, 0.60 W.

P-versus-V at fixed R is quadratic through zero; P-versus-R at fixed V is a decreasing inverse curve for positive R.

I = 12/(4 + 8) = 1 A through both. PA = I²RA = 4 W and PB = 8 W.

The model predicts B brighter because its larger resistance converts more power at the same current.

Both have 12 V. PA = 144/4 = 36 W and PB = 144/8 = 18 W.

The model predicts A brighter. Actual incandescent resistance changes with temperature, and visible brightness is not quantitatively comparable solely from power across different bulb designs.

Rp = (6 × 12)/(6 + 12) = 4 Ω; Req = 3 + 4 = 7 Ω. Itotal = 18/7 A.

The feed drop is 54/7 V, leaving 72/7 V across both branches. I6 = 12/7 A and I12 = 6/7 A; their sum is 18/7 A.

1/Rp′ = 1/6 + 1/12 + 1/12 = 1/3, so Rp′ = 3 Ω and Req′ = 6 Ω. Source current becomes 3 A.

The feed now drops 9 V, leaving 9 V across the group. Original 6 Ω current is 1.5 A, smaller than 12/7 A.

Total resistance = 2 + 9 + 1 = 12 Ω; I = 1 A. Vterminal = ε − Ir = 12 − 2 = 10 V. The load voltage is 9 V and the wire drop 1 V.

Open-circuit current is zero, so terminal voltage equals emf, 12 V. Wire resistance may be neglected when much smaller than other finite resistances; here it causes a stated 1 V drop and should be retained.

The ammeter is inserted in series with the load; the voltmeter is connected across the load’s two terminals. Ideal ammeter R = 0, ideal voltmeter R approaches infinity.

A nonideal ammeter adds series resistance and reduces current. A finite-resistance voltmeter adds a parallel load path, increasing demand through the upstream resistance and potentially reducing the load voltage. These are qualitative effects; no numerical loading calculation is required.

The source rise is +12 V; resistor drops are −6 V and −6 V. Each resistance is |ΔV|/I = 6/2 = 3 Ω.

The loop equation is 12 − (2)(3) − (2)(3) = 0. The graph returns to its starting height, as required for one potential at the starting point.

The source provides (0.50)(12) = 6.0 J; each resistor converts (0.50)(6) = 3.0 J.

Subtract 4 V from every graph height: 0 → 12 → 6 → 0 V. The drops and current remain the same.

The student crosses both separate outgoing parallel paths in one source-return traversal, which is not the stated physical loop. Correct equations are 15 − 2I0 − 3I1 = 0 and 15 − 2I0 − 6I2 = 0, with I0 = I1 + I2.

Subtract loops: 3I1 = 6I2, so I1 = 2I2 and I0 = 3I2. Then 15 − 6I2 − 6I2 = 0 gives I2 = 1.25 A, I1 = 2.50 A, I0 = 3.75 A.

The branch-to-branch loop is −3I1 + 6I2 = 0. It is obtained by subtracting the two source loops, so is a consistency check rather than an independent equation.

The graph is 0 → 15 → 7.5 → 0 V, since the feed drop is (3.75)(2) = 7.5 V and branch drop (1.25)(6) = 7.5 V.

IA = 4.0 × 10⁻³/(2.0 × 10⁻³) = 2.0 A. Junction rule gives 2.0 + 0.50 = 1.20 + ID, so ID = 1.30 A outward.

Charge-flow magnitude through D in 5 s is (1.30)(5.0) = 6.50 C.

Incoming total is 2.50 A; the reported outgoing total is 1.20 + 0.80 = 2.00 A. Net accumulation rate would be 0.50 C/s.

A growing charge distribution changes electric fields and redistributes currents, so that imbalance cannot describe the stipulated steady node.

1.4 = 0.8 + 1.0 + ID, so ID = −0.4 A relative to the assumed outward arrow.

D actually brings in 0.4 A. Incoming total 1.4 + 0.4 = 1.8 A equals outgoing 0.8 + 1.0 = 1.8 A.

Both carry 0.8 A because there is no alternative path or sustained accumulation between them.

Angles do not change a scalar flow rate, and resistor heating converts energy rather than consuming charge.

1/Ceq = 1/2 + 1/6 in μF⁻¹, so Ceq = 1.5 μF. Source-side charge Q = CeqV = 18 μC.

V2 = Q/2 = 9 V and V6 = Q/6 = 3 V. The initially neutral floating intermediate conductor has equal and opposite induced plate contributions, giving equal series Q. The equivalent 1.5 μF is less than 2 μF.

Ceq = 2 + 6 = 8 μF. Both capacitor voltages are 12 V, so Q2 = 24 μC and Q6 = 72 μC.

Total source-side charge is 24 + 72 = 96 μC = (8 μF)(12 V).

τ = RC = (5000)(200 × 10⁻⁶) = 1.0 s. Initially VC = 0, Q = 0, U = 0, and I = 10/5000 = 2.0 mA.

At long time I = 0, VC = 10 V, Q = 2.0 mC, and U = ½(200 × 10⁻⁶)(10)² = 10 mJ.

Q and VC rise from zero and flatten toward final values; |I| starts at its maximum and decreases toward zero; U rises toward 10 mJ. At one τ, charge is about 63% of its final value.

Doubling R doubles τ and halves initial current, stretching the approach to the same final charge, voltage, and energy.

Initially the voltage remains 10 V, so |I| = 2.0 mA, with direction opposite charging through the resistor. Charge magnitude and stored energy begin decreasing immediately.

Q, VC, |I|, and U all approach zero; charge is about 37% of its initial magnitude after one τ. Lost stored energy becomes resistor heating. At times much longer than τ use the zero-current, zero-voltage endpoint.

Assemble a low-voltage source, adjustable series control, ammeter in series with the sample and voltmeter across the sample. Vary voltage through several safe settings, measure paired sample voltage and current after a consistent short interval, and repeat settings in increasing and decreasing order. Keep sample dimensions and ambient conditions fixed; allow cooling between runs or monitor temperature. Plot I against sample ΔV and check whether a straight line through the origin describes repeated data within uncertainty.

Place the ammeter in series and voltmeter across the resistor. Use low currents and repeat paired I–V measurements after cooldown while keeping material and geometry fixed.

R = 1/0.020 = 50 Ω. Bounds reverse on reciprocation: 1/0.021 = 47.6 Ω to 1/0.019 = 52.6 Ω. ρ = RA/ℓ = 1.0 × 10⁻⁵ Ω·m.

Inspect residuals for curvature and compare ascending/descending low-voltage sweeps at controlled temperature. State the fixed-ρ idealization when applying one R.

P = IΔV = (0.080)(4.0) = 0.32 W; Δq = IΔt = 2.4 C.

Energy = PΔt = 9.6 J, also ΔqΔV = (2.4)(4) = 9.6 J. Thermal conversion can raise the resistor’s temperature and change real material resistivity, so later I–V points may differ.

At B: I1 = I2 + I5. At C: I3 + I5 = I4. Source-left loop: 12 − 2I1 − 4I2 = 0. Source-right: 12 − 4I3 − 2I4 = 0. Top bridge loop A→B→C→A: −2I1 − 4I5 + 4I3 = 0.

The bridge gives an alternative path at both B and C, so charge through R1 need not all pass through R2, and charge through R3 need not all pass through R4.

Use currents from voltage drops: (12 − VB)/2 = VB/4 + (VB − VC)/4, and (12 − VC)/4 + (VB − VC)/4 = VC/2. Simplify to 4VB − VC = 24 and −VB + 4VC = 12.

VB = 7.2 V and VC = 4.8 V give I1 = 2.4 A, I2 = 1.8 A, I3 = 1.8 A, I4 = 2.4 A, and I5 = 0.60 A.

Unused lower loop B→D→C→B gives −4I2 + 2I4 + 4I5 = −7.2 + 4.8 + 2.4 = 0. Source current is I1 + I3 = 4.2 A, so supplied power 50.4 W. Resistor powers are 11.52 + 12.96 + 12.96 + 11.52 + 1.44 = 50.4 W.

Insert an ideal ammeter in series with R5; connect an ideal voltmeter across B and C. Their ideal resistances are zero and infinite respectively.

A nonideal voltmeter adds another B–C current path and changes the network, so the measured potential difference and branch currents can shift. No numerical meter-loading solution is required.

Initially C has zero voltage, so B is at D’s 0 V. I1 = 12/2000 = 6.0 mA, I2 = 0, and IC = 6.0 mA.

At long time IC = 0. R1 and R2 form the remaining series loop, so I1 = I2 = 12/6000 = 2.0 mA. VB = I2R2 = 8.0 V.

Qf = C(8.0) = 800 μC and Uf = ½(100 × 10⁻⁶)(8)² = 3.2 mJ. The feed resistor’s 4 V drop prevents the capacitor from reaching full battery voltage.

The only discharge path is R2, so τ = (4000)(100 × 10⁻⁶) = 0.40 s. Initial voltage remains 8 V, giving |I2,0| = 8/4000 = 2.0 mA.

Both voltage and current magnitudes fall with flattening curves toward zero; about 37% initial charge remains at one τ. The initial 3.2 mJ of stored energy eventually converts to resistor heating.

Use a low-voltage supply, known R2 and rated C, switch, and high-resistance voltage sensor. Charge to a known initial voltage, switch to R2-only discharge, and record VC versus elapsed time. Read the time when voltage reaches about 37% of its initial value.

Control C, initial voltage, meter placement, and resistor temperature. Repeat switching trials; report a time range bracketed by adjacent samples crossing the landmark, adding switch-timing uncertainty and component tolerances. Repeat with doubled R2 to test a doubled landmark time.

For each trial, plot the measured voltage divided by its initial value against elapsed seconds. Give both axes scales, show time-sampling and voltage uncertainty, and locate the crossing of 0.37. Read the time interval consistent with the nearby readings rather than claiming precision finer than the sampling. Compare repeated crossing times with τ = R2C = 0.40 s.

The plotted fractions are 1.000, 0.613, 0.475, 0.375, 0.288, 0.225 and 0.138. The 0.40 s point is consistent with 0.37 within the voltage uncertainty. Report τ ≈ 0.4 s, with the crossing bracketed by 0.30–0.50 s readings, rather than asserting a fitted three-decimal time.