The problems are on the RC circuits worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Kirchhoff's junction rule conserves charge and his loop rule conserves energy, giving equations for every branch current. These worked answers correct the 20 problems of the AP Physics C worksheet on RC circuits, from junction rule to spot the error: each solution shows the formula, the substitution and the final answer with its unit.
Complete methods follow: the equation, the substitution and the final result with its unit in bold. When a current comes out negative, note what that means physically, and when a time constant is involved, check that your answer is a plausible multiple of τ.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:E&M Unit 11
Problem 1 Junction rule
1. The sum of currents entering a junction equals the sum leaving it.2. In = 5.0 A and out so far = 4.0 A, so 1.0 A leaves along the remaining wire.
Problem 2 Time constant
1. τ = RC = (1.0 × 10⁴)(22 × 10⁻⁶) = 0.22 s.2. Ω × F = (V/A)(C/V) = C/A = s.
Problem 3 Start and finish
1. I₀ = ε/R = 6.0/1500 = 4.0 mA.2. The capacitor is an open circuit when full, so I → 0.
3. Q = Cε = (100 × 10⁻⁶)(6.0) = 6.0 × 10⁻⁴ C.
Problem 4 Fractions charged
1. 1 − e⁻¹ = 0.632.2. 1 − e⁻³ = 0.950.
3. 1 − e⁻⁵ = 0.993.
Problem 5 Discharge fraction
1. q = Q₀e⁻² = 80(0.1353) = 10.8 μC.2. 86.5 % has gone.
Problem 6 Reading a charging graph
1. VC = 0.63 × 10 V = 6.3 V at 2.0 s, so τ = 2.0 s.2. R = τ/C = 2.0/(1000 × 10⁻⁶) = 2.0 kΩ.
Problem 7 Resistors in a loop
1. 12 − I(4.0 + 2.0) = 0 gives I = 2.0 A.2. V = IR = 2.0 × 4.0 = 8.0 V.
Problem 8 Two-loop circuit
1. I₃ = I₁ + I₂.2. Left loop: 10 − 1.0I₁ − 4.0I₃ = 0 gives 10 = 5I₁ + 4I₂. Right loop: 4.0 − 2.0I₂ − 4.0I₃ = 0 gives 4 = 4I₁ + 6I₂.
3. Eliminating I₂: 22 = 7I₁, so I₁ = 3.1 A, I₂ = (10 − 5 × 3.143)/4 = −1.4 A and I₃ = 1.7 A. Battery 2 is being charged.
4. Supplied: 10(3.143) + 4(−1.429) = 25.7 W. Dissipated: 3.143² × 1 + 1.429² × 2 + 1.714² × 4 = 9.88 + 4.08 + 11.76 = 25.7 W, which agree.
Problem 9 Steady state
1. A full capacitor has no further voltage change, so charge stops flowing in its branch.2. I = 12/(4000 + 2000) = 2.0 mA.
3. VC = IR₂ = 4.0 V and Q = CV = 20 μC.
Problem 10 Time to reach 8 V
1. τ = (2.0 × 10⁶)(3.3 × 10⁻⁶) = 6.6 s.2. 8 = 12(1 − e−t/τ) gives e−t/τ = 1/3, so t = τ ln 3 = 7.3 s.
Problem 11 Discharge current
1. τ = (470)(2200 × 10⁻⁶) = 1.03 s.2. I₀ = 9.0/470 = 19 mA.
3. I = 19.15 mA × e−2.0/1.034 = 19.15 × 0.1445 = 2.8 mA.
Problem 12 Half-time
1. Set q = Q₀/2: e−t/τ = 1/2, so t½ = τ ln 2.2. τ = (1.0 × 10⁵)(1.0 × 10⁻⁵) = 1.0 s, so t½ = 0.69 s.
3. 1/16 = (1/2)⁴ so four half-times: 2.8 s.
Problem 13 Verify the charging law
1. I = dq/dt = Cε(1/RC)e−t/RC = (ε/R)e−t/RC.2. IR + q/C = εe−t/RC + ε(1 − e−t/RC) = ε, so the loop equation holds.
Problem 14 Log graph of discharge
1. Gradient = (0.48 − 2.48)/4.0 = −0.50 s⁻¹.2. ln V = ln V₀ − t/τ, so τ = 1/0.50 = 2.0 s.
3. R = τ/C = 2.0/(470 × 10⁻⁶) = 4.3 kΩ.
Problem 15 Battery with internal resistance
1. I = ε/(R + r) = 12/6.0 = 2.0 A.2. V = ε − Ir = 12 − 1.0 = 11 V.
Problem 16 Energy in charging
1. U = ½Cε² = ½(100 × 10⁻⁶)(144) = 7.2 × 10⁻³ J.2. ∫I²R dt = (ε²/R)∫₀^∞ e−2t/RC dt = (ε²/R)(RC/2) = ½Cε² = 7.2 mJ.
3. R cancels: a larger R slows the current but lengthens the time, leaving the energy unchanged.
Problem 17 Divider charging
1. Vfinal = 12 × 20/(10 + 20) = 8.0 V.2. Req = 10 ∥ 20 = 6.67 kΩ, so τ = (6.67 × 10³)(10 × 10⁻⁶) = 0.067 s.
3. 6.0 = 8.0(1 − e−t/τ) gives e−t/τ = 0.25, so t = τ ln 4 = 0.092 s.
Problem 18 Leaky capacitor
1. R = ρd/A and C = κε₀A/d, so RC = ρκε₀: A and d cancel.2. τ = (1.0 × 10¹²)(4.0)(8.85 × 10⁻¹²) = 35 s.
Problem 19 Flash lamp
1. U₀ = ½(470 × 10⁻⁶)(625) = 0.15 J.2. P = V²/R = (V₀²/R)e−2t/τ; integrating from 0 to t gives (V₀²τ/2R)(1 − e−2t/τ), and V₀²τ/2R = ½CV₀² = U₀.
3. U₀(1 − e⁻²) = 0.1469(0.8647) = 0.13 J.
Problem 20 Spot the error
1. Each τ removes the same fraction (63 %) of what is left, so charging never completes in a fixed finite time.2. 1 − e−t/τ = 0.99 gives t = τ ln 100 = 0.22 × 4.605 = 1.0 s.
3. I = (6.0/10⁴)(0.01) = 6.0 μA.
