RC Circuits

Time constant: τ = RC τ = (100 × 10³ Ω)(10 × 10⁻⁶ F) τ = 1.0 s

Physical meaning of τ: • Time for capacitor to charge to 63.2% of final voltage (or discharge to 36.8%) • Characterizes how "fast" the circuit responds • After time t = 5τ, circuit is ~99% to equilibrium • Larger τ = slower charging/discharging

For this circuit: • τ = 1.0 second • After 1 second: V_C ≈ 7.6 V (63% of 12 V) • After 5 seconds: V_C ≈ 11.9 V (essentially fully charged)

During CHARGING (from Q = 0):

Charge: Q(t) = Q_max(1 - e^(-t/τ)) Q(t) = Cε(1 - e^(-t/RC)) Q(t) = (10 × 10⁻⁶)(12)(1 - e^(-t/1.0)) Q(t) = 1.2 × 10⁻⁴(1 - e^(-t)) coulombs

where Q_max = Cε = 120 μC

Current: I(t) = I₀e^(-t/τ) I(t) = (ε/R)e^(-t/RC) I(t) = (12/100,000)e^(-t/1.0) I(t) = 1.2 × 10⁻⁴ e^(-t) amperes

where I₀ = ε/R = 120 μA

Key features: • Q starts at 0, approaches Q_max exponentially • I starts at maximum (ε/R), decays to 0 • At t = 0: capacitor acts like wire (I max) • At t = ∞: capacitor acts like open circuit (I = 0)

For DISCHARGING: V(t) = V₀e^(-t/τ)

Given: V₀ = 12 V V(t) = 3 V τ = RC = (50 × 10³)(20 × 10⁻⁶) = 1.0 s

Step 1: Substitute into equation 3 = 12e^(-t/1.0)

Step 2: Solve for t 3/12 = e^(-t) 0.25 = e^(-t) ln(0.25) = -t -1.386 = -t t = 1.386 s

Alternatively: t = -τ ln(V/V₀) t = -1.0 ln(3/12) t = -1.0 ln(0.25) t = 1.39 seconds

The voltage drops from 12 V to 3 V (to 1/4 of original) in about 1.4 seconds, which is about 1.4τ.

Note: Voltage drops by half every 0.693τ (like radioactive half-life).

Step 1: Apply Kirchhoff's voltage law ε - V_R - V_C = 0 ε - IR - Q/C = 0

Step 2: Express I in terms of Q I = dQ/dt

ε - R(dQ/dt) - Q/C = 0

Step 3: Rearrange to standard form R(dQ/dt) = ε - Q/C dQ/dt = ε/R - Q/(RC) dQ/dt = (Cε - Q)/(RC)

Step 4: Separate variables dQ/(Cε - Q) = dt/(RC)

Step 5: Integrate both sides ∫dQ/(Cε - Q) = ∫dt/(RC) -ln(Cε - Q) = t/(RC) + constant

Step 6: Apply initial condition Q(0) = 0 -ln(Cε) = constant

Step 7: Solve for Q(t) -ln(Cε - Q) = t/(RC) - ln(Cε) ln(Cε - Q) - ln(Cε) = -t/(RC) ln[(Cε - Q)/Cε] = -t/τ (Cε - Q)/Cε = e^(-t/τ) Cε - Q = Cε e^(-t/τ)

Q(t) = Cε(1 - e^(-t/τ))

This is the charging equation!

(a) Power at t = τ: I(t) = I₀e^(-t/τ) At t = τ: I = I₀e^(-1) = I₀/e I₀ = ε/R = 10/(10⁶) = 10 μA I(τ) = 10/e ≈ 3.68 μA

P = I²R = (3.68 × 10⁻⁶)²(10⁶) = 13.5 μW

(b) Energy stored when fully charged: U_C = (1/2)CV² U_C = (1/2)(2 × 10⁻⁶)(10)² U_C = 100 μJ

(c) Energy from battery: U_battery = Qε = Cε² = (2 × 10⁻⁶)(10)² = 200 μJ

Energy dissipated as heat: U_heat = U_battery - U_C = 200 - 100 = 100 μJ

Remarkable result: Exactly HALF the energy from the battery is dissipated as heat in the resistor during charging, regardless of R! This is always true for RC circuits.