RC Circuits

Capacitor charging and discharging with differential equations

RC Circuits

Charging Capacitor

Circuit: battery (E\mathcal{E}), resistor (RR), capacitor (CC) in series.

Kirchhoff's loop rule: E−IR−QC=0\mathcal{E} - IR - \frac{Q}{C} = 0

E−RdQdt−QC=0\mathcal{E} - R\frac{dQ}{dt} - \frac{Q}{C} = 0

Differential equation: dQdt=ER−QRC\frac{dQ}{dt} = \frac{\mathcal{E}}{R} - \frac{Q}{RC}

Solution: Charging

Q(t)=CE(1−e−t/RC)Q(t) = C\mathcal{E}(1 - e^{-t/RC})

Current: I(t)=dQdt=ERe−t/RCI(t) = \frac{dQ}{dt} = \frac{\mathcal{E}}{R}e^{-t/RC}

Voltage across capacitor: VC(t)=QC=E(1−e−t/RC)V_C(t) = \frac{Q}{C} = \mathcal{E}(1 - e^{-t/RC})

Voltage across resistor: VR(t)=IR=Ee−t/RCV_R(t) = IR = \mathcal{E}e^{-t/RC}

Time constant: τ=RC\tau = RC

After time τ\tau:

  • Capacitor reaches (1−1/e)≈63%(1 - 1/e) \approx 63\% of final charge
  • Current drops to 1/e≈37%1/e \approx 37\% of initial

Discharging Capacitor

Initial charge Q0Q_0 on capacitor, no battery.

Loop rule: −IR−QC=0-IR - \frac{Q}{C} = 0

RdQdt+QC=0R\frac{dQ}{dt} + \frac{Q}{C} = 0

Solution: Q(t)=Q0e−t/RCQ(t) = Q_0e^{-t/RC}

Current: I(t)=−dQdt=Q0RCe−t/RCI(t) = -\frac{dQ}{dt} = \frac{Q_0}{RC}e^{-t/RC}

Voltage: VC(t)=V0e−t/RCV_C(t) = V_0e^{-t/RC}

Energy Considerations

Charging:

Energy supplied by battery: Wbattery=QE=CE2W_{battery} = Q\mathcal{E} = C\mathcal{E}^2

Energy stored in capacitor: UC=12CE2U_C = \frac{1}{2}C\mathcal{E}^2

Energy dissipated in resistor: UR=12CE2U_R = \frac{1}{2}C\mathcal{E}^2

(Half the energy is always dissipated as heat, independent of RR!)

Discharging:

All energy dissipated in resistor: UR=12CV02U_R = \frac{1}{2}CV_0^2

General RC Circuit

For any RC circuit, differential equation has form:

RCdVCdt+VC=VfinalRC\frac{dV_C}{dt} + V_C = V_{final}

General solution: VC(t)=Vfinal+(Vinitial−Vfinal)e−t/RCV_C(t) = V_{final} + (V_{initial} - V_{final})e^{-t/RC}

Multiple Capacitors

Capacitors in series or parallel can be replaced by equivalent capacitance, then analyze as simple RC circuit.

Effective time constant: τ=RCeq\tau = RC_{eq}

Applications

Timer circuits: Delay determined by RCRC

Filters: Block DC, pass AC (or vice versa)

Integrators/Differentiators: For signal processing

Defibrillators: Store energy, rapid discharge through heart

📚 Practice Problems

1Problem 1medium

❓ Question:

An RC circuit consists of R = 2.0 MΩ, C = 5.0 μF, and a battery ε = 12 V. The capacitor is initially uncharged. Find: (a) the time constant τ, (b) the charge on the capacitor at t = 5.0 s, and (c) the current at t = 5.0 s.

💡 Show Solution

Given:

  • R = 2.0 MΩ = 2.0 × 10⁶ Ω
  • C = 5.0 μF = 5.0 × 10⁻⁶ F
  • ε = 12 V
  • Q₀ = 0 (initially uncharged)
  • t = 5.0 s

(a) Time constant:

τ=RC=(2.0×106)(5.0×10−6)\tau = RC = (2.0 \times 10^6)(5.0 \times 10^{-6})

τ=10 s\tau = \boxed{10 \text{ s}}

(b) Charge at t = 5.0 s:

For charging capacitor: Q(t)=Qmax(1−e−t/τ)Q(t) = Q_{max}(1 - e^{-t/\tau})

where Qmax=Cε=(5.0×10−6)(12)=6.0×10−5Q_{max} = C\varepsilon = (5.0 \times 10^{-6})(12) = 6.0 \times 10^{-5} C

Q(5.0)=(6.0×10−5)(1−e−5.0/10)Q(5.0) = (6.0 \times 10^{-5})(1 - e^{-5.0/10})

Q(5.0)=(6.0×10−5)(1−e−0.5)Q(5.0) = (6.0 \times 10^{-5})(1 - e^{-0.5})

Q(5.0)=(6.0×10−5)(1−0.6065)Q(5.0) = (6.0 \times 10^{-5})(1 - 0.6065)

Q(5.0)=(6.0×10−5)(0.3935)=2.36×10−5 C=23.6 μCQ(5.0) = (6.0 \times 10^{-5})(0.3935) = \boxed{2.36 \times 10^{-5} \text{ C} = 23.6 \text{ μC}}

(c) Current at t = 5.0 s:

For charging: I(t)=I0e−t/τI(t) = I_0 e^{-t/\tau}

where I0=ε/R=12/(2.0×106)=6.0×10−6I_0 = \varepsilon/R = 12/(2.0 \times 10^6) = 6.0 \times 10^{-6} A

I(5.0)=(6.0×10−6)e−0.5I(5.0) = (6.0 \times 10^{-6})e^{-0.5}

I(5.0)=(6.0×10−6)(0.6065)=3.64×10−6 A=3.64 μAI(5.0) = (6.0 \times 10^{-6})(0.6065) = \boxed{3.64 \times 10^{-6} \text{ A} = 3.64 \text{ μA}}

2Problem 2medium

❓ Question:

An RC circuit consists of R = 2.0 MΩ, C = 5.0 μF, and a battery ε = 12 V. The capacitor is initially uncharged. Find: (a) the time constant τ, (b) the charge on the capacitor at t = 5.0 s, and (c) the current at t = 5.0 s.

💡 Show Solution

Given:

  • R = 2.0 MΩ = 2.0 × 10⁶ Ω
  • C = 5.0 μF = 5.0 × 10⁻⁶ F
  • ε = 12 V
  • Q₀ = 0 (initially uncharged)
  • t = 5.0 s

(a) Time constant:

τ=RC=(2.0×106)(5.0×10−6)\tau = RC = (2.0 \times 10^6)(5.0 \times 10^{-6})

τ=10 s\tau = \boxed{10 \text{ s}}

(b) Charge at t = 5.0 s:

For charging capacitor: Q(t)=Qmax(1−e−t/τ)Q(t) = Q_{max}(1 - e^{-t/\tau})

where Qmax=Cε=(5.0×10−6)(12)=6.0×10−5Q_{max} = C\varepsilon = (5.0 \times 10^{-6})(12) = 6.0 \times 10^{-5} C

Q(5.0)=(6.0×10−5)(1−e−5.0/10)Q(5.0) = (6.0 \times 10^{-5})(1 - e^{-5.0/10})

Q(5.0)=(6.0×10−5)(1−e−0.5)Q(5.0) = (6.0 \times 10^{-5})(1 - e^{-0.5})

Q(5.0)=(6.0×10−5)(1−0.6065)Q(5.0) = (6.0 \times 10^{-5})(1 - 0.6065)

Q(5.0)=(6.0×10−5)(0.3935)=2.36×10−5 C=23.6 μCQ(5.0) = (6.0 \times 10^{-5})(0.3935) = \boxed{2.36 \times 10^{-5} \text{ C} = 23.6 \text{ μC}}

(c) Current at t = 5.0 s:

For charging: I(t)=I0e−t/τI(t) = I_0 e^{-t/\tau}

where I0=ε/R=12/(2.0×106)=6.0×10−6I_0 = \varepsilon/R = 12/(2.0 \times 10^6) = 6.0 \times 10^{-6} A

I(5.0)=(6.0×10−6)e−0.5I(5.0) = (6.0 \times 10^{-6})e^{-0.5}

I(5.0)=(6.0×10−6)(0.6065)=3.64×10−6 A=3.64 μAI(5.0) = (6.0 \times 10^{-6})(0.6065) = \boxed{3.64 \times 10^{-6} \text{ A} = 3.64 \text{ μA}}

3Problem 3medium

❓ Question:

An RC circuit consists of R = 2.0 MΩ, C = 5.0 μF, and a battery ε = 12 V. The capacitor is initially uncharged. Find: (a) the time constant τ, (b) the charge on the capacitor at t = 5.0 s, and (c) the current at t = 5.0 s.

💡 Show Solution

Given:

  • R = 2.0 MΩ = 2.0 × 10⁶ Ω
  • C = 5.0 μF = 5.0 × 10⁻⁶ F
  • ε = 12 V
  • Q₀ = 0 (initially uncharged)
  • t = 5.0 s

(a) Time constant:

τ=RC=(2.0×106)(5.0×10−6)\tau = RC = (2.0 \times 10^6)(5.0 \times 10^{-6})

τ=10 s\tau = \boxed{10 \text{ s}}

(b) Charge at t = 5.0 s:

For charging capacitor: Q(t)=Qmax(1−e−t/τ)Q(t) = Q_{max}(1 - e^{-t/\tau})

where Qmax=Cε=(5.0×10−6)(12)=6.0×10−5Q_{max} = C\varepsilon = (5.0 \times 10^{-6})(12) = 6.0 \times 10^{-5} C

Q(5.0)=(6.0×10−5)(1−e−5.0/10)Q(5.0) = (6.0 \times 10^{-5})(1 - e^{-5.0/10})

Q(5.0)=(6.0×10−5)(1−e−0.5)Q(5.0) = (6.0 \times 10^{-5})(1 - e^{-0.5})

Q(5.0)=(6.0×10−5)(1−0.6065)Q(5.0) = (6.0 \times 10^{-5})(1 - 0.6065)

Q(5.0)=(6.0×10−5)(0.3935)=2.36×10−5 C=23.6 μCQ(5.0) = (6.0 \times 10^{-5})(0.3935) = \boxed{2.36 \times 10^{-5} \text{ C} = 23.6 \text{ μC}}

(c) Current at t = 5.0 s:

For charging: I(t)=I0e−t/τI(t) = I_0 e^{-t/\tau}

where I0=ε/R=12/(2.0×106)=6.0×10−6I_0 = \varepsilon/R = 12/(2.0 \times 10^6) = 6.0 \times 10^{-6} A

I(5.0)=(6.0×10−6)e−0.5I(5.0) = (6.0 \times 10^{-6})e^{-0.5}

I(5.0)=(6.0×10−6)(0.6065)=3.64×10−6 A=3.64 μAI(5.0) = (6.0 \times 10^{-6})(0.6065) = \boxed{3.64 \times 10^{-6} \text{ A} = 3.64 \text{ μA}}

4Problem 4medium

❓ Question:

A capacitor C = 100 μF is charged to V₀ = 50 V and then connected to a resistor R = 500 Ω (battery removed). Find: (a) the initial energy stored, (b) the time for the energy to decrease to 25% of its initial value, and (c) the total energy dissipated in the resistor as t → ∞.

💡 Show Solution

Given:

  • C = 100 μF = 1.0 × 10⁻⁴ F
  • V₀ = 50 V
  • R = 500 Ω
  • Discharging circuit

(a) Initial energy:

U0=12CV02=12(1.0×10−4)(50)2U_0 = \frac{1}{2}CV_0^2 = \frac{1}{2}(1.0 \times 10^{-4})(50)^2

U0=0.125 JU_0 = \boxed{0.125 \text{ J}}

(b) Time for U = 0.25U₀:

Energy in capacitor: U(t)=12C[V(t)]2=12C[V0e−t/τ]2=U0e−2t/τU(t) = \frac{1}{2}C[V(t)]^2 = \frac{1}{2}C[V_0 e^{-t/\tau}]^2 = U_0 e^{-2t/\tau}

Set U = 0.25U₀: 0.25U0=U0e−2t/τ0.25U_0 = U_0 e^{-2t/\tau} 0.25=e−2t/τ0.25 = e^{-2t/\tau} ln⁡(0.25)=−2tτ\ln(0.25) = -\frac{2t}{\tau}

Time constant: τ=RC=(500)(1.0×10−4)=0.05\tau = RC = (500)(1.0 \times 10^{-4}) = 0.05 s

t=−τln⁡(0.25)2=−(0.05)(−1.386)2t = -\frac{\tau \ln(0.25)}{2} = -\frac{(0.05)(-1.386)}{2}

t=0.0347 s=34.7 mst = \boxed{0.0347 \text{ s} = 34.7 \text{ ms}}

(c) Total energy dissipated:

As t → ∞, all energy in capacitor is dissipated in resistor:

Edissipated=U0=0.125 JE_{dissipated} = U_0 = \boxed{0.125 \text{ J}}

Check: ∫0∞I2R dt=∫0∞V02Re−2t/τdt=V02R⋅τ2=12CV02\int_0^\infty I^2 R \, dt = \int_0^\infty \frac{V_0^2}{R}e^{-2t/\tau} dt = \frac{V_0^2}{R} \cdot \frac{\tau}{2} = \frac{1}{2}CV_0^2 ✓

5Problem 5medium

❓ Question:

A capacitor C = 100 μF is charged to V₀ = 50 V and then connected to a resistor R = 500 Ω (battery removed). Find: (a) the initial energy stored, (b) the time for the energy to decrease to 25% of its initial value, and (c) the total energy dissipated in the resistor as t → ∞.

💡 Show Solution

Given:

  • C = 100 μF = 1.0 × 10⁻⁴ F
  • V₀ = 50 V
  • R = 500 Ω
  • Discharging circuit

(a) Initial energy:

U0=12CV02=12(1.0×10−4)(50)2U_0 = \frac{1}{2}CV_0^2 = \frac{1}{2}(1.0 \times 10^{-4})(50)^2

U0=0.125 JU_0 = \boxed{0.125 \text{ J}}

(b) Time for U = 0.25U₀:

Energy in capacitor: U(t)=12C[V(t)]2=12C[V0e−t/τ]2=U0e−2t/τU(t) = \frac{1}{2}C[V(t)]^2 = \frac{1}{2}C[V_0 e^{-t/\tau}]^2 = U_0 e^{-2t/\tau}

Set U = 0.25U₀: 0.25U0=U0e−2t/τ0.25U_0 = U_0 e^{-2t/\tau} 0.25=e−2t/τ0.25 = e^{-2t/\tau} ln⁡(0.25)=−2tτ\ln(0.25) = -\frac{2t}{\tau}

Time constant: τ=RC=(500)(1.0×10−4)=0.05\tau = RC = (500)(1.0 \times 10^{-4}) = 0.05 s

t=−τln⁡(0.25)2=−(0.05)(−1.386)2t = -\frac{\tau \ln(0.25)}{2} = -\frac{(0.05)(-1.386)}{2}

t=0.0347 s=34.7 mst = \boxed{0.0347 \text{ s} = 34.7 \text{ ms}}

(c) Total energy dissipated:

As t → ∞, all energy in capacitor is dissipated in resistor:

Edissipated=U0=0.125 JE_{dissipated} = U_0 = \boxed{0.125 \text{ J}}

Check: ∫0∞I2R dt=∫0∞V02Re−2t/τdt=V02R⋅τ2=12CV02\int_0^\infty I^2 R \, dt = \int_0^\infty \frac{V_0^2}{R}e^{-2t/\tau} dt = \frac{V_0^2}{R} \cdot \frac{\tau}{2} = \frac{1}{2}CV_0^2 ✓

6Problem 6medium

❓ Question:

A capacitor C = 100 μF is charged to V₀ = 50 V and then connected to a resistor R = 500 Ω (battery removed). Find: (a) the initial energy stored, (b) the time for the energy to decrease to 25% of its initial value, and (c) the total energy dissipated in the resistor as t → ∞.

💡 Show Solution

Given:

  • C = 100 μF = 1.0 × 10⁻⁴ F
  • V₀ = 50 V
  • R = 500 Ω
  • Discharging circuit

(a) Initial energy:

U0=12CV02=12(1.0×10−4)(50)2U_0 = \frac{1}{2}CV_0^2 = \frac{1}{2}(1.0 \times 10^{-4})(50)^2

U0=0.125 JU_0 = \boxed{0.125 \text{ J}}

(b) Time for U = 0.25U₀:

Energy in capacitor: U(t)=12C[V(t)]2=12C[V0e−t/τ]2=U0e−2t/τU(t) = \frac{1}{2}C[V(t)]^2 = \frac{1}{2}C[V_0 e^{-t/\tau}]^2 = U_0 e^{-2t/\tau}

Set U = 0.25U₀: 0.25U0=U0e−2t/τ0.25U_0 = U_0 e^{-2t/\tau} 0.25=e−2t/τ0.25 = e^{-2t/\tau} ln⁡(0.25)=−2tτ\ln(0.25) = -\frac{2t}{\tau}

Time constant: τ=RC=(500)(1.0×10−4)=0.05\tau = RC = (500)(1.0 \times 10^{-4}) = 0.05 s

t=−τln⁡(0.25)2=−(0.05)(−1.386)2t = -\frac{\tau \ln(0.25)}{2} = -\frac{(0.05)(-1.386)}{2}

t=0.0347 s=34.7 mst = \boxed{0.0347 \text{ s} = 34.7 \text{ ms}}

(c) Total energy dissipated:

As t → ∞, all energy in capacitor is dissipated in resistor:

Edissipated=U0=0.125 JE_{dissipated} = U_0 = \boxed{0.125 \text{ J}}

Check: ∫0∞I2R dt=∫0∞V02Re−2t/τdt=V02R⋅τ2=12CV02\int_0^\infty I^2 R \, dt = \int_0^\infty \frac{V_0^2}{R}e^{-2t/\tau} dt = \frac{V_0^2}{R} \cdot \frac{\tau}{2} = \frac{1}{2}CV_0^2 ✓

7Problem 7hard

❓ Question:

In an RC circuit with R = 1.5 kΩ, C = 20 μF, and ε = 9.0 V, the switch is closed at t = 0. Derive and evaluate: (a) the differential equation for Q(t), (b) the time when the voltage across the capacitor equals the voltage across the resistor, and (c) the rate of energy storage in the capacitor at this time.

💡 Show Solution

Given:

  • R = 1.5 kΩ = 1500 Ω
  • C = 20 μF = 2.0 × 10⁻⁵ F
  • ε = 9.0 V

(a) Differential equation:

Kirchhoff's voltage law: ε−VR−VC=0\varepsilon - V_R - V_C = 0 ε−IR−QC=0\varepsilon - IR - \frac{Q}{C} = 0

Since I=dQ/dtI = dQ/dt: ε−RdQdt−QC=0\varepsilon - R\frac{dQ}{dt} - \frac{Q}{C} = 0

Rearranging: dQdt+QRC=εR\boxed{\frac{dQ}{dt} + \frac{Q}{RC} = \frac{\varepsilon}{R}}

This is first-order linear ODE with solution: Q(t)=Cε(1−e−t/RC)Q(t) = C\varepsilon(1 - e^{-t/RC})

(b) Time when V_C = V_R:

VC=QC=ε(1−e−t/τ)V_C = \frac{Q}{C} = \varepsilon(1 - e^{-t/\tau})

VR=IR=εe−t/τV_R = IR = \varepsilon e^{-t/\tau}

Set equal: ε(1−e−t/τ)=εe−t/τ\varepsilon(1 - e^{-t/\tau}) = \varepsilon e^{-t/\tau} 1−e−t/τ=e−t/τ1 - e^{-t/\tau} = e^{-t/\tau} 1=2e−t/τ1 = 2e^{-t/\tau} e−t/τ=0.5e^{-t/\tau} = 0.5 t=τln⁡(2)t = \tau \ln(2)

where τ=RC=(1500)(2.0×10−5)=0.03\tau = RC = (1500)(2.0 \times 10^{-5}) = 0.03 s

t=(0.03)ln⁡(2)=0.0208 s=20.8 mst = (0.03)\ln(2) = \boxed{0.0208 \text{ s} = 20.8 \text{ ms}}

(c) Rate of energy storage:

Energy in capacitor: UC=Q22CU_C = \frac{Q^2}{2C}

dUCdt=QC⋅dQdt=VC⋅I\frac{dU_C}{dt} = \frac{Q}{C} \cdot \frac{dQ}{dt} = V_C \cdot I

At t = 20.8 ms: VC=4.5V_C = 4.5 V, I=VR/R=4.5/1500=0.003I = V_R/R = 4.5/1500 = 0.003 A

dUCdt=(4.5)(0.003)=0.0135 W=13.5 mW\frac{dU_C}{dt} = (4.5)(0.003) = \boxed{0.0135 \text{ W} = 13.5 \text{ mW}}

8Problem 8hard

❓ Question:

In an RC circuit with R = 1.5 kΩ, C = 20 μF, and ε = 9.0 V, the switch is closed at t = 0. Derive and evaluate: (a) the differential equation for Q(t), (b) the time when the voltage across the capacitor equals the voltage across the resistor, and (c) the rate of energy storage in the capacitor at this time.

💡 Show Solution

Given:

  • R = 1.5 kΩ = 1500 Ω
  • C = 20 μF = 2.0 × 10⁻⁵ F
  • ε = 9.0 V

(a) Differential equation:

Kirchhoff's voltage law: ε−VR−VC=0\varepsilon - V_R - V_C = 0 ε−IR−QC=0\varepsilon - IR - \frac{Q}{C} = 0

Since I=dQ/dtI = dQ/dt: ε−RdQdt−QC=0\varepsilon - R\frac{dQ}{dt} - \frac{Q}{C} = 0

Rearranging: dQdt+QRC=εR\boxed{\frac{dQ}{dt} + \frac{Q}{RC} = \frac{\varepsilon}{R}}

This is first-order linear ODE with solution: Q(t)=Cε(1−e−t/RC)Q(t) = C\varepsilon(1 - e^{-t/RC})

(b) Time when V_C = V_R:

VC=QC=ε(1−e−t/τ)V_C = \frac{Q}{C} = \varepsilon(1 - e^{-t/\tau})

VR=IR=εe−t/τV_R = IR = \varepsilon e^{-t/\tau}

Set equal: ε(1−e−t/τ)=εe−t/τ\varepsilon(1 - e^{-t/\tau}) = \varepsilon e^{-t/\tau} 1−e−t/τ=e−t/τ1 - e^{-t/\tau} = e^{-t/\tau} 1=2e−t/τ1 = 2e^{-t/\tau} e−t/τ=0.5e^{-t/\tau} = 0.5 t=τln⁡(2)t = \tau \ln(2)

where τ=RC=(1500)(2.0×10−5)=0.03\tau = RC = (1500)(2.0 \times 10^{-5}) = 0.03 s

t=(0.03)ln⁡(2)=0.0208 s=20.8 mst = (0.03)\ln(2) = \boxed{0.0208 \text{ s} = 20.8 \text{ ms}}

(c) Rate of energy storage:

Energy in capacitor: UC=Q22CU_C = \frac{Q^2}{2C}

dUCdt=QC⋅dQdt=VC⋅I\frac{dU_C}{dt} = \frac{Q}{C} \cdot \frac{dQ}{dt} = V_C \cdot I

At t = 20.8 ms: VC=4.5V_C = 4.5 V, I=VR/R=4.5/1500=0.003I = V_R/R = 4.5/1500 = 0.003 A

dUCdt=(4.5)(0.003)=0.0135 W=13.5 mW\frac{dU_C}{dt} = (4.5)(0.003) = \boxed{0.0135 \text{ W} = 13.5 \text{ mW}}

9Problem 9hard

❓ Question:

In an RC circuit with R = 1.5 kΩ, C = 20 μF, and ε = 9.0 V, the switch is closed at t = 0. Derive and evaluate: (a) the differential equation for Q(t), (b) the time when the voltage across the capacitor equals the voltage across the resistor, and (c) the rate of energy storage in the capacitor at this time.

💡 Show Solution

Given:

  • R = 1.5 kΩ = 1500 Ω
  • C = 20 μF = 2.0 × 10⁻⁵ F
  • ε = 9.0 V

(a) Differential equation:

Kirchhoff's voltage law: ε−VR−VC=0\varepsilon - V_R - V_C = 0 ε−IR−QC=0\varepsilon - IR - \frac{Q}{C} = 0

Since I=dQ/dtI = dQ/dt: ε−RdQdt−QC=0\varepsilon - R\frac{dQ}{dt} - \frac{Q}{C} = 0

Rearranging: dQdt+QRC=εR\boxed{\frac{dQ}{dt} + \frac{Q}{RC} = \frac{\varepsilon}{R}}

This is first-order linear ODE with solution: Q(t)=Cε(1−e−t/RC)Q(t) = C\varepsilon(1 - e^{-t/RC})

(b) Time when V_C = V_R:

VC=QC=ε(1−e−t/τ)V_C = \frac{Q}{C} = \varepsilon(1 - e^{-t/\tau})

VR=IR=εe−t/τV_R = IR = \varepsilon e^{-t/\tau}

Set equal: ε(1−e−t/τ)=εe−t/τ\varepsilon(1 - e^{-t/\tau}) = \varepsilon e^{-t/\tau} 1−e−t/τ=e−t/τ1 - e^{-t/\tau} = e^{-t/\tau} 1=2e−t/τ1 = 2e^{-t/\tau} e−t/τ=0.5e^{-t/\tau} = 0.5 t=τln⁡(2)t = \tau \ln(2)

where τ=RC=(1500)(2.0×10−5)=0.03\tau = RC = (1500)(2.0 \times 10^{-5}) = 0.03 s

t=(0.03)ln⁡(2)=0.0208 s=20.8 mst = (0.03)\ln(2) = \boxed{0.0208 \text{ s} = 20.8 \text{ ms}}

(c) Rate of energy storage:

Energy in capacitor: UC=Q22CU_C = \frac{Q^2}{2C}

dUCdt=QC⋅dQdt=VC⋅I\frac{dU_C}{dt} = \frac{Q}{C} \cdot \frac{dQ}{dt} = V_C \cdot I

At t = 20.8 ms: VC=4.5V_C = 4.5 V, I=VR/R=4.5/1500=0.003I = V_R/R = 4.5/1500 = 0.003 A

dUCdt=(4.5)(0.003)=0.0135 W=13.5 mW\frac{dU_C}{dt} = (4.5)(0.003) = \boxed{0.0135 \text{ W} = 13.5 \text{ mW}}

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