$\mathcal{E} - R\frac{dQ}{dt} - \frac{Q}{C} = 0$

Differential equation: $\frac{dQ}{dt} = \frac{\mathcal{E}}{R} - \frac{Q}{RC}$

Solution: Charging

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

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

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

Voltage across resistor: $V_R(t) = IR = \mathcal{E}e^{-t/RC}$

Time constant: $\tau = RC$

After time $\tau$:

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

Discharging Capacitor

Initial charge $Q_0$ on capacitor, no battery.

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

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

Solution: $Q(t) = Q_0e^{-t/RC}$

Current: $I(t) = -\frac{dQ}{dt} = \frac{Q_0}{RC}e^{-t/RC}$

Voltage: $V_C(t) = V_0e^{-t/RC}$

Energy Considerations

Charging:

Energy supplied by battery: $W_{battery} = Q\mathcal{E} = C\mathcal{E}^2$

Energy stored in capacitor: $U_C = \frac{1}{2}C\mathcal{E}^2$

Energy dissipated in resistor: $U_R = \frac{1}{2}C\mathcal{E}^2$

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

Discharging:

All energy dissipated in resistor: $U_R = \frac{1}{2}CV_0^2$

General RC Circuit

For any RC circuit, differential equation has form:

$RC\frac{dV_C}{dt} + V_C = V_{final}$

General solution: $V_C(t) = V_{final} + (V_{initial} - V_{final})e^{-t/RC}$

Multiple Capacitors

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

Effective time constant: $\tau = RC_{eq}$

Applications

Timer circuits: Delay determined by $RC$

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

Integrators/Differentiators: For signal processing

Defibrillators: Store energy, rapid discharge through heart

$\tau = \boxed{10 \text{ s}}$

(b) Charge at t = 5.0 s:

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

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

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

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

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

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

(c) Current at t = 5.0 s:

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

where $I_0 = \varepsilon/R = 12/(2.0 \times 10^6) = 6.0 \times 10^{-6}$ A

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

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