Electric Circuits & Capacitors

Capacitance

$C = \frac{Q}{V}$ Unit: farad (F). $1\ \text{F} = 1\ \text{C/V}$.

For a parallel-plate capacitor: $C = \frac{\epsilon_0 A}{d}$ where $A$ is plate area, $d$ is separation, and $\epsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$.

Capacitors in Parallel

$C_{\text{eq}} = C_1 + C_2 + C_3 + \cdots$ Same voltage across each; charges add.

Capacitors in Series

$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \cdots$ Same charge on each; voltages add.

Energy Stored in a Capacitor

$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$

Dielectrics

Inserting a dielectric (insulating material) increases capacitance: $C = \kappa C_0$ where $\kappa$ is the dielectric constant ($\kappa > 1$).

RC Circuits

Charging: $V_C(t) = \mathcal{E}(1 - e^{-t/RC})$ $I(t) = \frac{\mathcal{E}}{R}e^{-t/RC}$

Discharging: $V_C(t) = V_0 e^{-t/RC}$ $I(t) = -\frac{V_0}{R}e^{-t/RC}$

Time constant: $\tau = RC$. After $5\tau$, the capacitor is essentially fully charged/discharged.

Problem: A $10\ \mu\text{F}$ capacitor is charged to $100\ \text{V}$. How much energy is stored? If a $5\ \mu\text{F}$ capacitor is placed in series with it, what is $C_{\text{eq}}$?

Solution:

Energy: $U = \frac{1}{2}CV^2 = \frac{1}{2}(10 \times 10^{-6})(100)^2 = 0.05\ \text{J} = 50\ \text{mJ}$

Series: $\frac{1}{C_{\text{eq}}} = \frac{1}{10} + \frac{1}{5} = \frac{3}{10}$, so $C_{\text{eq}} = 10/3 \approx 3.33\ \mu\text{F}$.

• Capacitance: $C = Q/V$; for parallel plates, $C = \epsilon_0 A/d$.
• Series: $1/C_{\text{eq}} = \sum 1/C_i$. Parallel: $C_{\text{eq}} = \sum C_i$.
• Energy stored: $U = \frac{1}{2}CV^2$.
• Dielectrics increase capacitance by factor $\kappa$.
• RC time constant $\tau = RC$ governs charging/discharging rates.