Electric Circuits

Ohm's Law and Power

$V = IR, \quad P = IV = I^2R = V^2/R$

Kirchhoff's Rules

1. Junction Rule: $\sum I_{\text{in}} = \sum I_{\text{out}}$
2. Loop Rule: $\sum \Delta V = 0$ around any closed loop

RC Circuits — Charging

Applying the loop rule: $\mathcal{E} - IR - V_C = 0$, where $V_C = Q/C$ and $I = dQ/dt$: $\mathcal{E} = R\frac{dQ}{dt} + \frac{Q}{C}$

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

RC Circuits — Discharging

$Q(t) = Q_0 e^{-t/RC}$ $I(t) = -\frac{Q_0}{RC}e^{-t/RC}$ $V_C(t) = V_0 e^{-t/RC}$

Time Constant

$\tau = RC$

• At $t = \tau$: charge reaches $\approx 63\%$ of max (charging) or drops to $\approx 37\%$ (discharging).
• At $t = 5\tau$: effectively complete ($>99\%$).

Series and Parallel Resistors

• Series: $R_{\text{eq}} = \sum R_i$
• Parallel: $1/R_{\text{eq}} = \sum 1/R_i$

EMF and Internal Resistance

$V_{\text{terminal}} = \mathcal{E} - Ir$

Problem: Derive the charging equation for an RC circuit from the differential equation.

Solution:

Loop rule: $\mathcal{E} - R\frac{dQ}{dt} - Q/C = 0$

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

Let $u = Q - C\mathcal{E}$, so $du = dQ$ and the equation becomes: $\frac{du}{dt} = -\frac{u}{RC}$

Solution: $u = u_0 e^{-t/RC}$. With $Q(0) = 0$: $u_0 = -C\mathcal{E}$.

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

• RC charging: $Q = C\mathcal{E}(1-e^{-t/RC})$, derived from the loop-rule differential equation.
• RC discharging: $Q = Q_0 e^{-t/RC}$.
• Time constant $\tau = RC$ governs the exponential behavior.
• Kirchhoff's rules + calculus = complete circuit analysis.