The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Direct-Current Circuits – Lecture 8 (University Physics II)

Overview

Direct-current circuits combine the two components studied so far — resistors and capacitors — driven by a source of electromotive force (a battery). This lecture assembles the tools for analyzing any such circuit: how resistors and capacitors combine in series and in parallel, Kirchhoff's junction and loop rules for networks too tangled to reduce by inspection, the meaning of a battery's emf and its internal resistance, electrical power, and the time-dependent behavior of a resistor–capacitor (RC) circuit. A recurring theme is that resistors and capacitors follow parallel logic, with current playing the role for resistors that charge plays for capacitors.

Key Concepts

Ohm's law. The voltage across a resistor is $V = IR$ . This holds for ordinary conductors (metals near room temperature); it fails for materials like semiconductors, whose resistance changes strongly with conditions.
Capacitor voltage. Capacitance is charge per unit voltage, $C = Q/V$ , so the voltage across a capacitor is $V = Q/C$ . Capacitance is a fixed property of the device itself.
Series connection. Elements lie along a single conducting path, so the same current flows through each resistor (and the same charge sits on each capacitor). Equivalent values:

\begin{aligned} R_\text{eq} &= R_1 + R_2 + \cdots \\ \frac{1}{C_\text{eq}} &= \frac{1}{C_1} + \frac{1}{C_2} + \cdots \end{aligned}

Parallel connection. Elements share the same two nodes, so the same voltage appears across each. Equivalent values:

\begin{aligned} \frac{1}{R_\text{eq}} &= \frac{1}{R_1} + \frac{1}{R_2} + \cdots \\ C_\text{eq} &= C_1 + C_2 + \cdots \end{aligned}

Intuition check. Resistors in series impede flow more, so $R_\text{eq}$ is larger than any one of them; resistors in parallel offer extra paths, so $R_\text{eq}$ is smaller than the smallest. Capacitors behave oppositely (series smaller, parallel larger).
Kirchhoff's junction rule. At any junction the currents in equal the currents out — a statement of conservation of charge: $\sum I = 0$ (taking inflows positive and outflows negative).
Kirchhoff's loop rule. Around any closed loop the potential changes sum to zero, $\sum \Delta V = 0$ — a consequence of the electric force being conservative. Bookkeeping: crossing a resistor with the current drops the potential by $IR$ ; crossing it against the current raises it; crossing a battery from $-$ to $+$ adds its emf, and the reverse subtracts it.
Emf and internal resistance. A battery's electromotive force $\mathcal{E}$ is not a force but an energy supplied per unit charge (a voltage). A real battery has an internal resistance $r$ , so its terminal voltage is $V = \mathcal{E} - Ir$ when it drives current. If a stronger source pushes current backward through it, the battery is being charged.
Electrical power. The power delivered to (or by) an element is $P = IV$ , and using Ohm's law $P = I^2 R = V^2/R$ . Power dissipated in a resistor leaves as heat.
RC circuits. With a resistor and capacitor in series, the loop rule gives $\mathcal{E} = \dfrac{Q}{C} + IR$ . The product $\tau = RC$ (the time constant) has units of time and sets how fast the capacitor charges or discharges.
Charging (capacitor starts empty). The charge climbs toward its final value $Q_f = \mathcal{E}C$ while the current decays from its initial value $I_0 = \mathcal{E}/R$ :

\begin{aligned} Q(t) &= Q_f\left(1 - e^{-t/RC}\right) \\ I(t) &= I_0\, e^{-t/RC} \end{aligned}

Discharging (capacitor starts with charge $Q_0$ , battery removed). Both quantities decay with the same exponential form as charging; here the initial current is negative, $I_0 = -Q_0/RC$ , since the charge now flows off the capacitor (opposite to the charging direction):

\begin{aligned} Q(t) &= Q_0\, e^{-t/RC} \\ I(t) &= I_0\, e^{-t/RC} \end{aligned}

Worked Example

An 18 V battery (negligible internal resistance) drives a 4 Ω resistor in series with a parallel pair of 6 Ω and 3 Ω resistors. Find the equivalent resistance and the current in each resistor.

First collapse the parallel pair, using $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$ :

\begin{aligned} R_p &= \frac{R_1 R_2}{R_1 + R_2} \\ &= \frac{(6\,\Omega)(3\,\Omega)}{6\,\Omega + 3\,\Omega} = 2\,\Omega \end{aligned}

That 2 Ω is in series with the 4 Ω resistor, and the total current follows from Ohm's law:

\begin{aligned} R_\text{eq} &= 4\,\Omega + R_p = 4\,\Omega + 2\,\Omega = 6\,\Omega \\ I &= \frac{\mathcal{E}}{R_\text{eq}} = \frac{18\,\text{V}}{6\,\Omega} = 3\,\text{A} \end{aligned}

All 3 A pass through the 4 Ω resistor. The parallel section drops $V_p = I R_p = (3\,\text{A})(2\,\Omega) = 6\,\text{V}$ , and that same 6 V sits across both parallel resistors:

\begin{aligned} I_{6\,\Omega} &= \frac{V_p}{6\,\Omega} = \frac{6\,\text{V}}{6\,\Omega} = 1\,\text{A} \\ I_{3\,\Omega} &= \frac{V_p}{3\,\Omega} = \frac{6\,\text{V}}{3\,\Omega} = 2\,\text{A} \end{aligned}

The two branch currents sum to $1\,\text{A} + 2\,\text{A} = 3\,\text{A}$ , matching the current into the junction — a check on the junction rule, and confirmation that the smaller resistor carries the larger share of the current.

Summary

The voltage across a resistor is $V = IR$ ; across a capacitor it is $V = Q/C$ .
In series, current (resistors) or charge (capacitors) is shared and $R_\text{eq} = \sum R$ , $\tfrac{1}{C_\text{eq}} = \sum \tfrac{1}{C}$ ; in parallel, voltage is shared and $\tfrac{1}{R_\text{eq}} = \sum \tfrac{1}{R}$ , $C_\text{eq} = \sum C$ .
Series resistance exceeds any single resistor; parallel resistance is less than the smallest.
Kirchhoff's junction rule (conservation of charge) and loop rule ( $\sum \Delta V = 0$ ) analyze circuits that don't reduce by simple series/parallel steps.
A battery's emf $\mathcal{E}$ is an energy per unit charge; internal resistance $r$ lowers the terminal voltage to $\mathcal{E} - Ir$ , and power is $P = IV = I^2R = V^2/R$ .
In an RC circuit both $Q(t)$ and $I(t)$ vary exponentially with time constant $\tau = RC$ : charging builds toward $Q_f = \mathcal{E}C$ from an initial current $I_0 = \mathcal{E}/R$ , and discharging decays from $Q_0$ .