Direct-Current Circuits: Electric Current, Resistance, Power, and Kirchhoff's Rules

Nature and Fundamentals of Electric Current

• Electric currents consist of electric charges in motion; specifically, current requires a net motion of charge carriers in a definite direction.

• For steady currents established in a conducting loop, if an amount of charge $\Delta q$ passes through a hypothetical plane (such as $aa'$) during a time interval $\Delta t$, the electric current $i$ across that plane is defined as: $i = \frac{\Delta q}{\Delta t}$

• In the infinitesimal differential limit, instantaneous current is given by: $i = \frac{dq}{dt}$

• The SI unit of electric current is the ampere ($\text{A}$): $1\,\text{A} = 1\,\text{C/s}$

• One ampere corresponds to one coulomb of charge passing across the cross-sectional plane per second ($\text{Coulomb}/\text{second} = \text{Ampere}$).

• Current is a scalar physical quantity. Although arrows are drawn along conducting paths to signify current, these directional markers represent the flow along the path rather than spatial vector quantities.

• Continuity of current across varying cross sections:

  • In a steady-state conductor, the current $i$ is identical across every cross section, regardless of variations in cross-sectional area or planar orientation.

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  • Whether evaluated through a narrow transverse cross section $aa'$, a wide cross section $bb'$, or an oblique cut $cc'$, the net charge passing per unit time remains constant.

• Direction of current and charge carrier sign conventions:

  • Current arrows are drawn by convention in the direction in which positive charge carriers would move.

  • In typical metallic conductors, physical charge carriers are negatively charged conduction electrons moving in the direction opposite to the conventional current arrow.

  • Bending, twisting, or coiling a wire does not alter the magnitude or continuity of the current traversing it.

• Conservation of charge at junctions:

  • Because electric charge is strictly conserved and charge cannot accumulate indefinitely at a junction, the magnitudes of branch currents must balance the incoming current.

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  • For a conductor carrying current $i_0$ that divides at junction $a$ into branches carrying $i_1$ and $i_2$: $i_0 = i_1 + i_2$

Resistance, Resistivity, and Ohm's Law

• Definition of electrical resistance:

  • The resistance $R$ between two points of a conductor is determined by applying an electrical potential difference $V$ across those points and measuring the resulting current $i$: $R = \frac{V}{i}$

  • The SI unit of electrical resistance is the ohm ($\Omega$): $1\,\Omega = 1\,\text{V/A}$

  • A circuit element whose specific functional purpose is to introduce a defined resistance is called a resistor.

• Distinction between resistance and resistivity:

  • Resistance $R$ is an extrinsic property of a specific object, determined by both its geometry and its composition.

  • Resistivity $\rho$ is an intrinsic macroscopic property of a material, independent of the dimensions of the sample.

  • For a uniform conductor of length $L$ and cross-sectional area $A$, resistance and resistivity are related by: $R = \rho \frac{L}{A}$ $\rho = \frac{R A}{L}$

  • The SI unit of resistivity is the ohm-meter ($\Omega \cdot \text{m}$).

• Temperature dependence of resistivity:

  • The resistivity of conducting materials varies approximately linearly with temperature over moderate ranges according to the relation: $\rho = \rho_0 [1 + \alpha(T - T_0)]$

  • The parameter $\alpha$ is the temperature coefficient of resistivity of the material.

  • The reference temperature $T_0$ is standardized at room temperature: $T_0 = 293\,\text{K}$

  • $\rho_0$ represents the measured resistivity at reference temperature $T_0$.

• Resistivities of representative substances at $293\,\text{K}$:

  • Insulators:

  • Quartz (fused): $7.5 \times 10^{17}\,\Omega \cdot \text{m}$

  • Rubber: $1 \times 10^{13} \text{ to } 100 \times 10^{13}\,\Omega \cdot \text{m}$

  • Glass: $1 \times 10^{9} \text{ to } 10{,}000 \times 10^{9}\,\Omega \cdot \text{m}$

  • Semiconductors:

  • Silicon (pure): $0.10 \text{ to } 60\,\Omega \cdot \text{m}$

  • Germanium: $0.001 \text{ to } 0.5\,\Omega \cdot \text{m}$

  • Conductors:

  • Lead: $22 \times 10^{-8}\,\Omega \cdot \text{m}$

  • Iron: $9.71 \times 10^{-8}\,\Omega \cdot \text{m}$

  • Tungsten: $5.6 \times 10^{-8}\,\Omega \cdot \text{m}$

  • Aluminum: $2.65 \times 10^{-8}\,\Omega \cdot \text{m}$

  • Gold: $2.20 \times 10^{-8}\,\Omega \cdot \text{m}$

  • Copper: $1.68 \times 10^{-8}\,\Omega \cdot \text{m}$

  • Copper (annealed): $1.72 \times 10^{-8}\,\Omega \cdot \text{m}$

  • Silver: $1.59 \times 10^{-8}\,\Omega \cdot \text{m}$

• Ohm's Law and ohmic behavior:

  • Ohm's law asserts that the current $i$ traversing a conducting device is directly proportional to the applied potential difference $V$: $i \propto V \implies V = iR$

  • A conducting device obeys Ohm's law if its resistance $R$ remains completely independent of the magnitude and polarity of the applied potential difference $V$.

  • A conducting material obeys Ohm's law if its resistivity $\rho$ is independent of the magnitude and direction of the applied electric field $\mathbf{E}$.

  • Experimental test procedure:

  • Apply potential difference $V$ across a component while varying magnitude and polarity, measuring current $i$.

  • Polarity convention: $V$ is defined positive when the left terminal is at a higher potential than the right terminal; the resulting left-to-right current is assigned a positive sign.

  • Reverse polarity (right terminal at higher potential) is assigned a negative potential, producing current with a negative sign.

• Worked example — Ranking conductors by current:

  • Three cylindrical copper conductors are subjected to the same potential difference $V$ across their lengths:

  • Conductor (a): length $L$, face area $A$

  • Conductor (b): length $1.5L$, face area $\frac{A}{2}$

  • Conductor (c): length $\frac{L}{2}$, face area $\frac{A}{2}$

  • Expressing resistance for each conductor: $R_a = \rho \frac{L}{A}$ $R_b = \rho \frac{1.5L}{A/2} = 3\rho \frac{L}{A} = 3 R_a$ $R_c = \rho \frac{L/2}{A/2} = \rho \frac{L}{A} = R_a$

  • Using $i = \frac{V}{R}$: $i_a = \frac{V}{R_a}$ $i_b = \frac{V}{3 R_a} = \frac{1}{3} i_a$ $i_c = \frac{V}{R_a} = i_a$

  • Ranking of currents from greatest to least: $i_a = i_c > i_b$

Electrical Power and Resistive Dissipation

• Energy transfer mechanisms in electric circuits:

  • An external conducting path connects the terminals of an emf device, establishing a steady current $i$.

  • In a time interval $\Delta t$, the quantity of charge $\Delta q$ transported through the circuit is: $\Delta q = i \Delta t$

  • As charge $\Delta q$ moves through an external potential decrease $V$, its electric potential energy decreases by magnitude $\Delta U$: $\Delta U = \Delta q V = i V \Delta t$

• Formulation of electric power:

  • The rate of electric energy transfer, defined as electric power $P$, is: $P = \frac{\Delta U}{\Delta t} = i V$

  • For a purely resistive device of resistance $R$, applying Ohm's law ($V = iR$ and $i = \frac{V}{R}$) gives the rate of resistive dissipation (Joule heating): $P = i^2 R$ $P = \frac{V^2}{R}$

• Worked example — Energy dissipation in Nichrome heating wire:

  • A uniform Nichrome (nickel-chromium-iron alloy) heating wire has resistance $R = 72\,\Omega$.

  • Situation 1: A potential difference of $120\,\text{V}$ is applied across the full length: $P_1 = \frac{V^2}{R} = \frac{(120\,\text{V})^2}{72\,\Omega} = \frac{14{,}400\,\text{V}^2}{72\,\Omega} = 200\,\text{W}$

  • Situation 2: The wire is cut in half, and $120\,\text{V}$ is applied across each half:

  • Because resistance scales linearly with length, each half has resistance: $R_{\text{half}} = \frac{R}{2} = \frac{72\,\Omega}{2} = 36\,\Omega$

  • Energy dissipation rate across each half: $P_{\text{half}} = \frac{V^2}{R_{\text{half}}} = \frac{(120\,\text{V})^2}{36\,\Omega} = \frac{14{,}400\,\text{V}^2}{36\,\Omega} = 400\,\text{W}$

  • Combined rate of energy dissipation for both halves: $P_{\text{total}} = 400\,\text{W} + 400\,\text{W} = 800\,\text{W}$

• Worked example — Charge transport calculation:

  • An electrical device rated at $P = 18.0\,\text{W}$ has $V = 9.00\,\text{V}$ maintained across it. The operating duration is $\Delta t = 4.00\,\text{h}$.

  • Current through the device: $i = \frac{P}{V} = \frac{18.0\,\text{W}}{9.00\,\text{V}} = 2.00\,\text{A}$

  • Time interval in seconds: $\Delta t = 4.00\,\text{h} \times 3600\,\text{s/h} = 14{,}400\,\text{s}$

  • Total charge passing through the device: $\Delta q = i \Delta t = 2.00\,\text{A} \times 14{,}400\,\text{s} = 28{,}800\,\text{C} = 2.88 \times 10^4\,\text{C}$

Electromotive Force and Real Circuit Models

• Fundamental principles of electromotive force (emf):

  • Maintaining a steady current through a resistive conductor requires maintaining an electric potential difference between its terminals.

  • To achieve a continuous current, a "charge pump" must do internal work on charge carriers to transport them against the internal electrostatic field from low potential to high potential.

  • Devices that supply emf include chemical batteries, electric generators, photovoltaic solar cells, and fuel cells.

  • Diagrammatic symbol: The emf $\mathcal{E}$ is indicated by an arrow pointing from the negative terminal to the positive terminal, marked with a small open circle at its tail to distinguish it from current arrows.

• Potential changes in real batteries with internal resistance:

  • An ideal battery maintains an internal potential difference equal to rated emf $\mathcal{E}$ under all operating loads.

  • A real battery possesses internal resistance $r$ arising from internal materials and chemical pathways.

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  • Tracking electric potential through a closed circuit containing a real battery and an external resistor $R$:

  • Reference potential at negative terminal $a$ is $V_a$.

  • Moving across the internal emf source elevates potential by $+\mathcal{E}$.

  • Traversing internal resistance $r$ in the direction of current drops potential by $-ir$.

  • Terminal voltage $V_b - V_a$ across the external battery terminals is: $V_b - V_a = \mathcal{E} - ir$

  • Traversing external resistor $R$ drops potential by $-iR$, returning potential to $V_a$.

  • Summing potential changes around the complete loop: $\mathcal{E} - ir - iR = 0$

  • Solving for circuit current $i$: $i = \frac{\mathcal{E}}{R + r}$

Resistor Combinations and Network Reductions

• Resistors in series:

  • In a series connection, all components are connected end-to-end such that the current through every resistor is identical: $i_1 = i_2 = \dots = i_n = i$

  • The total potential difference across the series bank is the sum of the individual potential drops: $V = V_1 + V_2 + \dots + V_n = i R_1 + i R_2 + \dots + i R_n$

  • The equivalent resistance $R_{\text{eq}}$ is: $R_{\text{eq}} = \sum_{j=1}^n R_j$

  • The equivalent resistance of a series connection is strictly greater than any individual resistance in the combination.

• Resistors in parallel:

  • In a parallel connection, components share common terminal nodes, making the potential difference across every resistor identical: $V_1 = V_2 = \dots = V_n = V$

  • Total current entering the parallel bank splits among the individual branches: $i = i_1 + i_2 + \dots + i_n = \frac{V}{R_1} + \frac{V}{R_2} + \dots + \frac{V}{R_n}$

  • The equivalent resistance $R_{\text{eq}}$ satisfies: $\frac{1}{R_{\text{eq}}} = \sum_{j=1}^n \frac{1}{R_j}$

  • For two resistors in parallel: $R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}$

  • The equivalent resistance of a parallel connection is strictly less than the smallest individual resistance in the combination.

• Lightbulb operating characteristics in series vs. parallel:

  • Three lightbulbs rated at $120\,\text{V}$ with rated powers $60\,\text{W}$, $75\,\text{W}$, and $200\,\text{W}$:

  • Internal filament resistance calculated via $R = \frac{V^2}{P_{\text{rated}}}$: $R_{60} = \frac{(120\,\text{V})^2}{60\,\text{W}} = 240\,\Omega$ $R_{75} = \frac{(120\,\text{V})^2}{75\,\text{W}} = 192\,\Omega$ $R_{200} = \frac{(120\,\text{V})^2}{200\,\text{W}} = 72\,\Omega$

  • The $60\,\text{W}$ bulb has the greatest filament resistance ($240\,\Omega$).

  • Series connection behavior: The current $i$ is identical through all three bulbs. Power dissipation follows $P = i^2 R$. Because power dissipation scales directly with resistance, the $60\,\text{W}$ bulb dissipates the most power and glows brightest, followed by the $75\,\text{W}$ bulb, while the $200\,\text{W}$ bulb glows dimmest ($P_{60} > P_{75} > P_{200}$).

  • Parallel connection behavior: Each bulb experiences the full line voltage of $120\,\text{V}$, dissipating power according to $P = \frac{V^2}{R}$. The $200\,\text{W}$ bulb dissipates the most power and glows brightest ($P_{200} > P_{75} > P_{60}$).

  • Three lightbulbs rated at $100\,\text{V}$ with rated powers $25\,\text{W}$, $75\,\text{W}$, and $150\,\text{W}$ connected in parallel:

  • The bulbs experience the same potential difference ($V \approx 100\,\text{V}$).

  • Current drawn by each bulb follows $i = \frac{P}{V}$: $i_{25} = \frac{25\,\text{W}}{100\,\text{V}} = 0.25\,\text{A}$ $i_{75} = \frac{75\,\text{W}}{100\,\text{V}} = 0.75\,\text{A}$ $i_{150} = \frac{150\,\text{W}}{100\,\text{V}} = 1.50\,\text{A}$

  • The $150\,\text{W}$ bulb draws the most current.

  • Filament resistances follow $R = \frac{V^2}{P}$: $R_{25} = \frac{(100\,\text{V})^2}{25\,\text{W}} = 400\,\Omega$ $R_{75} = \frac{(100\,\text{V})^2}{75\,\text{W}} \approx 133.3\,\Omega$ $R_{150} = \frac{(100\,\text{V})^2}{150\,\text{W}} \approx 66.7\,\Omega$

  • The $150\,\text{W}$ bulb has the least resistance.

• Practical design example — Three-way incandescent lightbulb:

  • A three-way lightbulb contains two independent filaments rated at $75\,\text{W}$ and $100\,\text{W}$ under $120\,\text{V}$, controlled by switches $S_1$ and $S_2$.

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  • Setting 1 (Low): Switch $S_1$ open, switch $S_2$ closed; current energizes only the $75\,\text{W}$ filament ($P = 75\,\text{W}$).

  • Setting 2 (Medium): Switch $S_1$ closed, switch $S_2$ open; current energizes only the $100\,\text{W}$ filament ($P = 100\,\text{W}$).

  • Setting 3 (High): Both switches $S_1$ and $S_2$ closed; both filaments operate in parallel, yielding combined dissipation: $P = 75\,\text{W} + 100\,\text{W} = 175\,\text{W}$

• Complex network reduction techniques:

  • Example 1 — Symmetric Bridge Network:

  • Four $1\,\Omega$ resistors form a diamond bridge between nodes $a$ and $b$, with a central bridge resistor of $5\,\Omega$ between intermediate nodes $c$ and $d$.

  • By bridge symmetry, the potential at node $c$ equals the potential at node $d$ ($V_c = V_d$).

  • Because $\Delta V_{cd} = 0$, zero current passes through the central $5\,\Omega$ resistor ($i_{cd} = 0$).

  • Removing the central resistor reduces the network to two identical parallel branches of two series resistors: $R_{\text{top}} = 1\,\Omega + 1\,\Omega = 2\,\Omega$ $R_{\text{bottom}} = 1\,\Omega + 1\,\Omega = 2\,\Omega$

  • Total equivalent resistance between $a$ and $b$: $R_{ab} = \frac{2\,\Omega \times 2\,\Omega}{2\,\Omega + 2\,\Omega} = 1\,\Omega$

  • Example 2 — Sequential Ladder Network Reduction:

  • A ten-resistor network between terminals $A$ and $B$ is evaluated systematically from the furthest branch:

    • Series reduction: $R_8 = 10\,\Omega$ in series with $R_{10} = 2\,\Omega$ yields $R_{8,10} = 12\,\Omega$.

    • Parallel reduction: $R_{8,10} = 12\,\Omega$ in parallel with $R_9 = 6\,\Omega$ yields $R_A = \frac{12 \times 6}{12 + 6} = 4\,\Omega$.

    • Series reduction: $R_A = 4\,\Omega$ in series with $R_7 = 8\,\Omega$ yields $R_{A,7} = 12\,\Omega$.

    • Parallel reduction: $R_{A,7} = 12\,\Omega$ in parallel with $R_6 = 6\,\Omega$ yields $R_B = \frac{12 \times 6}{12 + 6} = 4\,\Omega$.

    • Series reduction: $R_B = 4\,\Omega$ in series with $R_5 = 4\,\Omega$ yields $R_{B,5} = 8\,\Omega$.

    • Parallel reduction: $R_{B,5} = 8\,\Omega$ in parallel with $R_4 = 8\,\Omega$ yields $R_C = \frac{8 \times 8}{8 + 8} = 4\,\Omega$.

    • Series reduction: $R_C = 4\,\Omega$ in series with $R_3 = 4\,\Omega$ yields $R_{C,3} = 8\,\Omega$.

    • Parallel reduction: $R_{C,3} = 8\,\Omega$ in parallel with $R_2 = 8\,\Omega$ yields $R_D = \frac{8 \times 8}{8 + 8} = 4\,\Omega$.

    • Final series reduction: $R_1 = 6\,\Omega$ in series with $R_D = 4\,\Omega$ yields total equivalent resistance: $R_{\text{EQ}} = 6\,\Omega + 4\,\Omega = 10\,\Omega$

Kirchhoff's Rules and Multiloop Circuit Analysis

• Kirchhoff's junction rule:

  • The algebraic sum of all currents meeting at any junction in a circuit equals zero: $\sum I = 0$

  • Equivalently, the sum of currents entering a junction equals the sum leaving: $\sum I_{\text{in}} = \sum I_{\text{out}}$

  • Physical origin: Derived from the conservation of electric charge.

• Kirchhoff's loop rule:

  • The algebraic sum of all potential differences around any closed circuit loop equals zero: $\sum_{\text{closed loop}} \Delta V = 0$

  • Physical origin: Derived from the conservation of energy in an electrostatic field.

• Sign conventions for loop traversal:

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  • Traversing a resistor in the assigned direction of current: $\Delta V = -IR$

  • Traversing a resistor opposite to the assigned direction of current: $\Delta V = +IR$

  • Traversing an ideal source of emf in the direction of the emf (from $-$ to $+$): $\Delta V = +\mathcal{E}$

  • Traversing an ideal source of emf opposite to the direction of the emf (from $+$ to $-$): $\Delta V = -\mathcal{E}$

• General problem-solving methodology:

  • Draw a circuit diagram and label all known emf values, resistances, and junctions.

  • Assign an arbitrary direction to the current in each distinct circuit branch.

  • If a calculated branch current is negative, its magnitude is correct and its actual physical direction is opposite to the assumed direction.

  • Write $N - 1$ independent junction equations for a circuit with $N$ principal junctions.

  • Apply the loop rule to independent closed loops until the number of independent equations equals the number of unknown branch currents.

  • Solve the system of simultaneous linear equations for the unknown currents.

• Worked example 1 — Two-loop multiloop circuit with two sources:

  • A circuit has three branches meeting at junction $A$: a left branch containing a $15\,\text{V}$ battery and a $100.0\,\Omega$ resistor; a center branch containing a $100.0\,\Omega$ resistor; and a right branch containing a $9.0\,\text{V}$ battery opposing current $I_2$ and a $100.0\,\Omega$ resistor.

  • Junction rule at node $A$: $I_1 - I_2 - I_3 = 0$

  • Loop 1 equation (clockwise traversal of left loop, letting $R = 100.0\,\Omega$): $15\,\text{V} - I_3 R - I_1 R = 0$

  • Loop 2 equation (clockwise traversal of right loop, letting $R = 100.0\,\Omega$): $-9.0\,\text{V} - I_2 R + I_3 R = 0$

  • Solving simultaneously yields: $I_1 = 0.070\,\text{A}$ $I_2 = -0.010\,\text{A}$ $I_3 = 0.080\,\text{A}$

  • Interpretation: Current $I_2$ has an actual magnitude of $0.010\,\text{A}$ directed opposite to the assumed arrow direction.

• Worked example 2 — Single-loop circuit with opposing emf devices:

  • A single loop consists of battery $\mathcal{E}_1 = 6.0\,\text{V}$, resistor $R_1 = 8.0\,\Omega$, battery $\mathcal{E}_2 = 12\,\text{V}$ (opposing $\mathcal{E}_1$), and resistor $R_2 = 10\,\Omega$.

  • Assuming clockwise current $I$ and traversing clockwise from node $a$: $6.0\,\text{V} - I R_1 - 12\,\text{V} - I R_2 = 0$ $6.0\,\text{V} - 8.0 I - 12\,\text{V} - 10 I = 0$ $-6.0 - 18 I = 0 \implies I = -0.33\,\text{A}$

  • Power delivered to each resistor: $P_{R_1} = I^2 R_1 = (-0.333\,\text{A})^2 \times 8.0\,\Omega = 0.89\,\text{W}$ $P_{R_2} = I^2 R_2 = (-0.333\,\text{A})^2 \times 10\,\Omega = 1.1\,\text{W}$

  • Power delivered by the net driving $12\,\text{V}$ battery: $P = I \mathcal{E}_2 = (0.333\,\text{A}) \times 12\,\text{V} = 4.0\,\text{W}$

• Worked example 3 — Reduction and branch currents in a multiloop circuit:

  • An ideal battery of $\mathcal{E} = 12\,\text{V}$ powers resistors $R_1 = 20\,\Omega$, $R_2 = 20\,\Omega$, $R_3 = 30\,\Omega$, and $R_4 = 8.0\,\Omega$, where $R_2$ and $R_3$ are connected in parallel.

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  • Equivalent resistance of parallel pair $R_{23}$: $R_{23} = \frac{R_2 R_3}{R_2 + R_3} = \frac{20\,\Omega \times 30\,\Omega}{20\,\Omega + 30\,\Omega} = \frac{600}{50} = 12\,\Omega$

  • Total equivalent resistance of circuit: $R_{\text{eq}} = R_1 + R_{23} + R_4 = 20\,\Omega + 12\,\Omega + 8.0\,\Omega = 40\,\Omega$

  • Current through the battery ($i_1$): $i_1 = \frac{\mathcal{E}}{R_{\text{eq}}} = \frac{12\,\text{V}}{40\,\Omega} = 0.30\,\text{A}$

  • Potential difference across parallel bank $R_{23}$: $V_{23} = i_1 R_{23} = 0.30\,\text{A} \times 12\,\Omega = 3.6\,\text{V}$

  • Branch current $i_2$ through $R_2$: $i_2 = \frac{V_{23}}{R_2} = \frac{3.6\,\text{V}}{20\,\Omega} = 0.18\,\text{A}$

  • Branch current $i_3$ through $R_3$: $i_3 = \frac{V_{23}}{R_3} = \frac{3.6\,\text{V}}{30\,\Omega} = 0.12\,\text{A}$

  • Verification via junction rule: $i_1 = i_2 + i_3 = 0.18\,\text{A} + 0.12\,\text{A} = 0.30\,\text{A}$

Capacitors in Circuits and Duality with Resistors

• Capacitors in parallel combinations:

  • When capacitors are connected in parallel, the potential difference across every capacitor is identical: $V_1 = V_2 = \dots = V$

  • The total stored charge $Q$ equals the sum of the charges on each capacitor: $Q = Q_1 + Q_2 + Q_3 + \dots = C_1 V + C_2 V + C_3 V + \dots$

  • The equivalent capacitance $C_{\text{eq}}$ is: $C_{\text{eq}} = \sum_{j=1}^n C_j$

  • The SI unit of capacitance is the farad ($\text{F}$): $1\,\text{F} = 1\,\text{C/V}$

• Capacitors in series combinations:

  • When capacitors are connected in series, the magnitude of charge on each capacitor is identical: $Q_1 = Q_2 = \dots = Q$

  • The total applied potential difference is the sum of the potential drops across each capacitor: $V = V_1 + V_2 + V_3 + \dots = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} + \dots$

  • The equivalent capacitance $C_{\text{eq}}$ satisfies: $\frac{1}{C_{\text{eq}}} = \sum_{j=1}^n \frac{1}{C_j}$

• Comparative mathematical summary of resistors versus capacitors:

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  • Resistors in series:

  • Equivalent resistance: $R_{\text{eq}} = \sum_{j=1}^n R_j$

  • Governing invariant: Identical current passes through all resistors.

  • Resistors in parallel:

  • Equivalent resistance: $\frac{1}{R_{\text{eq}}} = \sum_{j=1}^n \frac{1}{R_j}$

  • Governing invariant: Identical potential difference appears across all resistors.

  • Capacitors in series:

  • Equivalent capacitance: $\frac{1}{C_{\text{eq}}} = \sum_{j=1}^n \frac{1}{C_j}$

  • Governing invariant: Identical charge resides on all capacitors.

  • Capacitors in parallel:

  • Equivalent capacitance: $C_{\text{eq}} = \sum_{j=1}^n C_j$

  • Governing invariant: Identical potential difference appears across all capacitors.