Current and Resistance
Microscopic and macroscopic physics of electric current, current density, the classical Drude model of electron drift, temperature-dependent resistivity, Electromotive Force, Kirchhoff's circuit rules, multi-loop networks, galvanometers, ammeters, voltmeters, potentiometers, RC charging and discharging transients, and thermoelectric phenomena (Seebeck, Peltier, Thomson).
§4.1 Electric Current, Current Density, and the Microscopic Drude Model
1. Electric Current and Current Density
The instantaneous electric current $I$ is defined as the net charge passing through a surface per unit time: $$I = \frac{dq}{dt}$$ SI Unit: **Ampere (A)** ($1 \text{ A} = 1 \text{ C/s}$). Current is a macroscopic scalar quantity. The microscopic vector characterizing charge flow at every spatial point is the Current Density $\vec{J}$: $$I = \int_S \vec{J} \cdot d\vec{A}$$ For uniform current across a normal cross-sectional area $A$: $$J = \frac{I}{A} \quad (\text{Units: A/m}^2)$$2. Drift Velocity of Charge Carriers
In a conductor with free charge carrier density $n$ (electrons/m³) each carrying charge $q = -e$: In time increment $dt$, carriers advance by length $dx = v_d dt$, where $v_d$ is the average **drift velocity**. The charge traversing cross-section $A$ is $dq = n q (A v_d dt)$. Hence: $$I = n q A v_d = n e A v_d \implies \vec{J} = n q \vec{v}_d = -n e \vec{v}_d$$ Striking Numerical Fact: While electrical signals propagate at near the speed of light ($c \sim 3 \times 10^8\text{ m/s}$), the physical drift velocity of electrons in copper under typical domestic currents is astonishingly slow: $$v_d \sim 10^{-4} \text{ m/s} = 0.1 \text{ mm/s}$$ An electron takes roughly three hours to travel one meter down a copper wire!3. The Classical Drude Model of Electrical Conduction
Paul Drude (1900) modeled conduction electrons as an ideal classical gas undergoing random thermal collisions with positive ionic cores in a crystal lattice. Between collisions, electrons accelerate under electric field $\vec{E}$: $$\vec{a} = \frac{-e \vec{E}}{m}$$ Let $\tau$ be the mean relaxation time (average time between collisions). The average drift velocity acquired is: $$\vec{v}_d = \vec{a} \tau = -\frac{e \tau}{m} \vec{E}$$ Substituting into current density: $$\vec{J} = -n e \vec{v}_d = \left( \frac{n e^2 \tau}{m} \right) \vec{E}$$ Defining electrical conductivity $\sigma$: $$\vec{J} = \sigma \vec{E} \quad (\text{Microscopic Ohm's Law})$$ where: $$\sigma = \frac{n e^2 \tau}{m}, \quad \rho = \frac{1}{\sigma} = \frac{m}{n e^2 \tau}$$ This demonstrates that Ohm's law arises directly from frequent momentum-relaxing collisions.§4.2 Resistance, Resistivity, and Temperature Dependence
1. Macroscopic Ohm's Law and Resistance
Consider a conductor of uniform length $L$ and cross-sectional area $A$ carrying current $I$ under potential difference $V$. Using $E = V/L$ and $J = I/A$ in $\vec{J} = \sigma \vec{E}$: $$\frac{I}{A} = \sigma \frac{V}{L} = \frac{1}{\rho} \frac{V}{L} \implies V = I \left( \frac{\rho L}{A} \right)$$ We define Electrical Resistance ($R$): $$R = \frac{V}{I} = \frac{\rho L}{A}$$ SI Unit: **Ohm ($\Omega$)** ($1 \ \Omega = 1 \text{ V/A}$). Resistivity ($\rho$) is an intrinsic material property (Units: $\Omega\cdot\text{m}$).2. Temperature Variation of Resistivity
As temperature rises, thermal lattice vibrations (phonons) increase in amplitude, scattering conduction electrons more frequently and reducing collision time $\tau$. Over moderate temperature intervals: $$\rho(T) = \rho_0 [1 + \alpha (T - T_0)]$$ $$R(T) = R_0 [1 + \alpha (T - T_0)]$$ where $\alpha$ is the temperature coefficient of resistivity (K⁻¹ or °C⁻¹).- Metals ($\alpha > 0$): Resistivity increases with temperature (e.g., copper $\alpha \approx +0.0039\text{ K}^{-1}$).
- Semiconductors ($\alpha < 0$): In silicon and germanium, higher temperatures thermally excite vastly more covalent electrons into the conduction band, increasing carrier density $n$ exponentially ($n \propto e^{-E_g/2k_B T}$). Thus, resistivity drops sharply with temperature!
- Superconductors: Below a critical temperature $T_c$, electrical resistance vanishes completely ($R \equiv 0$).
§4.3 Electromotive Force, Terminal Voltage, and Kirchhoff's Laws
1. Electromotive Force (EMF, $\mathcal{E}$)
An Electromotive Force $\mathcal{E}$ is any non-electrostatic mechanism (chemical in batteries, mechanical/magnetic in dynamos, thermal in thermocouples) that does work on charge: $$\mathcal{E} = \frac{dW_{\text{non-elec}}}{dq}$$ A real voltage source possesses internal resistance $r$. When delivering load current $I$, the terminal potential difference $V$ across the battery is: $$V = \mathcal{E} - I r$$ If the source is open-circuited ($I = 0$), $V = \mathcal{E}$.2. Kirchhoff's Circuit Laws
Gustav Kirchhoff (1845) formulated two fundamental conservation laws for multi-loop electrical networks:- Kirchhoff's Current Law (KCL / Junction Rule): The algebraic sum of all electric currents entering any junction node is identically zero: $$\sum_{k} I_k = 0$$ Physical Basis: Direct consequence of the conservation of electric charge ($\frac{\partial\rho}{\partial t} = 0$).
- Kirchhoff's Voltage Law (KVL / Loop Rule): The algebraic sum of all potential differences (EMFs and resistive $IR$ drops) around any closed circuit loop is zero: $$\sum_{k} \mathcal{E}_k - \sum_{k} I_k R_k = 0$$ Physical Basis: Direct consequence of the conservation of energy in a conservative electrostatic field ($\oint \vec{E} \cdot d\vec{r} = 0$).
§4.4 Electrical Measuring Instruments: Galvanometer, Ammeter, Voltmeter, and Potentiometer
1. The Moving-Coil Galvanometer
A galvanometer detects minute currents. A coil of $N$ turns and resistance $R_g$ suspended in a radial magnetic field experiences deflecting torque $\tau = N I A B$. Balanced by torsional spring restoring torque $\tau_s = C \theta$: $$I = \left(\frac{C}{N A B}\right) \theta = K \theta$$ The deflection angle $\theta$ is directly proportional to current. Full-scale deflection current is denoted $I_g$ (typically $50\ \mu\text{A} - 1\text{ mA}$).2. Conversion of Galvanometer to an Ammeter
An ammeter must connect in series and possess extremely low resistance to avoid perturbing circuit current. A low-resistance resistor called a shunt resistor ($R_s$) is connected in parallel with the galvanometer: $$I_s R_s = I_g R_g \implies (I - I_g) R_s = I_g R_g$$ $$R_s = \frac{I_g R_g}{I - I_g}$$3. Conversion of Galvanometer to a Voltmeter
A voltmeter must connect in parallel and possess extremely high resistance so it draws negligible current from the circuit. A large multiplier resistor ($R_m$) is connected in series with the galvanometer: $$V = I_g (R_g + R_m) \implies R_m = \frac{V}{I_g} - R_g$$4. The Slide-Wire Potentiometer
A potentiometer measures unknown EMF $\mathcal{E}_x$ without drawing any current at balance (null deflection), providing the theoretical ideal of an infinite-impedance voltmeter: $$\frac{\mathcal{E}_x}{\mathcal{E}_0} = \frac{l_x}{l_0}$$ where $l_x$ is the balancing length for the test cell and $l_0$ for the standard cell.§4.5 RC Circuits: Charging and Discharging Transients
1. Charging an RC Circuit
Consider a series circuit with battery $\mathcal{E}$, resistor $R$, capacitor $C$, and switch closed at $t = 0$. By Kirchhoff's voltage law: $$\mathcal{E} - i R - \frac{q}{C} = 0$$ Since $i = \frac{dq}{dt}$: $$R \frac{dq}{dt} + \frac{q}{C} = \mathcal{E} \implies \frac{dq}{dt} = -\frac{q - C\mathcal{E}}{RC}$$ Integrating with initial condition $q(0) = 0$: $$q(t) = C\mathcal{E} \left( 1 - e^{-t/RC} \right) = Q_0 \left( 1 - e^{-t/\tau} \right)$$ where $\tau = R C$ is the **capacitive time constant** (Units: seconds, $\Omega \cdot \text{F} = \text{s}$). Differentiating charge gives the decaying charging current: $$i(t) = \frac{dq}{dt} = \frac{\mathcal{E}}{R} e^{-t/\tau} = I_0 e^{-t/\tau}$$2. Discharging an RC Circuit
Disconnecting the battery and closing the loop across $R$: $$-i R - \frac{q}{C} = 0 \implies R \frac{dq}{dt} + \frac{q}{C} = 0$$ $$q(t) = Q_0 e^{-t/\tau}, \quad i(t) = -\frac{Q_0}{RC} e^{-t/\tau} = -I_0 e^{-t/\tau}$$3. The 50% Energy Paradox in Capacitor Charging
During charging to final voltage $V$:- Total energy delivered by the battery: $$W_{\text{battery}} = \int_0^\infty \mathcal{E} i(t) dt = \mathcal{E} \int_0^\infty dq = \mathcal{E} Q_0 = C\mathcal{E}^2$$
- Final electrostatic energy stored in capacitor: $$U_C = \frac{1}{2} C\mathcal{E}^2$$
- Total Joule thermal energy dissipated in resistor: $$W_{\text{heat}} = \int_0^\infty i^2 R \, dt = \int_0^\infty \left(\frac{\mathcal{E}}{R} e^{-t/RC}\right)^2 R \, dt = \frac{\mathcal{E}^2}{R} \int_0^\infty e^{-2t/RC} dt = \frac{1}{2} C\mathcal{E}^2$$
§4.6 Thermoelectricity: Seebeck, Peltier, and Thomson Effects
1. The Seebeck Effect (1821)
Thomas Johann Seebeck discovered that when two dissimilar conducting wires $A$ and $B$ are joined at two junctions maintained at different temperatures $T_1$ and $T_2$, an open-circuit **thermoelectric EMF** $\mathcal{E}_{AB}$ is established: $$\mathcal{E}_{AB} = \int_{T_1}^{T_2} S_{AB}(T) \, dT$$ where $S_{AB} = S_A - S_B$ is the differential Seebeck coefficient (Thermoelectric Power) in $\mu\text{V/K}$. Over modest temperature ranges: $$\mathcal{E} = a (T_h - T_c) + \frac{1}{2} b (T_h - T_c)^2$$ The neutral temperature ($T_n$) is the hot-junction temperature where EMF reaches its maximum ($\frac{d\mathcal{E}}{dT} = 0 \implies T_n = -a/b$). Beyond the inversion temperature ($T_i = 2T_n - T_c$), the polarity of the EMF reverses.2. The Peltier Effect (1834)
Jean Charles Athanase Peltier discovered the exact thermodynamic inverse of the Seebeck effect: When an electric current $I$ is driven through a junction between two dissimilar conductors, heat is either absorbed or released at the junction (over and above irreversible Joule heating): $$\frac{dQ_{\text{Peltier}}}{dt} = \Pi_{AB} I$$ where $\Pi_{AB}$ is the Peltier coefficient (Volts). Reversing current direction reverses heating to cooling! This enables solid-state thermoelectric coolers (Peltier coolers) used in satellite sensors, PCR machines, and silent refrigeration.3. The Thomson Effect and Kelvin Relations
William Thomson (Lord Kelvin, 1854) applied thermodynamics to prove that heat is reversibly absorbed or evolved when current passes along an individual homogeneous conductor having a temperature gradient $dT/dx$: $$\frac{dQ_{\text{Thomson}}}{dx} = \mu I \frac{dT}{dx}$$ Kelvin derived the celebrated **Kelvin (Onsager) Relations**: $$\Pi_{AB} = T \cdot S_{AB}, \quad \mu_A - \mu_B = T \frac{dS_{AB}}{dT}$$ Connecting all three thermoelectric effects into a unified thermodynamic framework.Rigorous Analytical & Numerical Solved Problems
Comprehensive step-by-step mathematical proofs, dimensional evaluations, and calculations matching B.Sc. Honors university examinations.
A cylindrical copper wire of diameter $D = 2.05\text{ mm}$ (12 AWG gauge) carries a steady direct current $I = 15.0\text{ A}$. Copper has density $\rho = 8960\text{ kg/m}^3$, atomic mass $M = 63.55\text{ g/mol}$, and provides one conduction electron per atom. Avogadro's number $N_A = 6.022 \times 10^{23}\text{ mol}^{-1}$, and elementary charge $e = 1.602 \times 10^{-19}\text{ C}$.\n(a) Determine the free electron number density $n$ in copper,\n(b) Calculate the current density $J$ in the wire, and\n(c) Find the electron drift speed $v_d$ and the time required for an electron to travel $L = 3.00\text{ m}$ along the wire.
Copper contains roughly $8.49 \times 10^{28}$ conduction electrons per cubic meter.
The current density is $4.54 \times 10^6$ A/m².
Electrons drift at a sluggish 0.33 mm per second, taking nearly 2.5 hours to traverse 3 meters of wire.
A two-loop circuit contains two real DC batteries: Battery 1 has EMF $\mathcal{E}_1 = 12.0\text{ V}$ and internal resistance $r_1 = 1.00\ \Omega$; Battery 2 has EMF $\mathcal{E}_2 = 6.00\text{ V}$ and internal resistance $r_2 = 1.00\ \Omega$. The batteries are connected in parallel across a common load resistor $R_L = 10.0\ \Omega$, with each branch containing an additional resistor: $R_1 = 3.00\ \Omega$ in branch 1 and $R_2 = 2.00\ \Omega$ in branch 2.\n(a) Write the Kirchhoff current and loop equations for the network,\n(b) Solve for branch currents $I_1$ and $I_2$, and the load current $I_L$, and\n(c) Find the potential difference across the load resistor and the power delivered to it.
Applying KVL to each independent loop produces two linear simultaneous equations.
Because $I_2 < 0$, Battery 2 is actually being charged backwards by the stronger 12V Battery 1!
The load resistor sustains 7.32 V and dissipates 5.35 W of electrical power.
A series RC circuit consists of a battery $\mathcal{E} = 100.0\text{ V}$, a resistor $R = 50.0\text{ k}\Omega$, and an uncharged capacitor $C = 20.0\ \mu\text{F}$. The switch is closed at $t = 0$.\n(a) Determine the capacitive time constant $\tau$ and the initial current $I_0$,\n(b) Calculate the time required for the capacitor to charge to $90.0\%$ of its final maximum voltage, and\n(c) Calculate the total electrical energy delivered by the battery, the final energy stored in the capacitor, and the total Joule thermal heat dissipated in the resistor as $t \to \infty$.
The time constant is exactly 1.00 s and initial peak current is 2.00 mA.
Reaching 90% full voltage requires $2.30$ time constants.
The battery supplies 0.200 J; exactly 0.100 J (50%) is stored in the capacitor and 0.100 J (50%) is converted to heat in the resistor.