Free electrons · Electric field · $v_d = \frac{eE\tau}{m}$
In a conductor, free electrons move randomly at high speeds. When an electric field is applied, they gain a small net velocity opposite to the field — this is drift velocity. It is tiny (~mm/s) but responsible for all electric current.
Fig 1.1 — Free electrons drift opposite to E⃗; each e⁻ acquires drift velocity $v_d$
Derivation
Let $E$ be the electric field, $m$ the electron mass, $e$ the electron charge, $\tau$ the relaxation time (avg. time between collisions).
①
Force on each electron due to field (opposite to $\vec{E}$):
$$F = eE \quad \Rightarrow \quad \text{acceleration } a = \frac{eE}{m}$$
②
After each collision the electron starts fresh (zero drift). In time $\tau$ it acquires velocity:
$$v_d = u + a\tau = 0 + \frac{eE}{m}\cdot\tau$$
③
This average acquired velocity is the drift velocity:
★ Result
$$v_d = \frac{eE\tau}{m}$$
Direction: opposite to $\vec{E}$ | Typical value: $\sim 10^{-4}$ m/s — surprisingly slow!
④
Current in terms of drift velocity — in time $\Delta t$, electrons in length $v_d\Delta t$ cross area $A$:
$$I = nAev_d$$
where $n$ = number density of free electrons.
Why so slow yet light is instant? The electric field propagates at speed of light. Every electron in the wire starts drifting almost simultaneously — that's why the bulb lights up instantly.
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02
Derivation of Ohm's Law
From drift velocity → $V = IR$ · Microscopic origin of resistance
Ohm's Law states that at constant temperature, the current through a conductor is directly proportional to the voltage across it. Here we derive it from first principles using drift velocity.
Fig 2.1 — Cylindrical conductor of length $L$, cross-section $A$, carrying current $I$ under voltage $V$
Derivation
①
Electric field inside conductor: $E = \dfrac{V}{L}$
Current through the conductor:
$$I = nAev_d = nAe\cdot\frac{eV\tau}{mL} = \frac{ne^2\tau A}{mL}\cdot V$$
④
Since $n, e, \tau, m, A, L$ are all constants at a given temperature, let:
$$R = \frac{mL}{ne^2\tau A} \quad \Rightarrow \quad I = \frac{V}{R}$$
★ Result
$$V = IR \qquad \text{where } R = \frac{mL}{ne^2\tau A} = \frac{\rho L}{A}$$
Resistivity: $\rho = \dfrac{m}{ne^2\tau}$ — depends only on material & temperature, not shape. |
Conductivity: $\sigma = \dfrac{1}{\rho} = \dfrac{ne^2\tau}{m}$.
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03
Vector Form of Ohm's Law
Current density $\vec{J}$ · $\vec{J} = \sigma\vec{E}$
The scalar $V = IR$ works for a whole conductor. For point-by-point analysis inside materials, we use the vector form — relating current density $\vec{J}$ to field $\vec{E}$.
Fig 3.1 — $\vec{J}$ and $\vec{E}$ are parallel inside an isotropic conductor; both in the same direction
Derivation
①
Define current density $\vec{J}$ = current per unit cross-sectional area (a vector, direction = flow of positive charge):
$$J = \frac{I}{A}$$
②
From drift velocity: $I = nAev_d$ and $v_d = \dfrac{eE\tau}{m}$, so:
$$J = \frac{I}{A} = nev_d = ne\cdot\frac{eE\tau}{m} = \frac{ne^2\tau}{m}\,E$$
This is the true, general Ohm's Law — valid point by point inside any material. The familiar $V = IR$ is just its macroscopic consequence for a uniform conductor.
As temperature rises in a metal, lattice ions vibrate more, reducing relaxation time $\tau$ — so resistivity increases. Semiconductors behave opposite: more charge carriers free up, so resistivity falls.
Fig 4.1 — Metal: ρ rises linearly with T
Fig 4.2 — Semiconductor: ρ falls as T rises
Derivation
Since $\rho = \dfrac{m}{ne^2\tau}$, a change in $\tau$ (due to temperature) changes $\rho$. For metals, experimentally:
①
At temperature $T_0$, resistivity is $\rho_0$. A small increase in temperature $\Delta T = T - T_0$ causes a proportional increase in $\rho$:
$$\Delta\rho \propto \rho_0\,\Delta T \quad \Rightarrow \quad \Delta\rho = \rho_0\,\alpha\,\Delta T$$
②
Total resistivity at temperature $T$:
$$\rho_T = \rho_0 + \Delta\rho = \rho_0 + \rho_0\,\alpha\,(T-T_0)$$
★ Result
$$\rho_T = \rho_0\,\bigl[1 + \alpha(T - T_0)\bigr]$$
$\alpha$ = temperature coefficient of resistivity (unit: K⁻¹ or °C⁻¹)
Metals: $\alpha > 0$ (e.g. copper $\alpha \approx 3.9\times10^{-3}\,°C^{-1}$). |
Semiconductors & insulators: $\alpha < 0$ — resistance falls with temperature. |
Nichrome: very small $\alpha$ — used in heating elements for stability.
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05
Kirchhoff's Laws
KCL — Junction Law · KVL — Loop Law
Two fundamental rules for analysing any circuit — based on conservation of charge (KCL) and conservation of energy (KVL). Together they can solve any network, however complex.
Kirchhoff's Current Law (KCL) — Junction Rule
Fig 5.1 — KCL: currents entering (blue) = currents leaving (red) at any junction
Statement & Basis
KCL: The algebraic sum of all currents at any junction (node) in a circuit is zero.
★ KCL
$$\sum I = 0 \quad \text{at any junction}$$
Currents entering: positive | Currents leaving: negative (sign convention)
Based on: Conservation of charge — charge cannot pile up at a junction. Whatever flows in must flow out.
Kirchhoff's Voltage Law (KVL) — Loop Rule
Fig 5.2 — KVL: going around a loop, sum of all EMFs = sum of all IR drops
Statement & Basis
KVL: The algebraic sum of all EMFs and potential drops around any closed loop is zero.
★ KVL
$$\sum \varepsilon = \sum IR \quad \text{around any closed loop}$$
Or equivalently: $\sum \Delta V = 0$ around the loop
Based on: Conservation of energy — when you go round a complete loop and return to start, net work done = 0. |
Sign rule: EMF in direction of traversal = positive; IR drop in direction of current = negative.
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06
Balancing Condition — Wheatstone Bridge
$P/Q = R/S$ · No current through galvanometer
A Wheatstone bridge is a circuit of four resistors arranged in a diamond. It is balanced when no current flows through the galvanometer. Used to measure unknown resistance with great precision.
Fig 6.1 — Wheatstone bridge; at balance, galvanometer reads zero ($I_g = 0$)
Derivation of Balancing Condition
At balance, $I_g = 0$. So no current flows through galvanometer. Applying KCL:
①
Since $I_g = 0$: current through $P$ = current through $R$ = $I_1$, and current through $Q$ = current through $S$ = $I_2$.
②
Since $I_g = 0$, points B and C are at the same potential. Applying KVL to loop ABD:
$$I_1 P = I_2 Q \quad \cdots (1)$$
③
Applying KVL to loop BCD:
$$I_1 R = I_2 S \quad \cdots (2)$$
★ Result
$$\frac{P}{R} = \frac{Q}{S} \qquad \text{or equivalently} \qquad \frac{P}{Q} = \frac{R}{S}$$
This is the balancing condition of the Wheatstone bridge
Practical use: $P$, $Q$, $R$ are known resistances. Unknown $S$ is found as $S = \dfrac{QR}{P}$. |
Sensitivity: bridge is most sensitive when all four arms are equal ($P = Q = R = S$).
