Chapter 3: Current Electricity

Electrostatics dealt with charges at rest. Current electricity studies charges in steady, controlled motion — the basis of every wire, circuit, motor, and electronic device. We connect three levels of description:

1. Microscopic: electrons drifting under an applied field, giving rise to current density $\vec{J}$.
2. Macroscopic: Ohm's law $V = IR$, with $R$ characterizing a piece of material.
3. Circuit: Kirchhoff's laws, used to analyze networks of resistors and EMF sources.

By the end of the chapter you can analyze Wheatstone bridges, meter bridges, and potentiometers — the precision instruments of 19th-century physics, still the conceptual core of measurement labs today.

Concept Map

• 3.1 Electric current, current density, drift velocity, mobility
• 3.2 Ohm's law — microscopic and macroscopic forms
• 3.3 Resistivity & conductivity; temperature dependence; metals vs alloys vs semiconductors
• 3.4 Colour code of carbon resistors
• 3.5 Combination of resistors — series and parallel
• 3.6 EMF, internal resistance, terminal voltage
• 3.7 Cells in series and parallel
• 3.8 Kirchhoff's laws (KCL, KVL)
• 3.9 Wheatstone bridge — balance condition
• 3.10 Meter bridge
• 3.11 Potentiometer — EMF comparison and internal resistance
• 3.12 Electrical energy and power; heating effect

3.1 Electric Current, Current Density, Drift Velocity, Mobility

Definition — Current

Electric current $I$ is the rate of flow of charge across a cross-section:

Unit. Ampere ($1\text{ A} = 1\text{ C/s}$).

Current is a scalar (it has magnitude and a chosen direction along a wire, but does not transform as a vector). The conventional direction is the direction of $+$ charge flow (opposite to electron flow in metals).

Definition — Current Density

$\vec{J}$ is current per unit (cross-section) area, vectorially along the direction of positive flow:

Unit. $\text{A/m}^2$.

Derivation — Drift Velocity

In a metal, free electrons (number density $n$, charge $-e$, mass $m$) move randomly with thermal speeds $\sim 10^6\text{ m/s}$ but the net displacement is zero. Applying a field $\vec{E}$ tilts the distribution; the average velocity of electrons in the direction opposite to $\vec{E}$ is the drift velocity $\vec{v}_d$.

Step 1. Force on each electron: $\vec{F} = -e\vec{E}$, giving acceleration $\vec{a} = -e\vec{E}/m$.

Step 2. Each electron picks up velocity $\vec{v} = \vec{u} + \vec{a}t$ (where $\vec{u}$ is its random thermal velocity) until it collides. Between collisions, average free time is $\tau$ (the "relaxation time"). The average random velocity is zero, so:

Step 3. Magnitude:

For copper at room temp under $E\sim 1\text{ V/m}$: $v_d \sim 10^{-4}\text{ m/s}$ — about $4\text{ cm/min}$! Electron signals propagate at light speed because the field changes throughout the wire almost instantly.

Derivation — Relation $I = nAev_d$

In time $dt$, all electrons within a length $v_d\,dt$ of the cross-section will cross it.

• Volume swept: $A v_d\,dt$.
• Number of electrons: $nA v_d\,dt$.
• Charge: $-e\cdot nAv_d\,dt$, magnitude $eAnv_d\,dt$.
• Current (magnitude): $I = nAev_d$.

In vector form: $\vec{J} = ne\vec{v}_d$ (with the convention that $\vec{v}_d$ refers to the velocity of $+$ carriers, or equivalently $-\vec{v}_{\text{electron}}$). So

Mobility

The mobility $\mu$ of a carrier is the magnitude of its drift velocity per unit applied field:

Worked Example

A copper wire ($n = 8.5\times 10^{28}\text{ m}^{-3}$) of cross-section $1\text{ mm}^2$ carries $5\text{ A}$. Find $v_d$.

Pitfalls

• $v_d$ is tiny compared to thermal speed; the signal (field) propagates at $c$.
• $J$ is a vector; $I$ is a scalar (line integral).
• In semiconductors, two types of carriers (electrons and holes) contribute: $J = nev_e + pev_h$.

3.2 Ohm's Law

Microscopic Form

Combining $\vec{J} = ne\vec{v}_d$ and $\vec{v}_d = e\vec{E}\tau/m$ (with sign absorbed):

where the conductivity is

Macroscopic Form — Derivation $V = IR$

Consider a uniform conductor of length $L$, cross-section $A$, conductivity $\sigma$.

Step 1. Apply potential difference $V$ across the ends: $E = V/L$.

Step 2. Current density: $J = \sigma E = \sigma V/L$.

Step 3. Current: $I = JA = \sigma V A/L$, so

where the resistance is

Unit of $R$. Ohm ($\Omega = \text{V/A}$). Unit of $\rho$: $\Omega\cdot\text{m}$.

Limits of Validity

Ohm's law $V \propto I$ holds for ohmic materials (most metals, at constant temperature). It fails for:

• Non-ohmic devices (diodes, transistors, vacuum tubes).
• Materials at very high $E$ (where avalanche or breakdown occurs).
• At very high currents (Joule heating changes $R$).

Worked Example

A wire of length $1\text{ m}$, cross-section $1\text{ mm}^2$, has resistivity $1.7\times 10^{-8}\,\Omega\text{m}$. Resistance:

Pitfalls

• Ohm's law is an empirical relation, not a fundamental law of physics — many real materials violate it.
• $V = IR$ requires both $V$ and $I$ to be measured at the same instant for the same element.

3.3 Resistivity, Conductivity, Temperature Dependence

Temperature Dependence

For most metals over a moderate range:

where $\alpha$ is the temperature coefficient of resistivity (units $\text{K}^{-1}$).

For metals: $\alpha > 0$ (resistance increases with $T$, because relaxation time $\tau$ decreases as lattice vibrations grow).

For semiconductors: $\alpha < 0$ (more charge carriers thermally excited; $n$ rises faster than $\tau$ falls).

For alloys (e.g., manganin, constantan, nichrome): $\alpha$ is very small — these are used in standard resistors.

Resistivity Comparison

Worked Example

A copper coil has $R = 5.0\,\Omega$ at $20°\text{C}$. Find $R$ at $80°\text{C}$.

Pitfalls

• "Resistance" and "resistivity" are different — $R$ depends on geometry, $\rho$ does not.
• At very low $T$, some metals show superconductivity ($\rho \to 0$) — beyond linear law.
• Carbon resistors have negative $\alpha$ but are not classed as semiconductors here.

3.4 Carbon Resistor Colour Code

A standard carbon resistor has four colour bands:

• Band 1, 2: first two significant digits.
• Band 3: multiplier (power of 10).
• Band 4: tolerance ($\pm$ %).

Mnemonic: "B B ROY of Great Britain has a Very Good Wife".

Worked Example

A resistor has bands: Yellow, Violet, Orange, Gold.

• 4, 7, $\times 10^3$, $\pm 5\%$
• $R = 47\times 10^3 = 47\,\text{k}\Omega \pm 5\%$.

3.5 Combination of Resistors

Series — Derivation

Current $I$ is the same in all; voltages add:

The equivalent resistance is larger than the largest individual resistor.

Parallel — Derivation

Voltage $V$ is the same across all; currents add:

For two resistors in parallel: $R_{\text{eq}} = R_1R_2/(R_1+R_2)$. Equivalent resistance is smaller than the smallest.

Worked Example

Three resistors $2, 3, 6\,\Omega$.

• Series: $11\,\Omega$.
• Parallel: $1/R = 1/2 + 1/3 + 1/6 = 1 \Rightarrow R = 1\,\Omega$.
• Mixed ($2\,\Omega$ in series with $3\parallel 6$): $3\parallel 6 = 2\,\Omega$; total $= 4\,\Omega$.

Pitfalls

• Watch the polarity in series; in parallel ensure both ends share the same node pair.
• "Same current" in series, "same voltage" in parallel.

3.6 EMF and Internal Resistance

Definitions

The electromotive force (EMF) $\varepsilon$ of a cell is the energy supplied by the cell per unit charge as it pushes charge around the complete circuit — its open-circuit terminal voltage.

A real cell has an internal resistance $r$ (modelled as in series with the ideal EMF).

When current $I$ flows out of the $+$ terminal:

• On discharge ($I>0$): $V < \varepsilon$.
• On open circuit ($I=0$): $V = \varepsilon$.
• When charging the cell (forcing current into $+$ terminal): $V = \varepsilon + Ir > \varepsilon$.

Derivation — Closed-Circuit Current

Connect a cell ($\varepsilon$, $r$) to an external resistance $R$:

Terminal voltage: $V = IR = \varepsilon R/(R+r) = \varepsilon - Ir$.

Power Delivered to Load

Maximum power transfer: $dP/dR = 0$ gives $R = r$. Then $P_{\max} = \varepsilon^2/(4r)$ — half the EMF is dissipated inside the cell.

Worked Example

A $1.5\text{ V}$ cell with $r = 0.5\,\Omega$ drives a $4\,\Omega$ resistor. Find $I$, terminal voltage, and power dissipated in the load.

Pitfalls

• Internal resistance is not a physical resistor; it represents irreversibilities inside the cell.
• $\varepsilon$ is not directly the voltage you measure across a working cell — voltmeters read $V$, not $\varepsilon$ (except on open circuit).

3.7 Cells in Series and Parallel

Series — same direction

$n$ identical cells (each $\varepsilon, r$) and external $R$:

• $R\gg nr$: $I\approx n\varepsilon/R$ (gain by stacking voltage).
• $R\ll nr$: $I\approx \varepsilon/r$ (no improvement; using $n$ cells in series for tiny loads wastes them).

Parallel — same polarity

$m$ identical cells in parallel:

• Useful when $R\ll r$: $I\approx m\varepsilon/r$.
• Useful when $r$ is large (or you want to deliver large current to a small load).

Mixed (m rows, n cells in each row)

Total cells $N = mn$. Equivalent EMF $= n\varepsilon$, equivalent internal resistance $= nr/m$:

Maximum when $R = nr/m$ (matched), giving $I_{\max} = mn\varepsilon/(2nr) = m\varepsilon/(2r)$.

Cells in Parallel — Different EMFs (general)

Two cells $\varepsilon_1, r_1$ and $\varepsilon_2, r_2$ in parallel across $R$:

Pitfalls

• For cells opposing in series, EMFs subtract.
• Connecting unequal cells in parallel produces internal circulation currents.

3.8 Kirchhoff's Laws

KCL — Junction Rule

At any junction, the algebraic sum of currents is zero (charge conservation):

KVL — Loop Rule

Around any closed loop, the algebraic sum of EMFs and $IR$ drops is zero (energy conservation):

Sign Convention

Going around a loop in a chosen direction:

• Cross a resistor in the direction of current: $-IR$.
• Cross against the current: $+IR$.
• Cross a cell from $-$ to $+$ internally: $+\varepsilon$.
• Cross a cell from $+$ to $-$ internally: $-\varepsilon$.

Worked Example

Circuit with two cells: $\varepsilon_1 = 6\text{ V}, r_1 = 1\,\Omega$ in series with $R_1 = 2\,\Omega$; second loop has $\varepsilon_2 = 4\text{ V}, r_2 = 1\,\Omega$ and $R_2 = 3\,\Omega$; loops share resistor $R_3 = 4\,\Omega$.

Set up mesh currents $I_1, I_2$ in each loop (both clockwise). KVL loop 1:

KVL loop 2: $4 = -4I_1 + 8I_2$.

Solve: From the second, $I_1 = 2I_2 - 1$. Substitute: $6 = 7(2I_2 - 1) - 4I_2 = 10I_2 - 7$, so $I_2 = 1.3$ A. Then $I_1 = 1.6$ A. Current through $R_3$ is $I_1 - I_2 = 0.3$ A.

Pitfalls

• Set up consistent loop directions; don't mix sign conventions mid-problem.
• KCL is exact; KVL holds only in the static (or quasi-static) regime.

3.9 Wheatstone Bridge

Setup

Four resistors $P, Q, R, S$ in a diamond. A galvanometer $G$ bridges the diagonal; a cell drives current across the other diagonal. The bridge is balanced when no current flows through $G$.

Derivation — Balance Condition

Let the bridge be balanced. The currents in $P$ and $Q$ are equal ($I_1$); in $R$ and $S$ are equal ($I_2$); galvanometer carries zero current.

Step 1. Points $B$ and $D$ are at the same potential (no current through $G$).

Step 2. Drop across $P$ equals drop across $R$: $I_1 P = I_2 R$.

Step 3. Drop across $Q$ equals drop across $S$: $I_1 Q = I_2 S$.

Step 4. Divide:

Applications

• Precise measurement of an unknown resistance (one arm).
• Meter bridge and post-office box are practical implementations.

Worked Example

$P = 10\,\Omega$, $Q = 20\,\Omega$, $S = 30\,\Omega$. For balance, $R = PS/Q = 10\cdot 30/20 = 15\,\Omega$.

Pitfalls

• The cell's EMF and internal resistance do not affect the balance condition.
• The galvanometer's sensitivity matters for detecting balance, but not for the condition itself.

3.10 Meter Bridge

Principle

A meter bridge is a practical Wheatstone bridge with two arms ($P, Q$) replaced by sections of a uniform resistance wire of length $1\text{ m}$. Sliding a jockey along the wire varies the resistance ratio continuously.

Working

Let the bridge wire be split at the jockey point into lengths $\ell$ (from $A$ to jockey) and $(100-\ell)$ cm (jockey to $C$). Resistances are proportional to length (uniform wire):

At balance:

To find unknown $X$: put a known $S$ in one gap and $X$ in the other; find balance length $\ell$ — then $X = S\ell/(100-\ell)$.

Pitfalls

• The wire must be uniform in cross-section.
• End corrections (effective small lengths added at each end) are sometimes needed for precise work.

3.11 Potentiometer

Principle

A potentiometer uses a long, uniform resistance wire driven by a primary circuit (steady cell + rheostat). The potential drop along the wire is uniform per unit length:

called the potential gradient.

A test EMF connected to a fraction of this wire (via a galvanometer with no current at balance) is measured by the balance length — without drawing current from the test cell.

Comparison of Two EMFs

Connect cells $\varepsilon_1, \varepsilon_2$ in turn, finding balance lengths $\ell_1, \ell_2$:

Internal Resistance of a Cell

Connect the test cell to the potentiometer. Find balance length $\ell_1$ (open circuit, EMF = $\varepsilon$).

Now close the cell through a resistor $R$. The terminal voltage drops to $V = \varepsilon R/(R+r)$. Find new balance length $\ell_2$. Then:

Why a Potentiometer Beats a Voltmeter

A voltmeter draws some current from the test cell, so it reads $V$, not $\varepsilon$. The potentiometer at balance draws zero current — it is an ideal voltmeter. Sensitivity can be improved by lengthening the wire or reducing the gradient $k$.

Worked Example

A cell of unknown EMF gives balance at $80\text{ cm}$ when the potential gradient is $0.01\text{ V/cm}$. EMF $= 0.01\times 80 = 0.8\text{ V}$.

Pitfalls

• The driver cell EMF must exceed the test cell EMF — otherwise no balance exists.
• The galvanometer should deflect in opposite directions at the two ends; if not, leads are reversed.
• If sliding jockey is pressed too long, the wire heats — wait between measurements.

3.12 Electrical Energy and Power

Power Delivered to a Resistor

A current $I$ through a potential drop $V$ does work at rate

Unit. Watt ($1\text{ W} = 1\text{ J/s}$). All this power becomes heat (Joule heating).

Energy Consumed

$E = Pt$. The commercial unit is the kilowatt-hour:

Joule's Heating Law

Heat produced in time $t$:

Used in: incandescent bulbs (tungsten filament, $\rho$ rises with $T$); electric heaters (nichrome, high $\rho$, low $\alpha$); fuses (low melting alloy).

Worked Example

A $60\text{ W}, 220\text{ V}$ bulb is operated at $220\text{ V}$.

• Operating resistance: $R = V^2/P = 220^2/60 \approx 807\,\Omega$.
• Current: $I = P/V = 60/220 \approx 0.273\text{ A}$.
• Energy used in $10\text{ h}$: $E = 0.6\text{ kWh}$.

Maximum Power Transfer

(See 3.6.) $P_R$ is maximum when $R = r$, giving $P_{\max} = \varepsilon^2/(4r)$.

Pitfalls

• $P = V^2/R$ — at constant $V$, smaller $R$ means more power; at constant $I$, larger $R$ means more power.
• Bulbs rated at $V$, $P$: the resistance computed via $R = V^2/P$ applies at the rated voltage, where the filament is hot. At room temperature, $R$ is several times smaller.

Solved Problems

Problem 1 (Easy)

A wire of resistance $10\,\Omega$ is bent into a circle. Find the resistance between two diametrically opposite points.

Solution. The two semicircles each have resistance $5\,\Omega$, connected in parallel between the two endpoints: $R_{\text{eq}} = 5/2 = 2.5\,\Omega$.

Problem 2 (Easy)

A $2\text{ A}$ current flows through a wire of cross-section $10^{-6}\text{ m}^2$ with $n = 10^{29}\text{ m}^{-3}$. Find $v_d$.

Problem 3 (Medium)

Two cells of EMF $2\text{ V}$ each with internal resistance $1\,\Omega$ each are connected in parallel across a $3\,\Omega$ external resistance. Find the current through the external resistor.

$\varepsilon_{\text{eq}} = 2\text{ V}$, $r_{\text{eq}} = 0.5\,\Omega$. $I = 2/(3+0.5) = 4/7 \approx 0.571\text{ A}$.

Problem 4 (Medium)

In a Wheatstone bridge, $P = 100\,\Omega$, $Q = 10\,\Omega$, and the unknown $R$ balances when $S = 1\,\Omega$. Find $R$.

$P/Q = R/S \Rightarrow R = PS/Q = 100\cdot 1/10 = 10\,\Omega$.

Problem 5 (Medium)

A potentiometer wire is $400\text{ cm}$ long, driven by a $4\text{ V}$ cell with negligible internal resistance and a series rheostat that drops $2\text{ V}$. Find the potential gradient. A standard cell of $1.5\text{ V}$ — at what length does it balance?

Voltage across the wire $= 4 - 2 = 2\text{ V}$. Gradient: $k = 2/400 = 5\times 10^{-3}\text{ V/cm}$. Balance length $\ell = 1.5/k = 300\text{ cm}$.

Problem 6 (Hard) — Kirchhoff Analysis

A cell $\varepsilon = 12\text{ V}$ with $r = 1\,\Omega$ drives a network: $R_1 = 3\,\Omega$ in series with the parallel combination of $R_2 = 6\,\Omega$ and $R_3 = 12\,\Omega$. Find the current from the cell and the current through $R_3$.

$R_2 \| R_3 = 6\cdot 12/18 = 4\,\Omega$. Total external: $R_1 + 4 = 7\,\Omega$. Total: $8\,\Omega$. $I = 12/8 = 1.5\text{ A}$. Voltage across the parallel combo: $1.5\cdot 4 = 6\text{ V}$. So $I_{R_3} = 6/12 = 0.5\text{ A}$.

Problem 7 (Hard) — Temperature and Power

A heater is designed to consume $1\text{ kW}$ at $220\text{ V}$ when its filament is at operating temperature $T_{\text{op}}$. The room-temperature resistance is $30\,\Omega$. Find the operating temperature (assume $\alpha = 4\times 10^{-3}\text{ K}^{-1}$, room temp $30°\text{C}$).

Operating $R = V^2/P = 48.4\,\Omega$. Ratio $R/R_0 = 48.4/30 = 1.613 = 1 + \alpha\Delta T$, so $\Delta T = 0.613/0.004 = 153.3°\text{C}$. $T_{\text{op}} \approx 183°\text{C}$.

(In reality the filament is at $\sim 2500°\text{C}$; the linear law breaks down at such high temperatures.)

JEE/NEET Edge Cases

1. Drift velocity vs signal speed: $v_d \sim 10^{-4}\text{ m/s}$; signals at $\sim c$. Don't confuse.

2. Wire stretched to twice its length (volume constant): $L \to 2L$, $A \to A/2$, so $R \to 4R$.

3. Cells in series with opposing polarity: net EMF $= \varepsilon_1 - \varepsilon_2$. If $\varepsilon_2 > \varepsilon_1$, cell 1 gets charged.

4. Galvanometer with current $I_g$: full-scale deflection at $I_g$. Ammeter conversion: shunt $S = I_g G/(I - I_g)$. Voltmeter: series resistance $R = V/I_g - G$.

5. Equivalent resistance of an infinite ladder (each rung $R$): $R_\infty = R + R\|R_\infty$ leads to a self-consistent equation.

6. Maximum power transfer: occurs at $R_{\text{load}} = r_{\text{internal}}$; efficiency at that point is $50\%$ — not what you want in a power grid.

7. Bulb brightness: in parallel, the higher-rated (lower-$R$) bulb is brighter. In series, the lower-rated (higher-$R$) bulb is brighter.

8. Carbon resistor's $\alpha$ is negative — but the magnitude is small enough that the color code is read as a fixed value.

9. Superconductors: $\rho = 0$ below $T_c$, magnetic field is expelled (Meissner effect).

10. Mobility in semiconductors is much higher for electrons than for holes; this asymmetry is the basis of $n$- and $p$-type doping.

Quick Recap

• $I = dQ/dt$; $J = I/A$; $v_d = eE\tau/m$; $I = nAev_d$.
• Ohm: $V = IR$, $J = \sigma E$; $R = \rho L/A$.
• $\rho(T) = \rho_0(1 + \alpha\Delta T)$; $\alpha_{\text{metal}}>0$, $\alpha_{\text{semi}}< 0$.
• Series: $R = \sum R_i$. Parallel: $1/R = \sum 1/R_i$.
• EMF: $V = \varepsilon - Ir$ (discharge); $I = \varepsilon/(R+r)$.
• Wheatstone balance: $P/Q = R/S$.
• Meter bridge: $X = S\ell/(100-\ell)$.
• Potentiometer EMF comparison: $\varepsilon_1/\varepsilon_2 = \ell_1/\ell_2$.
• $P = VI = I^2R = V^2/R$; $1\text{ kWh} = 3.6\times 10^6\text{ J}$.

Formula Sheet