Ohm's Law (Macroscopic Form)

The voltage VV across a conductor is directly proportional to the current II flowing through it, provided temperature and other physical conditions remain constant: V=IRV = IR

where:

  • VV is the potential difference (voltage) in volts
  • II is the current in amperes
  • RR is the resistance in ohms (Ω\Omega)

Alternatively: I=VRI = \frac{V}{R}

Resistance and Resistivity

The resistance RR of a conductor depends on:

  1. Its length ll (longer conductors have more resistance)
  2. Its cross-sectional area AA (thicker conductors have less resistance)
  3. The material's resistivity ρρ (resistance per unit length per unit area)

R=ρlAR = \rho \frac{l}{A}

where ρρ is resistivity (measured in Ω·m)

Conductivity

Conductivity σ\sigma is the reciprocal of resistivity: σ=1ρ\sigma = \frac{1}{\rho}

Measured in (Ω·m)⁻¹ or Siemens per meter (S/m)

Its SI unit is S/m\text{S/m}.

Ohm's Law (Microscopic / Local Form)

At every point in a conductor, electric field EE and current density jj are related by: j=σEj = \sigma E

or equivalently, E=ρjE = \rho j

Connection between Macroscopic and Microscopic Forms

For a uniform conductor of length ll and area AA:

  • V=ElV = El
  • I=jAI = jA

So, V=El=ρjl=ρIAl=I(ρlA)V = El = \rho j l = \rho \frac{I}{A} l = I\left(\rho \frac{l}{A}\right)

Hence, V=IRV = IR with R=ρlAR = \rho \frac{l}{A}

[JEE Tip] Ohm's law holds both at the circuit level and at the microscopic level.

Drift Velocity from the Collision Model

Consider a single free electron in an electric field EE.

Force on electron: F=eEF = eE (where ee is magnitude of electron charge)

Acceleration: a=Fm=eEma = \frac{F}{m} = \frac{eE}{m}

Collision Model

Without collisions, an electron would continuously accelerate and velocity would increase indefinitely. However, electrons collide with lattice ions at regular intervals.

Mean time between collisions: τ\tau (relaxation time)

For a collision model:

  • Immediately after collision: electron has random velocity (no net direction)
  • Electron accelerates for time τ\tau under field EE
  • Just before collision: electron has acquired velocity component along field
  • Immediately after collision: velocity becomes random again
  • Cycle repeats

Drift Velocity

Velocity gained during time τ\tau: v=a⋅τ=eEm⋅τv = a \cdot \tau = \frac{eE}{m} \cdot \tau

This is the drift velocity: vd=eEτmv_d = \frac{eE\tau}{m}

Alternatively, using the relation v=μEv = \mu E where μ\mu is mobility: μ=eτm\mu = \frac{e\tau}{m}

Physical Interpretation

  • Larger τ\tau (fewer collisions): larger drift velocity for same field
  • Larger EE (stronger field): larger drift velocity
  • Larger mm (heavier particle): smaller drift velocity for same field

[NEET Important] The relaxation time τ\tau is the key microscopic parameter that determines conductivity. It depends on the crystal structure and temperature of the material, explaining why different materials have different conductivities.

Origin of Resistivity

Step 1: Relate drift velocity to current From Section 1, we know: I=neAvdI = neAv_d

Substituting vd=eEτmv_d = \frac{eE\tau}{m}: I=neA⋅eEτm=ne2AτEmI = neA \cdot \frac{eE\tau}{m} = \frac{ne^2A\tau E}{m}

Step 2: Find relationship between I and E I=ne2Aτm⋅EI = \frac{ne^2A\tau}{m} \cdot E

Comparing with Ohm's law V=IRV = IR where V=E⋅lV = E \cdot l: E⋅l=R⋅IE \cdot l = R \cdot I

Rearranging: R=EI/lR = \frac{E}{I/l}

But current density j=I/Aj = I/A, so: I=jAI = jA

Therefore: R=EjA/l=EljAR = \frac{E}{jA/l} = \frac{El}{jA}

Step 3: Derive resistivity formula From R=ρlAR = \rho \frac{l}{A}, we have: ρ=RAl=EjA/l⋅Al=Ej\rho = R \frac{A}{l} = \frac{E}{jA/l} \cdot \frac{A}{l} = \frac{E}{j}

From Step 1: I=ne2Aτm⋅EI = \frac{ne^2A\tau}{m} \cdot E, so: j=I/A=ne2τm⋅Ej = I/A = \frac{ne^2\tau}{m} \cdot E

Therefore: ρ=Ej=Ene2τmE=mne2τ\rho = \frac{E}{j} = \frac{E}{\frac{ne^2\tau}{m}E} = \frac{m}{ne^2\tau}

Final Resistivity Formula: ρ=mne2τ\boxed{\rho = \frac{m}{ne^2\tau}}

Conductivity Formula: σ=1ρ=ne2τm\sigma = \frac{1}{\rho} = \frac{ne^2\tau}{m}

Key Insights:

  1. ρ∝1n\rho \propto \frac{1}{n}: More free electrons → lower resistivity (better conductor)
  2. ρ∝1τ\rho \propto \frac{1}{\tau}: Shorter relaxation time → higher resistivity (more collisions increase resistance)
  3. ρ\rho independent of applied field or voltage (for ohmic conductors)
  4. This formula explains why metals (high nn, long τ\tau) are good conductors
  5. And why insulators (low nn) are poor conductors

[JEE Tip] The derivation ρ=mne2τ\rho = \frac{m}{ne^2\tau} is one of the most important results in this chapter. It connects macroscopic resistance with microscopic electron properties. Understanding this derivation is essential for JEE.

Why Ohm's Law Holds?

From our derivation: I=ne2Aτm⋅Eorj=ne2τm⋅EI = \frac{ne^2A\tau}{m} \cdot E \quad \text{or} \quad j = \frac{ne^2\tau}{m} \cdot E

This is exactly the form of Ohm's law j=σEj = \sigma E where: σ=ne2τm\sigma = \frac{ne^2\tau}{m}

For ohmic conductors:

  • nn (carrier density): relatively constant
  • τ\tau (relaxation time): determined by material structure, varies slightly with field
  • Ratio σ\sigma or E/jE/j: approximately constant

Therefore, j∝Ej \propto E (Ohm's law holds)

Why is it linear?

The key is that relaxation time τ\tau in the collision model is assumed independent of electron velocity or field strength (for weak fields). This independence makes the j-E relationship linear.

Temperature Dependence

At higher temperature, atoms vibrate more, increasing collision frequency (decreasing τ\tau):

ρ(T)=ρ0[1+α(T−T0)]\rho(T) = \rho_0[1 + \alpha(T - T_0)]

where:

  • ρ0\rho_0 is resistivity at reference temperature T0T_0
  • α\alpha is temperature coefficient of resistivity
  • For metals: α>0\alpha > 0 (resistivity increases with temperature)
  • For semiconductors: α<0\alpha < 0 (resistivity decreases with temperature)

Physical reason

  • Metals: More collisions due to lattice vibrations → higher ρ\rho
  • Semiconductors: More free electrons generated by thermal energy → lower ρ\rho (dominates)

[NEET Important] Understanding temperature effects on resistance is crucial for practical applications. This is why device ratings specify operating temperature ranges.

Ohmic and Non-Ohmic Conductors

Ohmic Conductors

Conductors that obey Ohm's law exactly (V∝IV \propto I or constant RR)

Examples:

  • Metallic wires at constant temperature
  • Resistors (carbon, metal film)
  • Electrolyte solutions

Characteristics:

  • Linear V-I curve passes through origin
  • Slope of V-I graph = Resistance R
  • Resistance independent of applied voltage
  • Resistance changes only with temperature

Non-Ohmic Conductors

Conductors that do NOT follow Ohm's law (V∝̸IV \not\propto I or variable RR)

Examples:

  1. Semiconductors (diodes, transistors)
  • Resistance depends strongly on voltage/current
  • V-I curve is curved (exponential)
  • Used for controlling current direction
  1. Tungsten filament (incandescent bulb)
  • At low current: behaves nearly ohmic
  • At high current: filament heats, resistance increases (temperature effect dominates)
  • V-I curve is curved upward
  • Eventually burns out at very high currents
  1. Gases (gas discharge)
  • At low field: very high resistance (insulating)
  • At breakdown voltage: suddenly becomes conductor
  • V-I curve shows discontinuous jump
  • Used in neon signs, plasma displays

Analyzing V-I Curves

Linear (Ohmic) Curve

  • Straight line through origin
  • Slope = 1/R (constant)
  • Examples: copper wire, resistor

Curved (Non-Ohmic) Curves

  • Diode: nearly zero current until threshold, then exponential rise
  • Tungsten bulb: becomes steeper (resistance decreases as filament heats slightly, but continues following modified model)
  • Thermistor: opposite behavior - resistance increases strongly with temperature

[JEE Tip] For many practical devices, the V-I curve can be linearized around an operating point, making them behave approximately ohmic over a small range. This is important for amplifier circuit design.

Example 1: Resistance from Ohm's Law

A copper wire has a voltage of 2.5 V applied across it, and a current of 5 A flows through it. Calculate the resistance of the wire.

Given:

  • V=2.5 VV = 2.5\,\text{V}
  • I=5 AI = 5\,\text{A}

Formula: R=VIR = \frac{V}{I}

Solution: R=2.55=0.5 ΩR = \frac{2.5}{5} = 0.5\,\Omega

Answer: 0.5 Ω0.5\,\Omega

Example 2: Resistance from Resistivity

A nichrome wire has length 1.5 m and cross-sectional area 0.8 mm2^2. If the resistivity of nichrome is 1.0×10−6 Ω⋅m1.0 \times 10^{-6}\,\Omega\cdot\text{m}, find the resistance of the wire.

Given:

  • l=1.5 ml = 1.5\,\text{m}
  • A=0.8 mm2=8×10−7 m2A = 0.8\,\text{mm}^2 = 8 \times 10^{-7}\,\text{m}^2
  • ρ=1.0×10−6 Ω⋅m\rho = 1.0 \times 10^{-6}\,\Omega\cdot\text{m}

Formula: R=ρlAR = \rho \frac{l}{A}

Solution: R=1.0×10−6×1.58×10−7R = 1.0 \times 10^{-6} \times \frac{1.5}{8 \times 10^{-7}}

R=1.0×10−6×1.58×107R = 1.0 \times 10^{-6} \times \frac{1.5}{8} \times 10^{7}

R=1.5×1078×106=1.5×108=158=1.875 ΩR = \frac{1.5 \times 10^{7}}{8 \times 10^{6}} = \frac{1.5 \times 10}{8} = \frac{15}{8} = 1.875 \text{ Ω}

Answer: 1.875 Ω1.875\,\Omega

Note: Nichrome is used in heating elements because its high resistivity generates significant heat for a given current.

Example 3: Drift Velocity from Current

A silver wire of area 1 mm2^2 carries a current of 2 A. The electron density in silver is n=5.8×1028 m−3n = 5.8 \times 10^{28}\,\text{m}^{-3}. Find the drift velocity of electrons.

Given:

  • A=1 mm2=10−6 m2A = 1\,\text{mm}^2 = 10^{-6}\,\text{m}^2
  • I=2 AI = 2\,\text{A}
  • n=5.8×1028 m−3n = 5.8 \times 10^{28}\,\text{m}^{-3}
  • e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}

Formula: vd=IneAv_d = \frac{I}{neA}

Solution: vd=25.8×1028×1.6×10−19×10−6=2.15×10−4 m/sv_d = \frac{2}{5.8 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}} = 2.15 \times 10^{-4}\,\text{m/s}

Answer: 2.15×10−4 m/s2.15 \times 10^{-4}\,\text{m/s}

Example 4: Drift Velocity from Electric Field

An electric field of 10 V/m is applied across a conductor. The relaxation time is τ=3×10−14\tau = 3 \times 10^{-14} s and electron mass is m=9.109×10−31m = 9.109 \times 10^{-31} kg. Calculate the drift velocity.

Given:

  • E=10 V/mE = 10\,\text{V/m}
  • τ=3×10−14 s\tau = 3 \times 10^{-14}\,\text{s}
  • m=9.109×10−31 kgm = 9.109 \times 10^{-31}\,\text{kg}
  • e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}

Formula: vd=eEτmv_d = \frac{eE\tau}{m}

Solution: vd=1.6×10−19×10×3×10−149.109×10−31=5.27×10−2 m/sv_d = \frac{1.6 \times 10^{-19} \times 10 \times 3 \times 10^{-14}}{9.109 \times 10^{-31}} = 5.27 \times 10^{-2}\,\text{m/s}

Answer: 5.27×10−2 m/s5.27 \times 10^{-2}\,\text{m/s}

Example 5: Resistivity and Conductivity from Microscopic Parameters

For a conductor with electron density n=8.5×1028 m−3n = 8.5 \times 10^{28}\,\text{m}^{-3} and relaxation time τ=2.4×10−14\tau = 2.4 \times 10^{-14} s, calculate the resistivity and conductivity.

Given:

  • n=8.5×1028 m−3n = 8.5 \times 10^{28}\,\text{m}^{-3}
  • τ=2.4×10−14 s\tau = 2.4 \times 10^{-14}\,\text{s}
  • e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}
  • m=9.109×10−31 kgm = 9.109 \times 10^{-31}\,\text{kg}

Formula: ρ=mne2τ,σ=1ρ\rho = \frac{m}{ne^2\tau}, \qquad \sigma = \frac{1}{\rho}

Solution: ρ=9.109×10−318.5×1028×(1.6×10−19)2×2.4×10−14≈1.74×10−8 Ω⋅m\rho = \frac{9.109 \times 10^{-31}}{8.5 \times 10^{28} \times (1.6 \times 10^{-19})^2 \times 2.4 \times 10^{-14}} \approx 1.74 \times 10^{-8}\,\Omega\cdot\text{m} σ=11.74×10−8≈5.75×107 S/m\sigma = \frac{1}{1.74 \times 10^{-8}} \approx 5.75 \times 10^7\,\text{S/m}

Answer: ρ≈1.74×10−8 Ω⋅m\rho \approx 1.74 \times 10^{-8}\,\Omega\cdot\text{m}, σ≈5.75×107 S/m\sigma \approx 5.75 \times 10^7\,\text{S/m}

Example 6: Resistance of a Stretched Wire

A wire of length ll and area AA has resistance RR. It is stretched to double its length while maintaining constant volume. What is the new resistance?

Solution: Since volume remains constant, lA=l′A′lA = l'A' If l′=2ll' = 2l, then A′=A2A' = \frac{A}{2}

Now, R′=ρl′A′=ρ2lA/2=4ρlA=4RR' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = 4\rho \frac{l}{A} = 4R

Answer: 4R4R

Example 7: Resistance at Higher Temperature

A copper wire has resistance 50 Ω\Omega at 0∘0^\circC. The temperature coefficient of resistivity for copper is α=4.3×10−3 K−1\alpha = 4.3 \times 10^{-3}\,\text{K}^{-1}. Calculate the resistance at 100∘100^\circC.

Given:

  • R0=50 ΩR_0 = 50\,\Omega
  • ΔT=100 K\Delta T = 100\,\text{K}
  • α=4.3×10−3 K−1\alpha = 4.3 \times 10^{-3}\,\text{K}^{-1}

Formula: R=R0(1+αΔT)R = R_0(1 + \alpha \Delta T)

Solution: R=50[1+(4.3×10−3)(100)]=50(1+0.43)=71.5 ΩR = 50[1 + (4.3 \times 10^{-3})(100)] = 50(1 + 0.43) = 71.5\,\Omega

Answer: 71.5 Ω71.5\,\Omega

Example 8: Complete Microscopic-to-Macroscopic Connection

A uniform electric field of 0.5 V/cm is applied across a copper rod. The relaxation time for copper is τ=3×10−14\tau = 3 \times 10^{-14} s. Calculate: (a) the drift velocity of electrons, (b) the current density, (c) the conductivity.

Given:

  • E=0.5 V/cm=50 V/mE = 0.5\,\text{V/cm} = 50\,\text{V/m}
  • τ=3×10−14 s\tau = 3 \times 10^{-14}\,\text{s}
  • n=8.5×1028 m−3n = 8.5 \times 10^{28}\,\text{m}^{-3}
  • e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}
  • m=9.109×10−31 kgm = 9.109 \times 10^{-31}\,\text{kg}

(a) Drift velocity: vd=eEτm=1.6×10−19×50×3×10−149.109×10−31≈0.263 m/sv_d = \frac{eE\tau}{m} = \frac{1.6 \times 10^{-19} \times 50 \times 3 \times 10^{-14}}{9.109 \times 10^{-31}} \approx 0.263\,\text{m/s}

(b) Current density: j=nevd=8.5×1028×1.6×10−19×0.263≈3.57×109 A/m2j = nev_d = 8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 0.263 \approx 3.57 \times 10^9\,\text{A/m}^2 This is also 0.357 MA/cm20.357\,\text{MA/cm}^2

(c) Conductivity: σ=jE=3.57×10950≈7.14×107 S/m\sigma = \frac{j}{E} = \frac{3.57 \times 10^9}{50} \approx 7.14 \times 10^7\,\text{S/m}

Answer: (a) vd≈0.263 m/sv_d \approx 0.263\,\text{m/s} (b) j≈3.57×109 A/m2j \approx 3.57 \times 10^9\,\text{A/m}^2 (c) σ≈7.14×107 S/m\sigma \approx 7.14 \times 10^7\,\text{S/m}