ElectricityTopic #28 of 35

Current and Resistance

Electric current, Ohm's law, resistance, resistivity, and power dissipation.

Overview

Electric current is the flow of electric charge. Resistance is the opposition to current flow. These concepts are fundamental to understanding electric circuits.

Electric Current

Definition

Rate of charge flow:

I=dQdtI = \frac{dQ}{dt}

For steady current:

I=QtI = \frac{Q}{t}

Unit: Ampere (A) = C/s

Direction

  • Conventional current: direction positive charges would flow
  • Electron flow: opposite to conventional current
  • In metals, electrons carry the charge

Current Density

Current per unit area:

J=IAJ = \frac{I}{A}

Unit: A/m²

Drift Velocity

Average velocity of charge carriers:

vd=InAqv_d = \frac{I}{nAq}

Where:

  • nn = charge carrier density (carriers/m³)
  • AA = cross-sectional area
  • qq = charge per carrier

Relationship to Current

I=nAvdqI = nAv_d q

Note: Drift velocity is very slow (mm/s), but electrical signals travel near light speed!

Resistance

Ohm's Law

V=IRV = IR

Or:

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

Unit: Ohm (Ω) = V/A

Ohmic vs Non-Ohmic Materials

  • Ohmic: RR is constant (linear V-I relationship)
  • Non-ohmic: RR varies with VV or II (diodes, bulbs)

Resistivity

Property of the material:

R=ρLAR = \frac{\rho L}{A}

Where:

  • ρ\rho = resistivity (Ω·m)
  • LL = length
  • AA = cross-sectional area

Resistivity Values

Materialρ\rho (Ω·m) at 20°C
Silver1.59×1081.59 \times 10^{-8}
Copper1.68×1081.68 \times 10^{-8}
Aluminum2.65×1082.65 \times 10^{-8}
Tungsten5.6×1085.6 \times 10^{-8}
Iron9.7×1089.7 \times 10^{-8}
Carbon3.5×1053.5 \times 10^{-5}
Glass1010101410^{10} - 10^{14}

Conductivity

σ=1ρ\sigma = \frac{1}{\rho}

Unit: S/m (siemens per meter)

Temperature Dependence

Metals

ρ(T)=ρ0[1+α(TT0)]\rho(T) = \rho_0[1 + \alpha(T - T_0)]

Or:

R(T)=R0[1+α(TT0)]R(T) = R_0[1 + \alpha(T - T_0)]

Where α\alpha = temperature coefficient of resistivity

Metals: α>0\alpha > 0 (resistance increases with temperature)

Semiconductors

Resistance decreases with temperature (α<0\alpha < 0)

Electric Power

Power Dissipation

P=IV=I2R=V2RP = IV = I^2R = \frac{V^2}{R}

Unit: Watt (W) = J/s

Energy Dissipation

E=PtE = Pt

Unit: Joule (J) or kilowatt-hour (kWh)

1 kWh=3.6×1061 \text{ kWh} = 3.6 \times 10^6 J

Electromotive Force (EMF)

Energy supplied per unit charge by a source:

ε=Wq\varepsilon = \frac{W}{q}

Real Battery

V=εIrV = \varepsilon - Ir

Where rr = internal resistance

Examples

Example 1: Current Calculation

5 Coulombs of charge flow through a wire in 2 seconds.

I=Qt=52=2.5 AI = \frac{Q}{t} = \frac{5}{2} = 2.5 \text{ A}

Example 2: Resistance from Ohm's Law

A 12 V battery drives 3 A through a resistor.

R=VI=123=4 ΩR = \frac{V}{I} = \frac{12}{3} = 4 \text{ Ω}

Example 3: Wire Resistance

A copper wire (ρ=1.68×108\rho = 1.68 \times 10^{-8} Ω·m) is 100 m long with diameter 2 mm.

A=π(0.001)2=3.14×106 m2A = \pi(0.001)^2 = 3.14 \times 10^{-6} \text{ m}^2 R=ρLA=1.68×108×1003.14×106=0.535 ΩR = \frac{\rho L}{A} = \frac{1.68 \times 10^{-8} \times 100}{3.14 \times 10^{-6}} = 0.535 \text{ Ω}

Example 4: Power in Resistor

A 100 Ω resistor carries 0.5 A.

P=I2R=0.25×100=25 WP = I^2 R = 0.25 \times 100 = 25 \text{ W}

Or:

V=IR=0.5×100=50 VV = IR = 0.5 \times 100 = 50 \text{ V} P=IV=0.5×50=25 WP = IV = 0.5 \times 50 = 25 \text{ W}

Example 5: Temperature Effect

A tungsten filament has R=20R = 20 Ω at 20°C. Find RR at 2500°C. (α=4.5×103\alpha = 4.5 \times 10^{-3} /°C)

R=R0[1+α(TT0)]R = R_0[1 + \alpha(T - T_0)] R=20[1+4.5×103×2480]R = 20[1 + 4.5 \times 10^{-3} \times 2480] R=20×12.16=243 ΩR = 20 \times 12.16 = 243 \text{ Ω}

Example 6: Real Battery

A battery (ε=12\varepsilon = 12 V, r=0.5r = 0.5 Ω) is connected to a 5.5 Ω resistor.

I=εR+r=126=2 AI = \frac{\varepsilon}{R + r} = \frac{12}{6} = 2 \text{ A} Vterminal=εIr=122×0.5=11 VV_{\text{terminal}} = \varepsilon - Ir = 12 - 2 \times 0.5 = 11 \text{ V} Pdelivered=I2R=4×5.5=22 WP_{\text{delivered}} = I^2 R = 4 \times 5.5 = 22 \text{ W} Plost=I2r=4×0.5=2 WP_{\text{lost}} = I^2 r = 4 \times 0.5 = 2 \text{ W}

Example 7: Drift Velocity

Copper wire (n=8.5×1028n = 8.5 \times 10^{28} electrons/m³) with diameter 2 mm carries 10 A.

A=π(0.001)2=3.14×106 m2A = \pi(0.001)^2 = 3.14 \times 10^{-6} \text{ m}^2 vd=InAq=108.5×1028×3.14×106×1.6×1019v_d = \frac{I}{nAq} = \frac{10}{8.5 \times 10^{28} \times 3.14 \times 10^{-6} \times 1.6 \times 10^{-19}} vd=2.3×104 m/s=0.23 mm/sv_d = 2.3 \times 10^{-4} \text{ m/s} = 0.23 \text{ mm/s}

Superconductivity

  • Below critical temperature: R=0R = 0
  • Perfect conductor
  • Applications: MRI, particle accelerators