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=dtdQ
For steady current:
I=tQ
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=AI
Unit: A/m²
Drift Velocity
Average velocity of charge carriers:
vd=nAqI
Where:
- n = charge carrier density (carriers/m³)
- A = cross-sectional area
- q = charge per carrier
Relationship to Current
I=nAvdq
Note: Drift velocity is very slow (mm/s), but electrical signals travel near light speed!
Resistance
Ohm's Law
V=IR
Or:
R=IV
Unit: Ohm (Ω) = V/A
Ohmic vs Non-Ohmic Materials
- Ohmic: R is constant (linear V-I relationship)
- Non-ohmic: R varies with V or I (diodes, bulbs)
Resistivity
Property of the material:
R=AρL
Where:
- ρ = resistivity (Ω·m)
- L = length
- A = cross-sectional area
Resistivity Values
| Material | ρ (Ω·m) at 20°C |
|---|
| Silver | 1.59×10−8 |
| Copper | 1.68×10−8 |
| Aluminum | 2.65×10−8 |
| Tungsten | 5.6×10−8 |
| Iron | 9.7×10−8 |
| Carbon | 3.5×10−5 |
| Glass | 1010−1014 |
Conductivity
σ=ρ1
Unit: S/m (siemens per meter)
Temperature Dependence
Metals
ρ(T)=ρ0[1+α(T−T0)]
Or:
R(T)=R0[1+α(T−T0)]
Where α = temperature coefficient of resistivity
Metals: α>0 (resistance increases with temperature)
Semiconductors
Resistance decreases with temperature (α<0)
Electric Power
Power Dissipation
P=IV=I2R=RV2
Unit: Watt (W) = J/s
Energy Dissipation
E=Pt
Unit: Joule (J) or kilowatt-hour (kWh)
1 kWh=3.6×106 J
Electromotive Force (EMF)
Energy supplied per unit charge by a source:
ε=qW
Real Battery
V=ε−Ir
Where r = internal resistance
Examples
Example 1: Current Calculation
5 Coulombs of charge flow through a wire in 2 seconds.
I=tQ=25=2.5 A
Example 2: Resistance from Ohm's Law
A 12 V battery drives 3 A through a resistor.
R=IV=312=4 Ω
Example 3: Wire Resistance
A copper wire (ρ=1.68×10−8 Ω·m) is 100 m long with diameter 2 mm.
A=π(0.001)2=3.14×10−6 m2
R=AρL=3.14×10−61.68×10−8×100=0.535 Ω
Example 4: Power in Resistor
A 100 Ω resistor carries 0.5 A.
P=I2R=0.25×100=25 W
Or:
V=IR=0.5×100=50 V
P=IV=0.5×50=25 W
Example 5: Temperature Effect
A tungsten filament has R=20 Ω at 20°C. Find R at 2500°C. (α=4.5×10−3 /°C)
R=R0[1+α(T−T0)]
R=20[1+4.5×10−3×2480]
R=20×12.16=243 Ω
Example 6: Real Battery
A battery (ε=12 V, r=0.5 Ω) is connected to a 5.5 Ω resistor.
I=R+rε=612=2 A
Vterminal=ε−Ir=12−2×0.5=11 V
Pdelivered=I2R=4×5.5=22 W
Plost=I2r=4×0.5=2 W
Example 7: Drift Velocity
Copper wire (n=8.5×1028 electrons/m³) with diameter 2 mm carries 10 A.
A=π(0.001)2=3.14×10−6 m2
vd=nAqI=8.5×1028×3.14×10−6×1.6×10−1910
vd=2.3×10−4 m/s=0.23 mm/s
Superconductivity
- Below critical temperature: R=0
- Perfect conductor
- Applications: MRI, particle accelerators