Engineer's Notebook §1.1

Ohm's law

In a resistor at constant temperature, the current is proportional to the voltage across it. The constant is the resistance:

V = IR
V
voltage
V
I
current
A
R
resistance
Ω
G
conductance, 1/R
S
P
power
W
ρ
resistivity
Ω·m
L
conductor length
m
A
cross-sectional area
m²
Left, a battery driving current around a loop through a resistor. Right, the straight current–voltage line of the resistor with the operating point marked. V = 12 V R = 100 Ω I = 120 mA VI 50 0.5
Fig. 1.1-1The current–voltage line is straight, with slope 1/R. Axes run to 50 V and 0.5 A.
G = 1/R10 mS
P = VI1.44 W
I = V/R120 mA

Worked example

Uses the values in Fig. 1.1-1. Change them and the working updates.

Given
  • Vvoltage across the resistor12 V
  • Rresistance100 Ω
Find

The current I.

Solution
I = V/R = 12 V / 100 Ω = 0.12 A
Answer
I = 120 mA

The resistor dissipates P = VI = 1.44 W as heat.

Related cases

  • Series

    R = R1 + R2

  • Parallel

    1R = 1R1 + 1R2

  • Voltage divider

    Vout = V R2R1 + R2

  • Resistance of a wire

    R = ρLA

Assumptions
  • The resistor is ohmic: its resistance stays constant. That holds for resistors and metal wires at steady temperature, but not for diodes, or lamps as they warm up.
  • The source and connecting wires are ideal, with no resistance of their own.
  • Direct current, or AC through a pure resistance.

Derivation
  1. In a metal, free electrons drift at a speed proportional to the electric field, so the current density is J = σeE, where σe is the conductivity.
  2. Along a uniform wire, E = V/L and J = I/A, so I/A = σeV/L.
  3. Rearrange: V = I · L/(σeA) = IR, with R = ρL/A and ρ = 1/σe.