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Topic 6 of 15

Current & Resistance

When a potential difference is maintained across a conductor, charges flow continuously — we call this electric current. The relationship between current, voltage, and the material's resistance is Ohm's law. Understanding how resistance depends on material, geometry, and temperature is essential for every practical circuit, from household wiring to microchips.

Key Concepts

Electric Current

The rate of charge flow past a cross section:

I = \Delta Q/\Delta t

. Unit: ampere (A = C/s). Conventional current flows from positive to negative terminal externally — opposite to the direction electrons move.

Resistance and Resistivity

Resistance

R

(ohms, Ω) opposes current flow. It depends on the material's resistivity

\rho

(Ω·m), the length

L

, and cross-sectional area

A

R = \rho L/A

. Longer or thinner conductors have higher resistance.

Ohm's Law

For ohmic (linear) materials, the current is proportional to the applied voltage:

V = IR

, or equivalently

I = V/R

. The resistance

R

is constant regardless of

V

I

for ohmic devices. Many devices (diodes, bulbs) are non-ohmic.

Temperature Dependence

For most metals, resistivity increases linearly with temperature:

R = R_0(1 + \alpha\,\Delta T)

, where

\alpha

is the temperature coefficient of resistance (positive for metals, negative for semiconductors) and

\Delta T = T - T_0

Electric Power

The rate at which a resistor converts electrical energy to heat:

P = IV = I^2R = V^2/R

. Unit: watt (W). Energy dissipated over time

t

E = Pt

Key Equations

Electric current

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

Current in amperes equals charge (coulombs) flowing per second.

Resistance from geometry

R = \frac{\rho L}{A}

R in ohms; ρ = resistivity (Ω·m); L = length; A = cross-sectional area.

Ohm's Law

V = IR

Voltage across a resistor equals current times resistance. Valid for ohmic (linear) devices.

Electric power

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

Power dissipated in a resistor. Use whichever form matches what is given.

Temperature dependence

R = R_0\bigl(1 + \alpha\,\Delta T\bigr)

R₀ is resistance at reference temperature T₀; α is temperature coefficient (e.g., copper: α ≈ 3.9×10⁻³ /°C).

Worked Example

Resistance, Current, and Power in a Copper Wire

Problem

A copper wire ( $\rho = 1.72\times10^{-8}\,\Omega\cdot$ m) has length $L = 10$ m and diameter $d = 1.0$ mm. Find (a) its resistance, (b) the current if connected to a 1.5 V battery, and (c) the power dissipated.

Solution

(a) Cross-sectional area: $A = \pi(d/2)^2 = \pi(0.50\times10^{-3})^2 = 7.85\times10^{-7}$ m². Resistance:

R = \frac{\rho L}{A} = \frac{(1.72\times10^{-8})(10)}{7.85\times10^{-7}} = 0.219\,\Omega

(b) Current by Ohm's Law:

I = \frac{V}{R} = \frac{1.5}{0.219} = 6.85\text{ A}

P = IV = (6.85)(1.5) = 10.3\text{ W}

Answer R = 0.219 Ω; I = 6.85 A; P = 10.3 W.

Practice

Exercises

7 problems

1 of 7

A charge of $Q = 120$ C passes through a wire cross-section in $t = 2.0$ minutes. What is the current (in A)?

Your answer

2 of 7

A nichrome wire ( $\rho = 1.10\times10^{-6}\,\Omega\cdot$ m) is $L = 2.0$ m long with cross-sectional area $A = 0.50\times10^{-6}$ m². What is its resistance (in Ω)?

Your answer

3 of 7

A $30\,\Omega$ resistor is connected to a $9.0$ V battery. What current (in A) flows through it?

Unlock Exercise 3