State that conventional current is from positive to negative and that the flow of free electrons is from negative to positive

4.2.2 Electric Current

Objective

State that conventional current is defined as flowing from the positive terminal to the negative terminal of a source, and that the actual flow of free electrons in a metallic conductor is in the opposite direction (from negative to positive).

Key Concepts

• Conventional current direction is a historical convention.
• In metals, the charge carriers are free electrons, which carry a negative charge.
• The drift velocity of electrons is very small, but the electric field propagates at near‑light speed, giving the appearance of instantaneous current.
• Current (\$I\$) is defined as the rate of charge flow: \$I = \frac{\Delta Q}{\Delta t}\$ measured in amperes (A).

Why the Convention?

When the concept of electric current was first introduced, the nature of the charge carriers was unknown. Scientists chose the direction from positive to negative as the reference. Later it was discovered that in most conductors the mobile charge carriers are electrons, which move opposite to the chosen direction. The convention was retained for consistency.

Direction of Charge Flow

In a simple circuit with a battery:

1. Positive terminal of the battery is considered the source of conventional current.
2. Conventional current leaves the positive terminal, passes through the external circuit, and returns to the negative terminal.
3. Free electrons in the metallic wires move from the negative terminal toward the positive terminal, opposite to the conventional current.

Comparison of Directions

Mathematical Relationship

The magnitude of current is the same regardless of which direction is chosen; only the sign changes. If \$n\$ electrons each carry charge \$e = 1.60 \times 10^{-19}\,\text{C}\$ and move with drift speed \$v_d\$ through a cross‑sectional area \$A\$, the current magnitude is:

\$I = n e A v_d\$

When using conventional current, the sign is taken as positive for flow from + to –.

Example Problem

Problem: A copper wire of cross‑section \$2.0 \times 10^{-6}\,\text{m}^2\$ carries a current of \$3.0\,\$A. The free electron density in copper is \$8.5 \times 10^{28}\,\text{m}^{-3}\$. Determine the average drift speed of the electrons.

1. Use \$I = n e A vd\$ and solve for \$vd\$:

   \$v_d = \frac{I}{n e A}\$

2. Substitute the values:

   \$v_d = \frac{3.0}{(8.5 \times 10^{28})(1.60 \times 10^{-19})(2.0 \times 10^{-6})}\$

3. Calculate:

   \$v_d \approx 1.1 \times 10^{-4}\,\text{m s}^{-1}\$

The electrons drift very slowly (about 0.1 mm per second) opposite to the conventional current direction.

Summary

• Conventional current flows from positive to negative.
• In metallic conductors, free electrons move from negative to positive.
• The convention remains useful for circuit analysis and diagramming.
• Current magnitude is given by \$I = \Delta Q / \Delta t\$ and is independent of the chosen direction.