3.1 - Electric Current

What Current Is

Current is the rate at which electric charge passes through a cross-sectional area of a wire.

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

where: -

I

= current (amperes, A) -

\Delta q

= charge passing through (coulombs, C) -

\Delta t

= time interval (seconds, s)

The unit is the ampere: 1 A = 1 coulomb per second (C/s).

This is on the AP equation sheet exactly as

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

. The elementary charge,

e = 1.60 \times 10^{-19}

C, is in the sheet’s constants table, so you can turn a number of electrons

n

into a charge with

\Delta q = ne

What Causes Current?

Charges move through a circuit in response to an electric potential difference (voltage), which is also called electromotive force (emf) and written as

\mathcal{E}

Without a potential difference, there’s no net current. The potential difference sets up an electric field that pushes on the charge carriers and makes them drift in a particular direction.

Example: connect a lightbulb between two points that are both at 5.0 V. There’s no potential difference across the filament, so there’s no electric field along it to push the charge carriers, and the bulb stays dark. Connect it between a point at 5.0 V and a point at 0 V instead, and the 5.0 V potential difference drives a current through the filament from the 5.0 V side toward the 0 V side.

Zero Current Doesn’t Mean Zero Motion

If the current in a wire is zero, that means the net motion of the charge carriers is zero. The individual carriers are still zipping around randomly at high speeds because of thermal energy. They just have no net displacement in any particular direction.

Picture people in a crowded room moving around randomly: there’s plenty of motion, but no overall flow in any direction.

Direction of Current

Current isn’t a vector, but it does have a direction attached to it.

The direction only tells you which way charge moves along the wire. It isn’t tied to an

x

y

coordinate system, so currents don’t add like vectors. Example: a wire carrying 2.0 A comes in from the left, a wire carrying 3.0 A comes in from below, and they meet at a point where a single wire carries the charge away. The outgoing current is

2.0 + 3.0 = 5.0

A, not

\sqrt{2.0^2 + 3.0^2} \approx 3.6

A. The angle between the wires doesn’t matter, because all the charge that arrives has to leave (this is the junction rule in 3.7).

The conventional current direction is the direction positive charges would move.

Reality vs. Convention

Conventional current flows from the positive terminal to the negative terminal.

Actual electron flow goes the other way, with electrons (negative charges) moving from the negative terminal to the positive terminal.

Inside the battery, conventional current goes from the − terminal to the + terminal. The battery does work on the charges to raise them to a higher potential, which is what keeps the current going around the loop.

In most circuits the current is really electrons moving, but we still use the conventional direction for all our analysis and problem-solving. This convention got locked in before scientists knew electrons were the actual charge carriers.

Example: in a wire,

2.5 \times 10^{18}

electrons drift to the left past a point every 0.20 s. Find the current and its direction.

I = \frac{\Delta q}{\Delta t} = \frac{ne}{\Delta t} = \frac{(2.5 \times 10^{18})(1.60 \times 10^{-19} \text{ C})}{0.20 \text{ s}} = 2.0 \text{ A}

Answer: 2.0 A, directed to the right. The electrons are negative and move left, so the conventional current points the opposite way, the way positive charge would have to move to carry the same current.

Practice Problems

1. A wire carries a current of 2.5 A. How much charge passes through a cross-section of the wire in 4.0 seconds?

2. If

3.6 \times 10^{19}

electrons pass a point in a circuit in 3.0 seconds, what is the current?

3. In a wire with zero net current, explain how the electrons are moving.

4. A current of 0.50 A flows through a resistor for 2.0 minutes. How many electrons pass through the resistor in that time?

Solutions

Use the definition of current and rearrange for charge:

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

\Delta q = I \cdot \Delta t = (2.5 \text{ A})(4.0 \text{ s}) = 10 \text{ C}

Answer: 10 coulombs of charge pass through the wire.

First find the total charge from the number of electrons and the elementary charge

e = 1.6 \times 10^{-19}

\Delta q = n \cdot e = (3.6 \times 10^{19})(1.6 \times 10^{-19} \text{ C}) = 5.76 \text{ C}

Then use the current equation:

I = \frac{\Delta q}{\Delta t} = \frac{5.76 \text{ C}}{3.0 \text{ s}} = 1.92 \text{ A}

Answer: the current is 1.92 A.

3. Even with zero net current, the electrons aren’t sitting still. They’re in rapid, random thermal motion at high speeds. Their net displacement is zero, though, because there’s no applied potential difference to give them a preferred direction.

For every electron heading one way, another electron heads the opposite way, so the net flow of charge cancels to zero. The average velocity of all the carriers is zero, but their individual speeds are definitely not zero.

First convert the time to seconds:

\Delta t = 2.0 \text{ min} \times \frac{60 \text{ s}}{1 \text{ min}} = 120 \text{ s}

Find the total charge transferred:

\Delta q = I \cdot \Delta t = (0.50 \text{ A})(120 \text{ s}) = 60 \text{ C}

Find the number of electrons using the elementary charge:

n = \frac{\Delta q}{e} = \frac{60 \text{ C}}{1.6 \times 10^{-19} \text{ C}} = 3.75 \times 10^{20} \text{ electrons}

Answer:

3.75 \times 10^{20}

electrons pass through the resistor.

Common Mistakes

Confusing zero current with zero motion. Zero current means zero net motion, not that the charges are frozen in place. Individual charges still have random thermal motion.

Mixing up conventional current and electron flow. Conventional current is defined as the way positive charges would move, which is the opposite of the actual electron motion in typical circuits.

Treating current as a vector. Current has a direction but it’s a scalar, so it doesn’t follow the rules of vector addition and isn’t tied to coordinate systems.

Unit errors. Always convert time to seconds when you use $I = \frac{\Delta q}{\Delta t}$ , and remember that 1 A = 1 C/s.

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