## Electric Current

Current measures the flow of charge over a period of time. Current is not a substance. Current is a rate, like dollars per hour is a rate.

It can be helpful, and accurate, to think of electric current like water current. Free electrons flow through metals just like H₂O flows through pipes.

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

\(I\) = electric current [A, amps, Amperes]\(q\) = charge that passes a point on a wire [C, coulombs]

\(\Delta t\) = a period of time [s]

**Example:**Every minute 1.2 coulombs of charge passes through a typical LED.

What is the current in the LED?

## solution

$$I = \frac{q}{\Delta t}$$ $$I = \frac{1.2\, \mathrm{C}}{60\, \mathrm{s}}$$ $$ I= 0.02 \, \mathrm{\tfrac{C}{s}}$$**Example:**Two wires of cross-sectional area 1.6 mm² connect the terminals of a battery to the circuitry in a clock. After 0.04 seconds, 5.0 x 10¹⁴ electrons move to the right through the cross section. What is the magnitude and direction of the current?

## hint

Build a conversion fraction with the charge of an electron divided by one electron.

## solution

$$q = 5.0 \times 10^{14} \, e^{-} \left(\frac{1.6 \times 10^{-19} \, \mathrm{C}}{1 \, e^{-}}\right) = 0.00008\, \mathrm{C}$$$$I = \frac{q}{\Delta t}$$ $$I = \frac{0.00008\, \mathrm{C}}{0.04 \, \mathrm{s}}$$ $$I = 0.002 \, \mathrm{A} $$ $$\text{if electrons are moving to the right the current is to the left}$$

**Example:**If 0.320 mA of charge flow through a calculator, how many electrons pass through per second?

## solution

$$ I = \frac{q}{\Delta t}$$ $$ q = I\Delta t$$ $$ q = (0.320 \times 10^{-3}\, \mathrm{A}) (1\, \mathrm{s})$$$$ q = 0.000320 \, \mathrm{C} \left(\frac{1 \, e^{-}} {1.6 \times 10^{-19}\, \mathrm{C}}\right) $$ $$ q = 2 \times 10^{15} \, e^{-}$$

## Charge Speed and Direction

Electric current can be thought of as the average motion of a huge number of charges. As the charges travel they don't just move in a straight line. They drift through the material in a random walk as they bounce off other charges. Statistically more charge will move forward than backwards.

The forward speed of electrons in a typical wire is around 0.0002 m/s. If the speed of electrons in is so slow how can we explain the near instantaneous speed of communication on telegraph, phone, and internet lines?

The direction of current is defined as the direction of positive charge flow. Since electrons have a negative charge they always move in the opposite direction of the current. Why don't the nuclei move?

What direction is the current in these simulations?

A small imbalance in net charge produces an electric field. The electric field points in the same direction as the current.

Positive charge flows from positive(red) potential to negative(blue) potential. Negative charge does the opposite.

## Voltage (Potential Difference)

Voltage is a measure of the difference in electrical potential between two points. Voltage is defined so that negatively charged objects are pulled towards higher voltages, while positively charged objects are pulled towards lower voltages. You can review the notes on voltage from the last section

Common sources of differences in potential (voltage) are the two terminals in a wall socket and the positive and negative terminals on a battery.

In these simulations I added a "battery" that pushes the electrons down.

Areas with more negative charge have negative potential, shown as
blue.

Areas with more positive charge have positive potential, shown as
red.

The simulation on the right is a
**circuit**. A
**circuit** is a looping path that guides the electrons back to the start, like a race track.

## AC⚡DC

In
**DC (direct current)** circuits current is defined to flow out of the longer, positive end of the battery, but
electrons actually move in the opposite direction. Below is an animated
circuit diagram of a battery providing DC voltage and a resistor dropping the voltage:

In
**AC (alternating current)** circuits the current pushes the electrons forward and backwards about 60 times a second.
Current from a wall socket is AC, while current from a battery is DC.

## Ohm's Law

Ohm's Law says that the voltage in a conductor is proportional to the current. This means we can build an equation
where V = I times a constant, resistance. The
**resistance** of a material is the inverse of its conductivity.
Conductors have nearly zero resistance and insulators have very high resistance.

Semiconductive elements, called resistors, have medium resistance. They are added to a circuit for precise control over current. If the current is too high, the circuit will get hot. If the current is too low the circuit can't do its job.

# $$ V = IR$$

\(I\) = electric current [A, amps, ampere]\(V\) = voltage, a change in electric potential [V, volts]

*Technically ΔV, but everyone uses V for simplicity*

Sources: battery, solar panel, wall socket

\(R\) = resistance [Ω, ohms]

Sources: resistor, light, motor

Play with this circuit simulator for resistance. Adjust the wiggle of the resistor to see how it changes the voltage and current.

**Example:**Find the voltage for a resistor that has 300 Ω of resistance and a current of 0.05 A.

## solution

$$ V = IR$$ $$ V = (0.05\, \mathrm{A}) (300 \, \Omega)$$ $$ V = 15 \, \mathrm{V}$$**Example:**Find the current for a 100 Ω resistor with 2 V drop?

## solution

$$ V = IR$$ $$ I = \frac{V}{R}$$ $$ I = \frac{2 \, \mathrm{V}}{100 \, \Omega}$$ $$ I = 0.02\, \mathrm{A}$$**Example:**If I double the voltage in a circuit while keeping the resistance the same, what happens to the current? Use proportions.

## solution

$$ \text{R is constant}$$ $$ \text{V and I are directly proportional}$$ $${ \Uparrow \atop V} {\atop =} { \Uparrow \atop I} { \atop R} $$ $$ \text{doubling V will double I}$$**Example:**What value resistor would drop 1.5 V in a 0.001 A current?

## solution

$$ V = IR$$ $$ R = \frac{V}{I}$$ $$ R = \frac{1.5\, \mathrm{V}}{0.001\, \mathrm{A}}$$ $$ R = 1500 \, \Omega$$**Example:**What voltage would cause 0.02 C of charge to pass through a 10 Ω resistor every 10 s?

## solution

$$ I = \frac{q}{\Delta t}$$ $$ I = \frac{0.02}{10}$$ $$ I = 0.002 \, \mathrm{A}$$$$ V = IR $$ $$ V = (0.002)(10)$$ $$ V = 0.02 \, \mathrm{V}$$