Did you Know?
The various forms of the Ohm’s Law equations are used in many ways. In this section, you will see:
- A quick math trick so that you only have to remember one version of the equation
- An example of how it is used in an electronic design
- How it dictates the relationship of the units V, A, and Ω
Ohm’s Law Equation Memory Trick
Although there are numerous memory tricks for remembering the versions that solve for I and R, you can also just remember V = I x R and then divide by both sides to isolate either I or R. In other words, if you’re solving for I, divide R into both sides of V = I x R, and the result is I = V / R. Or, if you’re solving for R, divide both sides by I for R = V / I.
(View full size: ol-solve-for-i-or-r.mp4)
Ohm’s Law: “The current through a conductor between two points is directly proportional to the voltage across the two points.”
That translates directly to I = V / R, where (1 / R) is the “directly proportional” constant you can multiply by the voltage to calculate current. Ohm’s law uses the term “two points” to make it more general. Sure, a point at each lead of a resistor is two points, but it can also apply to points on long wires. A long wire, like a power line, has a very small resistance per length. The longer the wire, the greater the resistance.
Calculate Resistance to Get the Most Current
Earlier, you experimented with changing resistors to make the light dimmer or brighter. The smaller resistors allow more current to flow through the circuit, making the light brighter. One of the goals in a prototype or project, might be to make the light as bright as it can be. This can be done checking the current limitations, then selecting a resistor that will make the circuit conduct the most current within those limitations.
According to the Edge Connector & micro:bit pinout, the V2 module’s 3.3 V supply can source up to 270 mA of current. But, the LED’s maximum current is 20 mA, so that is the limiting factor. So, if you were designing a device and needed the brightest light possible, here is how you would use Ohm’s law to calculate the smallest resistor you can safely use (without damaging the LED by exceeding its current specification).
LEDs have a property called forward voltage, and it changes a little with current, but not much. So let’s assume the voltage drop across it at 20 mA will still be about 2.1 V, like we tested in Measure Voltage. That means, the voltage across the resistor will still be about 1.2 V because they still have to add up to 3.3 V. Again, that’s because Kirchhoff's voltage law (KVL) says that the voltages across the components have to add up to the voltage of the supply.
R = V / I
= 1.2 V / 0.020 A
= 60 Ω
Important: Only use a resistor that small in an LED circuit if you are drawing power from the 3.3 V and GND rails on your breadboard. A micro:bit I/O pin cannot even supply 5 mA to an LED circuit with a 220 Ω without its voltage sagging.These are rough estimates, and in product designs, derating is often applied to ensure none of the parts ever fail by being too close to their specification maximums or minimums. For example, you might end up repeating the resistor calculations using 15 mA to be on the safe side.
Unit Equalities from Ohm’s Law
Since the unit for voltage is V, the unit for current is A, and the unit for resistance is Ω, Ohm’s law also tells us how V, A, and Ω relate:
1 A = 1 V / Ω
1 V = 1 A x Ω
1 Ω = 1 V / A
Your Turn
- Use what you have learned to create a script that calculates resistance from current and voltage.
- Use a calculator to verify the results from your script.
- Make sure to name the script calculate_r_from_v_and_i, and save your script.
- Don’t worry about displaying omega Ω for the units. Just use “ohms” instead.