G'day, we're going to talk about Ohm's law and some practical examples today. Ohm's law is the relationship between voltage, current, and resistance in an electrical circuit. But don't worry, we're not going to dwell too much on the theory, we're going to be doing some practical examples.

Ohm's law, or V equals IR, is such a useful equation that we've put it on this handy cheat sheet to show the relationships. So V equals IR can also be rearranged into I equals V on R, and R equals V on I. And that can be like a little bit, a little bit too much to remember, so here's a trick. If you rearrange V, I, R into a triangle, you can just cover the quantity that you want to solve for. For example, if I wanted to find the voltage across a resistor, and I knew its resistance and current flowing through the resistor, I want to find voltage, so I cover the V, and I'm left with I times R. Likewise, if I wanted to find the current through a known resistance with a known voltage across it, I would cover I to solve for V divided by R. And finally, to find an unknown resistance with known voltage and known current, I cover the R, and I'm left with voltage divided by current.

And now for a practical example. Here I have a circuit with a 470 ohm resistor that has 5 volts across it. If I want to solve for the current flowing through that resistor, I want to solve for I, the current, and that is equal to voltage divided by resistance. I equals V on R equals, that's 5 volts over 470 ohms. If I've divided by 470, the current flowing through the resistor is 0.01 amps. That's equal to 10 milliamps. In this next example, we have a known resistance with some voltage being applied across it, and we're measuring 0.5 amps flowing through that resistor. Resistor by using, say, a current meter. If we wanted to know this voltage, we can cover the V, because we're solving for voltage, and we get I times R. V equals I R equals 0.5 amps times 10 ohms. 0.5 times 10 is, of course, 5 volts. And here we can solve for an unknown resistance. Let's say we have a 12 volt power supply, and we know that it's providing 5 amps. We can find the resistance by covering R, and we get R is equal to V on I. R equals V on I, that is equal to 12 volts over 5 amps. 12 divided by 5 is 2.4, so that resistor must be 2.4 ohms.

Okay, so all that's well and good. We can calculate relationships between voltage, current, resistance, but how do we actually use that in a really honest-to-goodness practical maker circuit? Let me show you how Ohm's law can be applied to a problem that makers need to solve every day of the week, and that is choosing a resistor to power an LED.

Here I have a circuit with 9 volts across a resistor and an LED, a light emitting diode going to ground. Now, you need a resistor when you're powering an LED, because if you just connected voltage across an LED, it would instantly break. We need to choose a resistor that limits the current through the LED.

Now, I know that most regular LEDs, it's best not to power them with 10 milliamps. So let's say that I max is 10 milliamps, and this is I. That's one requirement. We want to make sure we're not passing more than 10 milliamps. I also know that a red LED will have a forward voltage of about 1.8 volts. I just know that from experience, but you can also check with a multimeter. This has 1.6 volts when there's no current flowing through it. When there's some current, that climbs a little bit. So we know the voltage across the LED. 

Will be about 1.8. We know the voltage across the LED, and we know the total voltage. That means the voltage across the resistor will just be the difference. That means that this voltage will be 9 minus 1.8, which is 7.2. All right, we're getting close. We have a voltage across an unknown resistance, and we have a theoretical maximum current that we don't want to go over. So let's just use that to find the minimum resistance. We want to solve for this resistance, so we cover R, and R is equal to V on I. R is equal to V on I.

The voltage we're working with is 7.2 volts across that resistor, and our maximum desired current is 10 milliamps, which means that our minimum resistance is 7.2 divided by 0.01 amps. Our minimum resistance is 720 ohms, so that means that if we use a 720 ohm resistor in this circuit, we will get about 10 milliamps. So here it is. I've built the circuit. I don't have a 720 ohm resistor, but that's the minimum resistance. So I've made this circuit using a 9 volt battery and a 1k, a 1000 ohm resistor. I'll power on the LED, and everything seems to be in order here.

So I've chosen a 1k resistor. I know that my current is going to be less than 10 milliamps, which was the maximum, but what will it actually be? Well, Ohm's law comes in again. The actual current, cover up I, is V on R, the chosen resistance. I equals V on R. We know that the voltage across the resistor is still going to be 7.2. 7.2 volts over 1000 ohms. 7.2 divided by 1000 is 0.0072, or 7.2 milliamps.

So here we've used Ohm's law to find the minimum resistance that we need to use for an LED, and that's going to be some theoretical value. We got 720 ohms. I just pulled a 1k resistor out of my parts bin because they're really common, I was able to then find what the actual current going through the LED would be using that higher resistance. Let's see how close we are.

I have my multimeter out, set to current, and we're looking for 7 milliamps in this circuit. If I measure the current, we get 0.007 amps. 7 milliamps. Nice. Now on the back of this card is another relationship for power. That is power is equal to voltage times current, and so we use this triangle in exactly the same way. Just cover up what you want to solve for, and then you can read the rest of the equation.

So here's a practical example where you might need to calculate power. We often use voltage regulators in maker electronics that take one voltage, for example 5 volts, and regulate it down to another voltage, let's say 3.3 volts. If I have a 3.3 volt circuit, and I know that I'm going to draw about half an amp through a 3.3 volt regulator from 5 volts, then I can calculate the power that will be dissipated by the regulator.

That's really useful because voltage regulators are often rated for a maximum power, say you know like 1 watt, half a watt, 5 watts, etc. So if I want to know how much power is dissipated by the regulator, I take the voltage difference V equals 5 minus 3.3, so V will be equal to 1.7. I know that I is equal to 0.5 amps, and so P power, I cover up P to solve, and that's voltage times current, which is 1.7 times 0.5, that's 0.85 watts. So if I were going to build this circuit, I would want to choose a voltage regulator that could handle at least, let's say a watt. There's also some extended formula on the bottom of the power side, which is P equals I squared R, and P equals V squared on R. This can be really useful as a shortcut for calculating power without having to use Ohm's law first.

For example, returning quickly to our LED circuit, remember we had 9 volts supplying an LED through a 1000 ohm resistor, and that gave us 7 milliamps. Resistors also have a maximum power rating, that might be say 1 watt, half a watt, a quarter of a watt. So to make sure we're not putting too much power through this resistor, we could use the extended formula P equals I squared R, because we already solved for the 7 milliamps. P equals I squared R equals 7 milliamps squared times 1000 ohms, 0.007 squared times 1000 equals about 50 milliwatts, and that's about 1 20th of a watt. So a quarter Watt resistor would be totally fine for this.

Hang on a second now, we actually calculated this current by using Ohm's law first. If we didn't actually know this, we could get here from just knowing the voltage. So if we knew that that voltage was 1.8 volts, then we know that the remaining voltage is 7.2 volts, that's where we started before using Ohm's law. So we could also use the other power formula P equals V squared on R, which is 7.2 volts on 1000, 7.2 volts squared on 1000 ohms, 7.2 squared divide by 1000, and we get to the same figure, about 50 milliwatts.

So this extended formula are really useful for calculating power without having to first go through Ohm's law to find one of the unknown quantities. So there you have it, some practical examples using Ohm's law and some formula for power. We were able to specify a resistor for our LED so we wouldn't damage our LED, and we were able to make sure that our linear regulator wasn't drawing too much power so that it wouldn't get too hot.

Until next time, thanks for watching.