Two footprints in the sand are enough to tell where the third will fall.
⟨ transmission ends ⟩The resistor in First Light was 470 Ω, and it kept the LED from burning out. Why 470 Ω, and not 220 Ω or 1000 Ω? You could try each one and burn a few LEDs finding out. There is a better way. We are going to learn how to calculate the right resistor, and it is very simple.
To calculate the right resistor, we should first know what the current is, so we need an ammeter. Here is a typical desk ammeter, the kind we can use on our bench.
Again, at a real bench we usually use a multimeter instead, which has more options in one place. To keep this lesson simple, we use an ammeter for now.
On the bench below, the ammeter is joined between the cell’s + end and a 1000 Ω resistor. An ammeter must be part of the circuit and act like a wire, so all the current goes through it. Three more cells are in the tray, and each one adds 1.5 V. How much do you think the current will rise with each cell added?
The cells here each give 1.5 V. Real cells give a little less while a current flows.
Every cell you added moved the needle up by the same 1.5 mA, so four times the voltage gave four times the current.
We will learn how to read a resistor’s value in The Resistor. For now, we show each resistor’s value beside it. That value is called its resistance. On the bench below there are a few resistors, so try each of them. What do you think will happen to the current when the resistance is bigger? Will it go up or down?
The four cells give 6 V throughout. Only the resistor changes.
As you checked, at the same voltage, doubling the resistance gave half the current. On the first bench, with the same resistance, doubling the voltage gave double the current. That gives us an obvious relation between the three. It is that simple.
Divide the voltage by the resistance, and you get the current. This is Ohm’s law, named after Georg Ohm, who measured it in the 1820s.
Resistance has a unit of its own, the ohm, written Ω and named after Ohm as well. One ohm is the resistance that lets 1 A through when the voltage is 1 V.
Both lines match the ammeter. All those zeros get in the way, so a thousand ohms is called a kilohm, written kΩ.
With Ohm’s law, we can know the current before we even connect the circuit. On the bench below there are four cells, 6 V in total, and a 1.5 kΩ resistor. The black lead is not connected yet. Calculate the current and set it on the slider. Then connect the black lead and see if you were right.
1.5 kΩ is 1500 Ω. Until the black lead touches the resistor, the circuit is open, nothing flows and the needle rests at 0.
As soon as the circuit closed, the ammeter read 4 mA, and 6 divided by 1.5 is 4. You knew the current before any current flowed. That means you can size a circuit on paper, instead of finding out by burning parts.
You can also run the sum the other way. If you know the voltage and the current, dividing one by the other gives the resistance. The resistor below is hidden under a sleeve, but the cells give 6 V and the ammeter reads 8 mA. What is under the sleeve?
8.0 mA is 0.008 A. Resistors are sold only in certain values, and the hidden one is one of them.
When we know two of the three numbers, we can always calculate the third. For the voltage, we multiply the current by the resistance.
The LED does not follow the sum the way a resistor does. Whatever current flows through it, a red LED like this one keeps about 1.9 V of the battery’s voltage for itself. The resistor gets what is left. From a 9 V battery, that leaves 7.1 V for the resistor. This LED is comfortable at 15 mA.
Resistors are only made in certain values, so we use the nearest one, 470 Ω. Put in the one you calculated, then try the others too.
The 9 V battery has about 2 Ω inside it, and the currents here include it. The red LED has 1.9 V across it. A small LED like this one is usually rated for about 25 mA at most, and 47 Ω lets about 145 mA through. This ammeter’s scale runs to 50 mA, so at 145 mA its needle hits the end.
The 470 Ω resistor gave 15 mA, the same current as in First Light. The 47 Ω resistor is what you get if you make a mistake with the zeros and divide by 0.15 A instead of 0.015 A. It let about 145 mA through, far more than the LED can carry, and the LED burnt out. Do not try this with real parts. With 1 kΩ the LED was safe, but dim.
Four questions, if you want to check yourself.
A 9 V battery sits behind a 1.8 kΩ resistor. What current flows?
The voltage across a resistor is doubled. What happens to the current through it?
A 5 V supply feeds a red LED that keeps 2 V. Which resistor gives it 10 mA?
A part passes 3 mA with 6 V across it. What is its resistance?
On a real resistor, the value is shown as coloured stripes. How do you read them? That is the next lesson, The Resistor.
Lesson 4 done.
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