We’re going to take a break from the LEDs for a bit, don’t worry they’ll be back soon. First, though, we need to take the time to define and understand two very important concepts: series and parallel circuits.
Series Circuits
When we say things are in series in normal language, what we mean is that they come one after another. Likewise in electrical circuits, components in series are connected one after the other in a single, continuous path. We’ve already been working with series circuits, like our basic LED setup:
Here we would say the the LED and resistor are in series because they are part of the same loop connecting 5V to GND. We noticed in Lesson 3 that the voltage drops across the LED and resistor (2V and 3V, respectively) always added up to 5V. This should make sense: the LED and resistor “use up” the voltage provided by the power supply, what does up must come down. This property of electricity is called Kirchoff’s Voltage Law which is usually written like this:
The algebraic sum of all voltages in a loop must equal zero
By algebraic, we mean that we account not only for the amount of voltage, but also the sign (positive/negative). Power supplies provide “positive voltage”, while voltage drops are considered “negative voltage”.
VSS + VLED + VR1
= 5V + (-2V) + (-3V)
= 0V
In our LED circuit, because of the special nature of LEDs using up a fixed voltage no matter what (AKA its forward voltage), the voltage drops were always the same, but what happens if instead we are just dealing with resistors? Does the voltage drop across each one in some consistent, predictable way as well?
To find out, lets start with this circuit:
Here we have 2 220Ω resistors in series. Remember the 200Ω resistors have the RED-RED-BROWN color code. The first thing I want you to do is measure Itot (the amount of current flowing through the loop). Remember to measure current:
- put the multimeter in 200mA mode
- disconnect the circuit (between the two resistors is easiest)
- make sure you power supply is on
- pretending the two probes of the multimeter are just the two ends of a wire, use them to reconnect the circuit.
You should see something around 11mA, record your measurement in your workbook.
Same-Same But Different
Slight sidetrack, but it’s worth quickly discussing that from the electrons’ points of view, these two resistors just feel like one big resistor. This is known as equivalent resistance which is a single value of resistance that can replace any number of resistors in series without altering the values of the current or the voltage in the circuit. To figure out the equivalent resistor for our 2 220Ω resistors, we can use Ohm’s Law to figure it out. Rearranging for R, Ohm’s Law tells us:
R = V / I
R = 5V / 11mA
R = 5V / 0.011A
R = 440Ω
What’s interesting about this result is that 440Ω is exactly 220Ω + 220Ω, in other words the equivalent resistance of two or more resistors in series is simply the sum of their individual resistances.
Back on Track
Making sure to first switch your multimeter back into 20V mode, measure and record the voltage drops VR1 and VR2, and then verify Kirchoff’s Voltage Law by adding those two values together, record this result as well.
There are two things to note from this exercise. First, you should have found that VR1 + VR2 = ~5V. This should come as no suprise as that’s literally the phenomenon we were just talking about. You will also notice that both VR1 and VR1 are pretty much identical, ~2.5V. This is probably not surprising as it follows from reason: the resistors are equally difficult for the electrons to move through, so an equal amount of energy is lost in the process.
What do you think would happen to the voltage drops if R1 and R2 weren’t equal? Let’s find out with a quick adjustment to our circuit:
First, as a follow up to our discussion on equivalent resistance, let’s work backwards and see if we can predict the current in this circuit (Itot). First let’s calculate the equivalent resistor to our 220Ω and 470Ω series resistors:
Rtot = R1 + R2
Rtot = 220Ω + 1000Ω
Rtot = 1220Ω
Record this in your workbook. Using this result along with our known VSS of 5V, we can rearrange Ohm’s Law to solve for I:
I = V / R
I = 5V / 1220Ω
I = 0.004098…A
I = ~0.004A
I = ~4mA
Record this in your workbook, and go ahead and measure and record Itot and see how close our prediction of 4mA is. It likely won’t be exactly correct, remember that close enough is what we’re all about in electronics. A small discrepency is due to a few factors:
- we rounded the result of our calculation
- resistors are not exactly their official resistance, plus or minus 15% is normal
- your power supply isn’t providing exactly 5V
As long as your measurement was close, you’re probably on the right track.
Now back to Kirchoff’s Voltage Law. Again measure and record VR1 and VR2 and calculate and record VR1+VR2. Hopefully your measurements added up to 5V once again (try again or ask a friend for help if not), but this time the individual voltage drops were different.
You should have found that VR2 was quite a bit more than VR1, does that agree with what you assumed would happen? This tells us that when resistors are in series, the amount of voltage each one “uses up” is proportional to their resistance, compared to the other resistors. Put simply: bigger resistors hog more of the available voltage.
Looking even closer, you might notice that VR2 is almost exactly 4 times bigger than VR1, just as R2 (1000Ω) is just a bit more than 4 times bigger than R1 (220Ω). This is no coincidence (which are very rare in this field). The distribution of voltage drops across resistors in series is exactly proportional to their relative resistances.
This is some heavy stuff to ponder, so we’ll leave it at that for today. Next time we look at parallel circuits and the interesting way they mirror and invert many of the concepts we’ve studied so far.