DC Series Circuits:
In this article, we’re going to be looking at DC series circuits. We will cover voltage, current resistance, and power consumption as well as using a multimeter or Voltmeter and a chance for you to test your knowledge at the end.
When we connect electronic components in an electrical circuit we can connect them either in series or parallel or we can combine these to make a series parallel circuit. We’re going to start with a series type which is the most basic but we’ll cover the other types in other articles.
So if we place two electronic components in a line end-to-end, always some wires in between the knees are connected. In series circuits the electrons only have one path they can take so they will all flow through each of the electronic components. As you can see clearly in the image given below.
By the way in these circuits I use electron flow which is from negative to positive. You might be used to seeing conventional current which is from positive to negative. Electron flow is what’s actually occurring conventional was the original theory but it’s still taught because it’s easy to understand just be aware of the two on which one we’re using.
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Resistance in series circuits:
Each component will have a certain resistance the resistance opposes the voltage being applied we measure resistance in unit of ohms. In the circuit given above, you can see a resistor is connected in series with an LED “Light Emitting Diode”. The LED is turned ON using a 9V battery with the + and – signs printed. Why we needed to connect a resistor in series with this LED? How we know which value resistor should be selected? How we calculate this? These are the questions that you might be thinking about. Basically the LED is 2.5V and 20mA. If you directly control this LED using the 9V, the LED will be damaged in no time. Because the operating voltage is 2.5, while the supplied voltage is 9V, so as a result this LED will be damaged.
To safely operate this LED you have some options, like you can 2.5V voltage source, or you can use some more LEDs in the series, or you can use a resistor. I will continue with the resistor. You can easily calculate the value of the resistor if you know the LED voltage and current.
Led Voltage = 2.5V
Led Current = 20mA
Source Voltage = Battery Voltage = 9V
We can use the Ohm’s law to find the resistor value, that can safely drop the excess voltage, and this way the LED will remain safe.
V = IR
R = V / I
R = ( 9 – 2.5 ) / 20 mA
R = 325
The nearest value is the 330 Ohms. So to operate this LED using a 9V source, you will need to connect a 330 Ohm resistor in series.
You already know, resistance opposes the voltage being applied and the resistance is measured with the unit Ohm Ω.
In series circuits we find the total resistance for the circuit by simply adding together all the resistances.
We label each resistor with a capital R and number them R1, R2, and R3 etc. The reason we assign different numbers is, we can easily differentiate and easily point to the specific resistor. Anyhow, the total resistance is shown with a capital letter R and a subscript T which represents the resistance total or the total resistance. To calculate the total resistance of a series circuit is super easy we simply add together the resistance value of each resistor.
Let’s assume initially we have only one resistor R1 = 10 ohms connected with the battery + and – terminals. If I ask you, what is the total resistance, I am sure your answer will be 10 ohms. Now if I just go ahead and connected another resistor R2 = 5 ohms, now the total resistance will become 15 ohms as 10+5 = 15, what we did? We simply added the resistances of both the resistors. Now, if I go ahead and connect another resistor R3 = 5, no doubt the total resistor will become R1 + R2 + R3 = 10 + 5 + 5 = 20 ohms. You can add more resistors as you want, simply keep adding the resistors values to get the total resistance. Now that you know the basic concept of the total resistance and now you know how we find the total resistance, now let’s dig deeper. In reality the wires we are using to connect the resistors or other electronic components, these wires will also add some resistance, but usually for small length wires the resistance value is too small, that we normally neglect, but for longer wires make sure you take into account the wires resistance too. You can study about the wires voltage drop or cables voltage drop etc.
Current in series:
You know right from the beginning that current is the flow of electrons. it’s like water which flows through a pipe the higher the current the more electrons are flowing. We measure current in the unit of ampere’s but engineers tend to shorten it to just amps. We measure current by placing an ammeter into the circuit for the electrons to flow through this is like a water meter in the sense that water must pass through it for us to measure it. I have already explained the method of measuring the current using a Digitalmultimeter.
You know measuring voltage is simple as compared to measuring the current. For measuring the current the Ammeter or the multimeter should be connected in series. This way the current flowing through the circuit will also flow through the Digital Multimerter for the Ammeter connected. Of course the meter will add some resistance to the circuit under test but usually this resistance is too small and we usually ignore it.
Multimeters are really cool as they are using for measuring the voltage, resistance, continuity, short circuit, etc. So, I highly recommend you should get one fast. I have a provided the Amazon purchase link. In the image above you can see I am measuring the current flowing through the LED. A 400 ohm resistor is connected in series to limit the current and drop some voltage to keep the LED safe. Previously the value we calculated was 330 ohm, the reason I am using a slight bigger value resistor is to further ensure the smooth working of the LED. This will also increase the life span of the LED, as the LED will glow a bit less brighter, I am sure you won’t notice this.
We can calculate the total current in Amperes of the circuit by dividing the voltage by the resistance. This is the simple Ohms Law V = IR.
So if we connect a 10 ohm resistor to a 9v battery, 9 volts divided by 10 ohms gives us 0.9 amps. If we added another 5 ohm resistor to the circuit that gives us 15 ohms of resistance. So 9 volts divided by 15 ohms equals 0.6 amps and if we added another 5 ohm resistor that gives us 20 ohms of resistance. So 9 volts divided by 20 ohms equals 0.45 amps. So we can see that as we add more resistance to the circuit the current reduces so fewer electrons are flowing and that means we can do less work.
You can practically test this by connecting a larger value resistor, let’s say you can start with 800 ohm resistor, you will see the LED will glow a little dim, and if you further increased the value of the resistor a point will come when the LED won’t glow. We can also use resistors to protect components in the circuit if I use a 100 ohm resistor with a 9-volt battery the current will be around 0.09 amps or 90 milliamps and that will be too much that will blow the LED if I use a 450 ohm resistor the current will be around 0.02 amps or 20 milliamps so the LEDs should be okay if I use a 900 ohm resistor the current will be 0.01 amp or 10mA the LED will be very dim then.
The thing I am going to say make it a permanent part of your memory, store it for the rest of your life. In a series circuit the current is same while the voltage is different. While in parallel circuits the voltage is same while the current is different.
No matter in which part of the series circuit you add a digital multimeter or Ammeter you will always get the same value of current on all the meters connected in the circuit, as the current has only one path to flow.
Voltage in series:
Remember voltage is the pushing force that results in the of electrons which is just like pressure in a pipe, the higher the pressure the more water can flow. The higher the voltage the more electrons can flow we can see that by varying the voltage to a lamp as illustrated here the lamp increases its brightness as the voltage increases. You want to see this practically? Take a 9V or 12V lamp, bulb, anything. Start with the minimum voltage let’s say 1.5V which you can easily find in your TV remote controller, go ahead and fetch one. Try to power up the 9V bulb using the 1.5V battery cell. You will see the bulb be off or it may glow with really less brightness, now you can connect two cells in series this will make the voltage equal to 3V, now you will see the bulb will increase its brightness, keep increasing the voltage and a point will come when the bulb will glow at its full brightness. If you further increase the voltage and if you crossed its maximum rated voltage you will no doubt damage the bulb.
If you measure the voltage at the bulbs terminals you will find the same voltage available on the battery side noted if the wire is of smaller lengths, as I already explained wires has the resistance, and wire too drop some voltage. So for shorter wires the voltage drop can be negligible but for longer wires the voltage drop will definitely make some difference. To measure the voltage using a digital multimeter, connect the two testing leads of the multimeter with the opposite ends. If you connect the test leads with the same sides you won’t be able to see any voltage.
If we place a 9-volt battery into the circuit we apply a nine-volt to the circuit we can increase this by wiring the batteries in series. So if we place two 9-volt batteries in the circuit in series and we get 18 volts, three 9-volt batteries will give us 27 volts.
Let’s take a 9-volt battery and add an R1 resistance to the circuit if we use a multimeter to measure across the resistor we get a voltage reading of 9 volts.
If we add another 10 ohm resistor. We get a reading of 9 volts across the two resistors, but we get a reading of 4.5 volts if we measure across either of the resistors individually, so the resistors are dividing the voltage if we replace the R2 resistor with a 5 ohm resistor the total voltage would again be 9 volts and that’s what we see if we measure across the two resistors.
But if we measure across the 10 ohm resistor we see a voltage of 6 volts and if we measure across the 5 ohm resistor we see 3 volts. we’ll look at why that is just shortly if we added another resistor R3 with 5 ohms into the circuit we again get a total voltage drop of 9 volts across the three resistors across the R1 10 ohm resistor we read 4.5 volts across the R2 5 ohm resistor we read 2.25 volts and across the last R3 5 ohm resistor we again see 2.25 volts. We can combine these readings to find the voltage at different parts of the circuit for example if we measure from the battery across R1 we see 4.5 volts if we measure from the battery across R1 and R2 we get 6.75 volts because 4.5 plus 2.25 volts,
So unlike current where it’s the same throughout the circuit the voltage will be different throughout a series circuit this shows us that the voltage is reduced by each resistor. So the resistor creates a voltage drop. So, in a nutshell, no matter how many resistors you add in series the total voltage drop will always be equal to the supplied voltage, but the voltage measures across individual resistor may be different. If same value resistors are connected in series then the voltage will be equally different among all the resistors.
That’s the purpose of the resistor to reduce the voltage or the pressure.
What’s happening is a resistor creates a more difficult path with electrons to flow through and as they flow through they will collide with other electrons this collision will convert the energy into heat the same amount of electrons will enter and exit the resistor that will just have less energy or pressure as there’s been a voltage drop.
We can calculate the voltage drop across each resistor individually by multiplying the total current in the circuit by the resistance of each component.
Remember in a series circuit the current is the same anywhere in the circuit the total voltage drop will be the total of all the individual voltage drops combined the first circuit that was a 10 ohm resistor by itself the circuit had a current of 0.9 amps so 0.9 amps multiplied by 10 ohms equals 9 volts the voltage drop across the resistor is therefore 9 volts and that’s the same as the voltage source the second circuit has a 10 ohm and a 5 ohm resistor together so the first resistor the voltage drop is 0.6 m/s multiplied by 10 ohms which gives us 6 volts the second resistor was 5 ohms and the current was the same so 0.6 amps multiplied by 5 ohms equals to 3 volts the total voltage drop is therefore 6 volts plus 3 volts which gives us our new moles the third second has a 10 ohm and 2 5 ohm resistors the circuit her current of 0.45 amps so R 1 is 0 point 4 5 amps multiplied by 10 ohms which gives us 4.5 volts R 2 and R 3 will be 0.45 amps multiplied by 5 ohms which gives us 2.25 volts on each the total voltage drop is therefore 9 volts which is 4.5 plus 2.25 plus 2.25.
Power consumption in series circuits:
How do we measure power consumption of the circuit well we can use the following equations we can even use power which is Watts equals voltage squared divided by resistance or we can use power equals voltage multiplied by current.
How can a resistor consume power?
You might be wondering how can a resistor consume power well as the resistor is creating a voltage drop the electrons are losing some energy. Where is this energy going well the electrical energy is being converted into heat and if we look at some resistors under a thermal imaging camera we can see the heat is being generated so the power consumption is actually the heat being dissipated from the circuit.
So in this circuit the resistance is 10 ohms the battery is providing 9 volts the current is 0.9 amps and the circuit consumes 8.1 watts of power. How do we calculate that using method one 9 volts squared or 9 multiplied by 9 is 81 divided by 10 ohms is 8.1 watts, alternatively nine volts multiplied by 0.9 amps equals 8.1 watts in the circuit with a 10 ohm and the 5 ohm resistor the total resistance for 15 ohms and the current was 0.6 amps so 9 volts squared is 81 divided by 15 ohms is 5.4 watts or 9 volts multiplied by 0.6 m/s equals 5.4 watts in the circuit with a 10 ohm and the 2 5 ohm resistors the total circuit resistance was 20 ohms and the current was 0.45 amps so 9 volts squared is 81 divided by 20 ohms gives us a 4.05 watts or alternatively we can use 9 volts multiplied by 0.45 amps which equals 4.05 watts.
Ok so now it’s time for you to test your knowledge so this led can’t exceed a maximum current of 0.02 amps or 20 milliamps otherwise it will burn out.
So if we connected it to these resistors and a 9-volt battery what will the approximate current in the circuit be? Let me know in a comment.