# Series Circuits

Series Circuits are circuits which satisfy the following rules:

• The same current flows through each element, the current doesn't branch off
• They have one common terminal which is not connected to another current carrying component

The key concept is that the current is the same through series components.

A series circuit with several resistors is shown below, we say that the resistors are connected in series.

com1 and com2 are common terminals between R1, R2 and R2 and R3 hence all three resistors are in series and the same current flows through them all.

To identify series connections in more complex circuits, we just need to ask ourselves if the same current would flow through the 2 or more components and if the answer is yes, then they are connected in series.

## Resistors in Series

When resistors are connected in series, the total resistance is equal to the sum of the resistances.

Each resistor opposes current flow hence the net effect is simply the total resistance added together through the current path.

Adding more resistors in series increases the total resistance of the circuit which decreases the current by Ohm's Law.

The above can also be proven mathematically from Ohm's Law:

Hence the above circuit is equivalent to the following circuit below where a single equivalent resistor replaces them all.

In short: Resistance adds in series.

## Voltage in Series

Series connections of voltage sources can be represented by a single source with a voltage equal to the sum of the individual sources.

Regarding the voltage drops adding in series,  this is due to the total voltage supplied by the voltage sources being dropped across the circuit and so if we have say two voltage sources and two resistors it follows that the total voltage supplied by those sources will be dropped across the two resistors.

Shown below is a circuit with two voltage sources and two resistors connected in series:

To intuitively understand why voltage sources add in series, which is incredibly important in really understanding voltage, it helps to imagine performing voltage readings at various points in the circuit.

In the circuit above we have defined the negative terminal of voltage source Vs1 to be 0V and so it follows that on the positive terminal the voltage read would be 1V as a coulomb of charge here would have 1V of energy higher than the other terminal.

Now, if we measure up to the negative terminal of the other voltage source Vs2, we would still read 1V as we treat a wire as having a resistance of 0Ω (even in reality it is so small you will read very close to the same result even with a very high precision instrument).

Looking at the positive side though, what would we read? Remember that fundamentally a voltage source provides its rated voltage across its terminals and that voltage is always relative to another point, so if we take the negative terminal of V1 to be our reference of 0V then the reading would be 2V. The reason is that Vs2 provides a voltage aka potential difference of 1V across its terminals and so we must be 1V above the the negative terminal which was 1V.

The above circuit is also equivalent to the circuit below after summing the series voltage sources and resistors together:

In short: Voltage adds in series.

### Examples:

1. What is the total resistance between terminals a and b in the circuit below?

On inspection the same current flows through all resistors as R1 and R2 have one common terminal and R2 and R3 have one common terminal, the current cannot branch off.This means all three resistors are connected in series, hence the total resistance is equal to the sum of the 3:

2. a) What is the current I in the circuit below:

The simplest way to work through this is to first determine the equivalent simple series circuit with a single resistor and voltage source. We can find the equivalent total series resistance by adding together the two series resistors, R1 and R2: Next we can find the equivalent total voltage provided by the voltage sources by summing V1 and V2: Now to calculate the current I we can use Ohm's law using the equivalent resistance and voltage of the circuit we have just calculated:

b) What is the voltage drop across resistor R1?

With the series current known we can now use Ohm's law to calculate the voltage drop across R1:

c) What is the voltage drop across resistor R2?

We could calculate this in a similar way to b), however, there is a quicker way (alternative methods are also handy as they give us another method to verify our answer). Given that the voltage sources provide 10V and 1V is across R1 then the remaining voltage must be dropped across R2. We can express this mathematically as:

Todo: Add example of designing circuit with certain voltage and resistance using bag of parts.

3. A junior engineering student has a bag of parts containing 1k, 2k, 5k and 4.7k resistors as well as 1.8V, 2.2V and 2.8V batteries. The lab demonstrator asks the student to create a circuit which supplies 5V to an 8k load, how would this circuit look?

To answer the above let's recall the fundamental properties of series resistors and series voltage sources to help us construct this circuit. Recall that resistance and voltage sources simply add in series and hence the question becomes which combination of resistors and batteries will achieve the target values? How can we make 8k from 1k, 2k, 5k and 4.7k? Well 1k+2k+5k=8k computes so let's use them! How can we make 5V from 1.8V, 2.2V and 2.8V? 2.2V+2.8V=5V so that will also work. The circuit then becomes 1k,2k,5k all connected in series to a 2.2V and 2.8V battery as shown below:

### Questions:

1. a) Determine the series resistance in the circuit below:

5kΩ

b) What is the current flowing through the circuit?

660uA

c) What is the voltage across R1?

0.66V

d) What is the voltage across R2?

2.64V

2. a) Determine the resistance in the circuit below:

2kΩ

b) What is the current flowing through the circuit?

4.15mA

c) What is the voltage across R1?

4.15V

d) What is the voltage across R2?

2.075V

e) What is the voltage across R3?

2.075V

## Lab:

Let's build further intuition by constructing series circuits using LTSpice.

To begin with let's create a simple circuit with a voltage source and multiple resistors. We can start by adding a voltage source as shown below:

From here let's connect up a few resistors and add a ground connection:

Now we can run a transient simulation to measure all of the voltages and the series current with settings as below:

We can then left click just above the voltage source to see 5V shown as we expect given we have a 5V ideal voltage source:

Let's then probe after R1 and R2 to see all of the voltages in the same chart by performing two single left clicks at each of the nodes.

Finally let's measure the current by left clicking on any of the resistors or voltage source:

From the bottom left we can see that the current flowing through the circuit is 833.333uA.

We can then try the same circuit with a current source as below:

Now if we measure the current we expect to see 1mA flowing:

How about the voltage produced by the current source for this resistance? We can measure that by left clicking at the node connecting the current source and R1:

Turns out to be 6V. Which we expect as 1mA*6k=6V.

Finally, let's try adding multiple voltage sources in a circuit such as below:

We can then run the simulation again and confirm that we measure 8V (3V+5V) at the node connecting R1 and Vs2:

We can then measure the series current by clicking on either voltage source or resistor:

4mA which makes sense as 8V/4k = 4mA.

I encourage you to play around with other circuits to get a feel for how this all works!