Resistors are said to be connected in series when they are daisy chained together in a single line resulting in a common current flowing through them.

Individual resistors can be connected together in either a series connection, a parallel connection or combinations of both series and parallel, to produce more complex resistor networks whose equivalent resistance is the mathematical combination of the individual resistors connected together.

A resistor is not only a fundamental electronic component that can be used to convert a voltage to a current or a current to a voltage, but by correctly adjusting its value a different weighting can be placed onto the converted current and/or the voltage allowing it to be used in voltage reference circuits and applications.

Resistors in series or complicated resistor networks can be replaced by one single equivalent resistor, R_{EQ} or impedance, Z_{EQ} and no matter what the combination or complexity of the resistor network is, all resistors obey the same basic rules as defined by Ohm’s Law and Kirchoff’s Circuit Laws.

Resistors are said to be connected in “**Series**“, when they are daisy chained together in a single line. Since all the current flowing through the first resistor has no other way to go it must also pass through the second resistor and the third and so on. Then, resistors in series have a **Common Current** flowing through them as the current that flows through one resistor must also flow through the others as it can only take one path.

Then the amount of current that flows through a set of resistors in series will be the same at all points in a series resistor network. For example:

In the following example the resistors R_{1}, R_{2} and R_{3} are all connected together in series between points A and B with a common current, I flowing through them.

As the resistors are connected together in series the same current passes through each resistor in the chain and the total resistance, R_{T} of the circuit must be **equal** to the sum of all the individual resistors added together. That is

and by taking the individual values of the resistors in our simple example above, the total equivalent resistance, R_{EQ} is therefore given as:

R_{EQ} = R_{1} + R_{2} + R_{3} = 1kΩ + 2kΩ + 6kΩ = 9kΩ

So we see that we can replace all three individual resistors above with just one single “equivalent” resistor which will have a value of 9kΩ.

Where four, five or even more resistors are all connected together in a series circuit, the total or equivalent resistance of the circuit, R_{T} would still be the sum of all the individual resistors connected together and the more resistors added to the series, the greater the equivalent resistance (no matter what their value).

This total resistance is generally known as the **Equivalent Resistance** and can be defined as; “*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*“. Then the equation given for calculating total resistance of the circuit when connecting together resistors in series is given as:

R_{total} = R_{1} + R_{2} + R_{3} + ….. R_{n} etc.

Note then that the total or equivalent resistance, R_{T} has the same effect on the circuit as the original combination of resistors as it is the algebraic sum of the individual resistances.

If two resistances or impedances in series are equal and of the same value, then the total or equivalent resistance, R |

If two resistors or impedances in series are unequal and of different values, then the total or equivalent resistance, R |

One important point to remember about resistors in series networks to check that your maths is correct. The total resistance ( R_{T} ) of any two or more resistors connected together in series will always be **GREATER** than the value of the largest resistor in the chain. In our example above R_{T} = 9kΩ where as the largest value resistor is only 6kΩ.

The voltage across each resistor connected in series follows different rules to that of the series current. We know from the above circuit that the total supply voltage across the resistors is equal to the sum of the potential differences across R_{1} , R_{2} and R_{3} , V_{AB} = V_{R1} + V_{R2} + V_{R3} = 9V.

Using Ohm’s Law, the voltage across the individual resistors can be calculated as:

Voltage across R_{1} = IR_{1} = 1mA x 1kΩ = 1V

Voltage across R_{2} = IR_{2} = 1mA x 2kΩ = 2V

Voltage across R_{3} = IR_{3} = 1mA x 6kΩ = 6V

giving a total voltage V_{AB} of ( 1V + 2V + 6V ) = 9V which is equal to the value of the supply voltage. Then the sum of the potential differences across the resistors is equal to the total potential difference across the combination and in our example this is 9V.

The equation given for calculating the total voltage in a series circuit which is the sum of all the individual voltages added together is given as:

Then series resistor networks can also be thought of as “voltage dividers” and a series resistor circuit having *N* resistive components will have N-different voltages across it while maintaining a common current.

By using Ohm’s Law, either the voltage, current or resistance of any series connected circuit can easily be found and resistor of a series circuit can be interchanged without affecting the total resistance, current, or power to each resistor.

Using Ohms Law, calculate the equivalent series resistance, the series current, voltage drop and power for each resistor in the following resistors in series circuit.

All the data can be found by using Ohm’s Law, and to make life a little easier we can present this data in tabular form.

Resistance | Current | Voltage | Power |

R_{1} = 10Ω |
I_{1} = 200mA |
V_{1} = 2V |
P_{1} = 0.4W |

R_{2} = 20Ω |
I_{2} = 200mA |
V_{2} = 4V |
P_{2} = 0.8W |

R_{3} = 30Ω |
I_{3} = 200mA |
V_{3} = 6V |
P_{3} = 1.2W |

R_{T} = 60Ω |
I_{T} = 200mA |
V_{S} = 12V |
P_{T} = 2.4W |

Then for the circuit above, R_{T} = 60Ω, I_{T} = 200mA, V_{S} = 12V and P_{T} = 2.4W

We can see from the above example, that although the supply voltage is given as 12 volts, different voltages, or voltage drops, appear across each resistor within the series network. Connecting resistors in series like this across a single DC supply has one major advantage, different voltages appear across each resistor producing a very handy circuit called a **Voltage Divider Network**.

This simple circuit splits the supply voltage proportionally across each resistor in the series chain with the amount of voltage drop being determined by the resistors value and as we now know, the current through a series resistor circuit is common to all resistors. So a larger resistance will have a larger voltage drop across it, while a smaller resistance will have a smaller voltage drop across it.

The series resistive circuit shown above forms a simple voltage divider network were three voltages 2V, 4V and 6V are produced from a single 12V supply. Kirchoff’s Voltage Law states that “*the supply voltage in a closed circuit is equal to the sum of all the voltage drops (IR) around the circuit*” and this can be used to good effect.

The **Voltage Division Rule**, allows us to use the effects of resistance proportionality to calculate the potential difference across each resistance regardless of the current flowing through the series circuit. A typical “voltage divider circuit” is shown below.

The circuit shown consists of just two resistors, R_{1} and R_{2} connected together in series across the supply voltage V_{in}. One side of the power supply voltage is connected to resistor, R_{1}, and the voltage output, V_{out} is taken from across resistor R_{2}. The value of this output voltage is given by the corresponding formula.

If more resistors are connected in series to the circuit then different voltages will appear across each resistor in turn with regards to their individual resistance R (Ohms Law **IxR**) values providing different but smaller voltage points from one single supply.

So if we had three or more resistances in the series chain, we can still use our now familiar potential divider formula to find the voltage drop across each one. Consider the circuit below.

The *potential divider circuit* above shows four resistances connected together is series. The voltage drop across points A and B can be calculated using the potential divider formula as follows:

We can also apply the same idea to a group of resistors in the series chain. For example if we wanted to find the voltage drop across both R2 and R3 together we would substitute their values in the top numerator of the formula and in this case the resulting answer would give us 5 volts (2V + 3V).

In this very simple example the voltages work out very neatly as the voltage drop across a resistor is proportional to the total resistance, and as the total resistance, (R_{T}) in this example is equal to 100Ω or 100%, resistor R1 is 10% of R_{T}, so 10% of the source voltage V_{S} will appear across it, 20% of V_{S} across resistor R2, 30% across resistor R3, and 40% of the supply voltage V_{S} across resistor R4. Application of Kirchoff’s voltage law (KVL) around the closed loop path confirms this.

Now lets suppose that we want to use our two resistor potential divider circuit above to produce a smaller voltage from a larger supply voltage to power an external electronic circuit. Suppose we have a 12V DC supply and our circuit which has an impedance of 50Ω requires only a 6V supply, half the voltage.

Connecting two equal value resistors, of say 50Ω each, together as a potential divider network across the 12V will do this very nicely until we connect the load circuit to the network. This is because the loading effect of resistor R_{L} connected in parallel across R_{2} changes the ratio of the two series resistances altering their voltage drop and this is demonstrated below.

**Calculate the voltage drops across X and Y.**

a) Without R_{L} connected

b) With R_{L} connected

As you can see from above, the output voltage V_{out} without the load resistor connected gives us the required output voltage of 6V but the same output voltage at V_{out} when the load is connected drops to only 4V, (Resistors in Parallel).

Then we can see that a loaded voltage divider network changes its output voltage as a result of this loading effect, since the output voltage V_{out} is determined by the ratio of R_{1} to R_{2}. However, as the load resistance, R_{L} increases towards infinity (∞) this loading effect reduces and the voltage ratio of Vout/Vs becomes unaffected by the addition of the load on the output. Then the higher the load impedance the less is the loading effect on the output.

The effect of reducing a signal or voltage level is known as **Attenuation** so care must be taken when using a voltage divider network. This loading effect could be compensated for by using a potentiometer instead of fixed value resistors and adjusted accordingly. This method also compensates the potential divider for varying tolerances in the resistors construction.

A variable resistor, potentiometer or pot as it is more commonly called, is a good example of a multi-resistor voltage divider within a single package as it can be thought of as thousands of mini-resistors in series. Here a fixed voltage is applied across the two outer fixed connections and the variable output voltage is taken from the wiper terminal. Multi-turn pots allow for a more accurate output voltage control.

The **Voltage Divider Circuit** is the simplest way of producing a lower voltage from a higher voltage, and is the basic operating mechanism of the potentiometer.

As well as being used to calculate a lower supply voltage, the voltage divider formula can also be used in the analysis of more complex resistive circuits containing both series and parallel branches. The voltage or potential divider formula can be used to determine the voltage drops around a closed DC network or as part of a various circuit analysis laws such as Kirchoff’s or Thevenin’s theorems.

We have seen that **Resistors in Series** can be used to produce different voltages across themselves and this type of resistor network is very useful for producing a voltage divider network. If we replace one of the resistors in the voltage divider circuit above with a Sensor such as a thermistor, light dependant resistor (LDR) or even a switch, we can convert an analogue quantity being sensed into a suitable electrical signal which is capable of being measured.

For example, the following thermistor circuit has a resistance of 10KΩ at 25°C and a resistance of 100Ω at 100°C. Calculate the output voltage (Vout) for both temperatures.

At 25°C

At 100°C

So by changing the fixed 1KΩ resistor, R_{2} in our simple circuit above to a variable resistor or potentiometer, a particular output voltage set point can be obtained over a wider temperature range.

So to summarise. When two or more resistors are connected together end-to-end in a single branch, the resistors are said to be connected together in series. **Resistors in Series** carry the same current, but the voltage drop across them is not the same as their individual resistance values will create different voltage drops across each resistor as determined by Ohm’s Law ( V = IxR ). Then series circuits are voltage dividers.

In a series resistor network the individual resistors add together to give an equivalent resistance, ( R_{T} ) of the series combination. The resistors in a series circuit can be interchanged without affecting the total resistance, current, or power to each resistor or the circuit.

In the next tutorial about Resistors, we will look at connecting resistors together in parallel and show that the total resistance is the reciprocal sum of all the resistors added together and that the voltage is common to a parallel circuit.

Error! Please fill all fields.

i will like to have more like this. i mean people to help me in calculation about electricity bc i learn a lot through this page and i like it thanks.

Nice tutorial

small mistake in the graphic of the Voltage Divider Network, it reads 2 times VR1, bottom one should be VR2.

Excellent job!

Oops, thanks Ardlab, ðŸ™‚

Hello!

If you have two bulbs with the same resistance in series, they would have the same voltage drop across them, right? So then why is it that some bulbs are brighter than the others?

Incandescent filament lamps, or bulbs as you call them, can be considered to be wholly resistive loads, therefore all the power is dissipated in the form of heat and light. However, the resistive values of the tungsten-filaments when hot (hot resistance) are not always exactly the same for the same wattage lamps. When illuminated, it appears that the lamp is always ON, but in reality it is alternating between maximum and minimum brightness in response to the mains sinusoidal waveform at a rate of 100 or 120 times a second (2 max, and 2 min per cycle). It therefore takes time for the filament to respond, heat up and reach full brilliance before it reduces to minimum again, also affecting the brightness. There are many more reasons for the effect on a lamps brightness.

This is amazing site I think!

good explanation! Thanx alot!!

electrical engineering problems are good

I want to start the tutorials from the beginning. Do i have only to click on start here on the page and then complete by clicking next or there is a different arrangement of the topics?

The order of tutorials will be decided by what you are studying, AC, DC, Digital, etc, but you can start with passives first, if that helps.

Wow

In Resistors in Series Example No1 I took 12/60 and came up with .2. When i use that number, everything works out. However i could not get anything to work with 200 mA. So where did 200 mA come from? Does it have something to do with the decimal point?

1.0 Ampere = 1000 milli-amperes. Therefore, 0.2A = 200mA

What if resistors of different resistance were to be integrated in series, where would be the greater potential difference?in the lesser resistor, or greater?

V = IxR, therefore the largest voltage drop would occur across the largest resistance in the series chain.

hi, i wonder if making a series of 1/4 watts would lower the overload of the resistors? meaning if the resistor are 1/4 watts max, so can i add more of resistors to increase watts max in a series ??? i got 1/4 resistors and i am trying not to burn the resistor out and i am using 12 volt and it produce 5 watts and the amps is .42!! Any ideasï»¿

Each resistor has a power rating of 1/4 watts or 0.250 watts. P = V x I. If P = 0.250 and I = 0.42 amperes, then the maximum voltage drop across each resistor will therefore be 0.250/0.42 = 0.6 volts. Then you would need 20 series resistors each with a value of 1/20th the value of R or 1.42 ohms each.