# Resistors in Series and Parallel Formula Derivation

*Eugene is a qualified control/instrumentation engineer Bsc (Eng) and has worked as a developer of electronics & software for SCADA systems.*

## Formulas for Resistors in Series and Parallel

Resistors are ubiquitous components in electronic circuitry both in industrial and domestic consumer products. Often in circuit analysis, we need to work out the values when two or more resistors are combined. In this tutorial, we'll work out the formulas for resistors connected in series and parallel.

## Some Revision: A Circuit With One Resistor

In an earlier tutorial, you learned that when a single resistor with resistance R ohms was connected in a circuit with a voltage source V, the current I through the circuit was given by Ohm's Law:

### Ohms Law

*I* = *V*/*R*

**Example:** A 240 volt mains supply is connected to a heater with resistance of 60 ohms. What current will flow through the heater?

Current = *V*/*R* = 240/60 = 4 amps

## Derivation of Formula for Resistance When Resistors Are in Series

Now let's add a second resistor in series. Series means that the resistors are like links in a chain, one after another. We call the resistors R_{1} and R_{2}.

Because the resistors are linked together, the voltage source V causes the same current I to flow through both of them.

There will be a voltage drop or *potential difference* across both resistors.

Let the voltage drop measured across R_{1} be V_{1 }and let the voltage measured across R_{2} be V_{2} as shown in the diagram below.

From Ohm's Law, we know that for a circuit with a resistance R and voltage V:

*I = V/R*

Therefore rearranging the equation by multiplying both sides by R

*IR = V*

or switching around

*V = IR*

So for resistor R_{1}

*V _{1} = IR_{1}*

and for resistor R_{2}

*V _{2} = IR_{2}*

**Kirchoff's Voltage Law**

From Kirchoff's Voltage Law, we know that the sum of voltages around a closed loop in a circuit add up to zero. We decide on a convention, so voltage sources with arrows pointing clockwise from negative to positive are considered positive and voltage drops across resistors are negative. So in our example:

*V - V _{1 }- V_{2} = 0*

Rearranging

*V = V _{1} + V_{2 ................}*(I.e. the voltage V equals the sum of the drops across the resistors)

Substitute for V_{1 }and V_{2} calculated earlier

*V = IR _{1} + IR_{2} = I*(

*R*+

_{1}*R*)

_{2}Divide both sides by I

*V/I = R _{1} + R_{2}*

But from Ohm's Law, we know* I = V/R*, so rearranging:

*V/I = R* = total resistance of the circuit. Let's call it R_{total}

Therefore

**V/I = R _{total} = R_{1} + R_{2}**

In general if we have n resistors:

**R _{total} = R_{1} + R_{2} + ...... R_{n}**

So to get the total resistance of resistors connected in series, we just add all the values.

**Example 1:**

Five 10k resistors and two 100k resistors are connected in series. What is the combined resistance?

**Answer:**

Resistor values are often specified in kiloohm (abbreviated to "k") or megaohms (abbreviated to "M")

1 kiloohm or 1k = 1000 ohms

1 megaohm or 1M = 1000,000 ohms

So total resistance = sum of the resistances

= 5 x (10k) + 2 x (100k)

= 50k + 200k

= 250k or 250,000 ohms

**Example 2:**

Three 47 ohm, five 1.2k, four 100k and two 3.3M resistors are connected in series. What is the total resistance?

**Answer:**

We often replace the decimal point in resistor values by the multiplier to avoid misreading if e.g. the "dot" gets erased from the value printed on a component or in documents. So 1.2k becomes 1k2.

So total resistance = sum of the resistances

= 3 x 47 + 5 x 1k2 + 4 x 100k + 2 x 3M3

= 3 x 47 + 5 x 1200 + 4 x 100,000 + 2 x 3,300,000

= 141 + 6000 + 400,000 + 6,600,000

= 7,006,141 ohms

## Derivation of Formula for Resistance of Two Resistors in Parallel

Next we'll derive the expression for resistors in parallel. Parallel means all the ends of the resistors are connected together at one point and all the other ends of the resistors are connected at another point.

When resistors are connected in parallel, the current from the source is split between all the resistors instead of being the same as was the case with series connected resistors. However, the same voltage is now common to all resistors.

Let the current through resistor R_{1} be I_{1} and the current through R_{2} be I_{2}

The voltage drop across both R_{1} and R_{2 }is equal to the supply voltage V

Therefore from Ohm's Law

*I _{1} = V/R_{1}*

and

*I _{2} = V/R_{2}*

But from Kirchoff's Current Law, we know the current entering a node (connection point) is equal to the current leaving the node

Therefore

*I = I _{1} + I_{2}*

Substituting the values derived for I_{1} and I_{2 }gives us

*I = V/R _{1} + V/R*

_{2 }

*= V*(1*/R _{1} + *1

*/R*)

_{2}The lowest common denominator (LCD) of 1/R_{1} and 1/R_{2 }is R_{1}R_{2} so we can replace the expression (1/R_{1} + 1/R_{2}) by

*R _{2}/R_{1}R_{2}*+

*R*

_{1}/R_{1}R_{2}Switching around the two fractions

= *R _{1}/R_{1}R_{2}*+

*R*

_{2}/R_{1}R_{2 }and since the denominator of both fractions is the same

*= *(*R _{1 }+ R_{2}*)

*/R*

_{1}R_{2}Therefore

*I = V*(1*/R _{1} + *1

*/R*)

_{2}*= V*(

*R*)

_{1 }+ R_{2}*/R*

_{1}R_{2}Rearranging gives us

*V/I = R _{1}R_{2}/*(

*R*+

_{1 }*R*)

_{2}But from Ohm's Law, we know V/I = total resistance of the circuit. Let's call it R_{total}

Therefore

*R _{total }= R_{1}R_{2} / *(

*R*)

_{1 }+ R_{2}_{}

So for two resistors in parallel, the combined resistance is the product of the individual resistances divided by the sum of the resistances.

**Example:**

A 100 ohm resistor and a 220 ohm resistor are connected in parallel. What is the combined resistance?

**Answer:**

For two resistors in parallel we just divide the product of the resistances by their sum.

So total resistance = 100 x 220 / (100 + 220) = 22000/320 = 8.75 ohms

## Derivation of Formula for Resistance of Multiple Resistors in Parallel

If we have more than two resistors connected in parallel, the current I equals the sum of all the currents flowing through the resistors.

So for n resistors

*I = I _{1}*+

*I*+

_{2}*I*+

_{3}. ...........*I*

_{n}= *V/R _{1}*+

*V/R*+

_{2}*V/R*+

_{3}*............. V/R*

_{n}= *V*(1*/R _{1}*+ 1

*/R*1

_{2}+ V/R_{3 }...........*/R*)

_{n}Rearranging

*I/V = *(1*/R _{1} + *1

*/R*1

_{2}+ V/R_{3 }...........*/R*)

_{n}If *V/I = R _{total}* then

*I/V = *1*/R _{total} = *(1

*/R*1

_{1}+*/R*1

_{2}+ V/R_{3 }...........*/R*)

_{n}So our final formula is

**1/R _{total} = **

**(1**

**/R**_{1}+**1**

**/R**_{2}+ V/R_{3 }...........**1**

**/R**_{n}**)**

We could invert the right side of the formula to give an expression for R_{total} , however it's easier to remember the equation for the reciprocal of resistance.

So to calculate the total resistance, we calculate the reciprocals of all the resistances first, sum them together giving us the reciprocal of the total resistance. Then we take the reciprocal of this result giving us R_{total}

**Example:**

Calculate the combined resistance of three 100 ohm and four 200 ohm resistors in **parallel.**

**Answer:**

Lets call the combined resistance R.

So

1/R = 1/100 + 1/100 + 1/100 + 1/200 + 1/200 + 1/200 + 1/200

We can use a calculator to work out the result for 1/R by summing all the fractions and then inverting to find R, but lets try and work it out "by hand".

So

1/R = 1/100 + 1/100 + 1/100 + 1/200 + 1/200 + 1/200 + 1/200

= 3/100 + 4/200

To simplify a sum or difference of fractions we can use a lowest common denominator (LCD). The LCD of 100 and 200 in our example is 200

Therefore multiply the top and bottom of the first fraction by 2 giving:

1/R = 3/100 + 4/200 = (2 x 3) / (2 x 100) + 4/200

= 6 / 200 + 4/200

= (6 + 4)/200 = 10/200

and inverting gives R = 200 / 10 = 20 ohms. No calculator needed!

## Recommended Books

*Introductory Circuit Analysis* by Robert L Boylestad covers the basics of electricity and circuit theory and also more advanced topics such as AC theory, magnetic circuits and electrostatics. It's well illustrated and suitable for high school students and also first and second year electric or electronic engineering students. New and used versions of the hardcover 10th edition are available on Amazon. Later editions are also available.

## References

Boylestad, Robert L. (1968) *Introductory Circuit Analysis *(6th ed. 1990) Merrill Publishing Company, London, England.

*This article is accurate and true to the best of the author’s knowledge. Content is for informational or entertainment purposes only and does not substitute for personal counsel or professional advice in business, financial, legal, or technical matters.*

**© 2020 Eugene Brennan**

## Comments

**Eugene Brennan (author)** from Ireland on February 13, 2020:

Thanks Umesh!

**Umesh Chandra Bhatt** from Kharghar, Navi Mumbai, India on February 13, 2020:

Interesting. You have revised my memories of high school science classes. Well presented. Keep it up.