# 12.1 Resistors in series and parallel  (Page 2/17)

 Page 2 / 17

These energies must be equal, because there is no other source and no other destination for energy in the circuit. Thus, $\mathit{qV}={\mathit{qV}}_{1}+{\mathit{qV}}_{2}+{\mathit{qV}}_{3}$ . The charge $q$ cancels, yielding $V={V}_{1}+{V}_{2}+{V}_{3}$ , as stated. (Note that the same amount of charge passes through the battery and each resistor in a given amount of time, since there is no capacitance to store charge, there is no place for charge to leak, and charge is conserved.)

Now substituting the values for the individual voltages gives

$V={\text{IR}}_{1}+{\text{IR}}_{2}+{\text{IR}}_{3}=I\left({R}_{1}+{R}_{2}+{R}_{3}\right).$

Note that for the equivalent single series resistance ${R}_{\text{s}}$ , we have

$V={\text{IR}}_{\text{s}}.$

This implies that the total or equivalent series resistance ${R}_{\text{s}}$ of three resistors is ${R}_{\text{s}}={R}_{1}+{R}_{2}+{R}_{3}$ .

This logic is valid in general for any number of resistors in series; thus, the total resistance ${R}_{\text{s}}$ of a series connection is

${R}_{\text{s}}={R}_{1}+{R}_{2}+{R}_{3}+\text{.}\text{.}\text{.},$

as proposed. Since all of the current must pass through each resistor, it experiences the resistance of each, and resistances in series simply add up.

## Calculating resistance, current, voltage drop, and power dissipation: analysis of a series circuit

Suppose the voltage output of the battery in [link] is $\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}$ , and the resistances are ${R}_{1}=1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega$ , ${R}_{2}=6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega$ , and ${R}_{3}=\text{13}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega$ . (a) What is the total resistance? (b) Find the current. (c) Calculate the voltage drop in each resistor, and show these add to equal the voltage output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors.

Strategy and Solution for (a)

The total resistance is simply the sum of the individual resistances, as given by this equation:

$\begin{array}{lll}{R}_{\text{s}}& =& {R}_{1}+{R}_{2}+{R}_{3}\\ & =& 1.00 \Omega +6.00 \Omega +\text{13.0 Ω}\\ & =& \text{20.0 Ω.}\end{array}$

Strategy and Solution for (b)

The current is found using Ohm’s law, $V=\text{IR}$ . Entering the value of the applied voltage and the total resistance yields the current for the circuit:

$I=\frac{V}{{R}_{\text{s}}}=\frac{\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}}{\text{20}\text{.}\text{0}\phantom{\rule{0.25em}{0ex}}\Omega }=0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}.$

Strategy and Solution for (c)

The voltage—or $\text{IR}$ drop—in a resistor is given by Ohm’s law. Entering the current and the value of the first resistance yields

${V}_{1}={\mathit{IR}}_{1}=\left(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}\right)\left(1\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega \right)=0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{V}.$

Similarly,

${V}_{2}={\mathit{IR}}_{2}=\left(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}\right)\left(6\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega \right)=3\text{.}\text{60}\phantom{\rule{0.25em}{0ex}}\text{V}$

and

${V}_{3}={\mathit{IR}}_{3}=\left(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}\right)\left(\text{13}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega \right)=7\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}\text{V}.$

Discussion for (c)

The three $\text{IR}$ drops add to $\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}$ , as predicted:

${V}_{1}+{V}_{2}+{V}_{3}=\left(0\text{.}\text{600}+3\text{.}\text{60}+7\text{.}\text{80}\right)\phantom{\rule{0.25em}{0ex}}\text{V}=\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}.$

Strategy and Solution for (d)

The easiest way to calculate power in watts (W) dissipated by a resistor in a DC circuit is to use Joule’s law    , $P=\text{IV}$ , where $P$ is electric power. In this case, each resistor has the same full current flowing through it. By substituting Ohm’s law $V=\text{IR}$ into Joule’s law, we get the power dissipated by the first resistor as

${P}_{1}={I}^{2}{R}_{1}=\left(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}{\right)}^{2}\left(1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega \right)=0\text{.}\text{360}\phantom{\rule{0.25em}{0ex}}\text{W}.$

Similarly,

${P}_{2}={I}^{2}{R}_{2}=\left(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}{\right)}^{2}\left(6\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega \right)=2\text{.}\text{16}\phantom{\rule{0.25em}{0ex}}\text{W}$

and

${P}_{3}={I}^{2}{R}_{3}=\left(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}{\right)}^{2}\left(\text{13}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega \right)=4\text{.}\text{68}\phantom{\rule{0.25em}{0ex}}\text{W}.$

Discussion for (d)

Power can also be calculated using either $P=\text{IV}$ or $P=\frac{{V}^{2}}{R}$ , where $V$ is the voltage drop across the resistor (not the full voltage of the source). The same values will be obtained.

Strategy and Solution for (e)

The easiest way to calculate power output of the source is to use $P=\text{IV}$ , where $V$ is the source voltage. This gives

$P=\left(0\text{.}\text{600}\phantom{\rule{0.25em}{0ex}}\text{A}\right)\left(\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}\right)=7\text{.}\text{20}\phantom{\rule{0.25em}{0ex}}\text{W}.$

#### Questions & Answers

Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply

### Read also:

#### Get the best Algebra and trigonometry course in your pocket!

Source:  OpenStax, Introductory physics - for kpu phys 1100 (2015 edition). OpenStax CNX. May 30, 2015 Download for free at http://legacy.cnx.org/content/col11588/1.13
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Introductory physics - for kpu phys 1100 (2015 edition)' conversation and receive update notifications?     By Lakeima Roberts    By 