<< Chapter < Page Chapter >> Page >
I = I 1 + I 2 + I 3 . size 12{I=I rSub { size 8{1} } +I rSub { size 8{2} } +I rSub { size 8{3} } } {}

Substituting the expressions for the individual currents gives

I = V R 1 + V R 2 + V R 3 = V 1 R 1 + 1 R 2 + 1 R 3 . size 12{I= { {V} over {R rSub { size 8{1} } } } + { {V} over {R rSub { size 8{2} } } } + { {V} over {R rSub { size 8{3} } } } =V left ( { {1} over {R rSub { size 8{1} } } } + { {1} over {R rSub { size 8{2} } } } + { {1} over {R rSub { size 8{3} } } } right )} {}

Note that Ohm’s law for the equivalent single resistance gives

I = V R p = V 1 R p . size 12{I= { {V} over {R rSub { size 8{p} } } } =V left ( { {1} over {R rSub { size 8{p} } } } right )} {}

The terms inside the parentheses in the last two equations must be equal. Generalizing to any number of resistors, the total resistance R p size 12{R rSub { size 8{p} } } {} of a parallel connection is related to the individual resistances by

1 R p = 1 R 1 + 1 R 2 + 1 R . 3 + . ... size 12{ { {1} over {R rSub { size 8{p} } } } = { {1} over {R rSub { size 8{1} } } } + { {1} over {R rSub { size 8{2} } } } + { {1} over {R rSub { size 8{ "." 3} } } } + "." "." "." "." } {}

This relationship results in a total resistance R p size 12{R rSub { size 8{p} } } {} that is less than the smallest of the individual resistances. (This is seen in the next example.) When resistors are connected in parallel, more current flows from the source than would flow for any of them individually, and so the total resistance is lower.

Calculating resistance, current, power dissipation, and power output: analysis of a parallel circuit

Let the voltage output of the battery and resistances in the parallel connection in [link] be the same as the previously considered series connection: V = 12 . 0 V size 12{V="12" "." 0" V"} {} , R 1 = 1 . 00 Ω size 12{R rSub { size 8{1} } =1 "." "00" %OMEGA } {} , R 2 = 6 . 00 Ω size 12{R rSub { size 8{2} } =6 "." "00" %OMEGA } {} , and R 3 = 13 . 0 Ω size 12{R rSub { size 8{3} } ="13" "." 0 %OMEGA } {} . (a) What is the total resistance? (b) Find the total current. (c) Calculate the currents in each resistor, and show these add to equal the total current 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 for a parallel combination of resistors is found using the equation below. Entering known values gives

1 R p = 1 R 1 + 1 R 2 + 1 R 3 = 1 1 . 00 Ω + 1 6 . 00 Ω + 1 13 . 0 Ω . size 12{ { {1} over {R rSub { size 8{p} } } } = { {1} over {R rSub { size 8{1} } } } + { {1} over {R rSub { size 8{2} } } } + { {1} over {R rSub { size 8{3} } } } = { {1} over {1 "." "00" %OMEGA } } + { {1} over {6 "." "00" %OMEGA } } + { {1} over {"13" "." 0 %OMEGA } } } {}

Thus,

1 R p = 1.00 Ω + 0 . 1667 Ω + 0 . 07692 Ω = 1 . 2436 Ω . size 12{ { {1} over {R rSub { size 8{p} } } } = { {1 "." "00"} over { %OMEGA } } + { {0 "." "167"} over { %OMEGA } } + { {0 "." "0769"} over { %OMEGA } } = { {1 "." "244"} over { %OMEGA } } } {}

(Note that in these calculations, each intermediate answer is shown with an extra digit.)

We must invert this to find the total resistance R p size 12{R rSub { size 8{p} } } {} . This yields

R p = 1 1 . 2436 Ω = 0 . 8041 Ω . size 12{R rSub { size 8{p} } = { {1} over {1 "." "2436"} } %OMEGA =0 "." "8041 " %OMEGA } {}

The total resistance with the correct number of significant digits is R p = 0 . 804 Ω . size 12{R rSub { size 8{p} } =0 "." "804" %OMEGA } {}

Discussion for (a)

R p is, as predicted, less than the smallest individual resistance.

Strategy and Solution for (b)

The total current can be found from Ohm’s law, substituting R p size 12{R rSub { size 8{p} } } {} for the total resistance. This gives

I = V R p = 12.0 V 0.8041 Ω = 14 . 92 A . size 12{I= { {V} over {R rSub { size 8{p} } } } = { {"12" "." 0" V"} over {0 "." "804 " %OMEGA } } ="14" "." "92"" A"} {}

Discussion for (b)

Current I size 12{I} {} for each device is much larger than for the same devices connected in series (see the previous example). A circuit with parallel connections has a smaller total resistance than the resistors connected in series.

Strategy and Solution for (c)

The individual currents are easily calculated from Ohm’s law, since each resistor gets the full voltage. Thus,

I 1 = V R 1 = 12 . 0 V 1 . 00 Ω = 12 . 0 A . size 12{I rSub { size 8{1} } = { {V} over {R rSub { size 8{1} } } } = { {"12" "." 0" V"} over {1 "." "00 " %OMEGA } } ="12" "." 0" A"} {}

Similarly,

I 2 = V R 2 = 12 . 0 V 6 . 00 Ω = 2 . 00 A size 12{I rSub { size 8{2} } = { {V} over {R rSub { size 8{2} } } } = { {"12" "." 0" V"} over {6 "." "00 " %OMEGA } } =2 "." "00"" A"} {}

and

I 3 = V R 3 = 12 . 0 V 13 . 0 Ω = 0 . 92 A . size 12{I rSub { size 8{3} } = { {V} over {R rSub { size 8{3} } } } = { {"12" "." 0" V"} over {"13" "." "0 " %OMEGA } } =0 "." "92"" A"} {}

Discussion for (c)

The total current is the sum of the individual currents:

I 1 + I 2 + I 3 = 14 . 92 A . size 12{I rSub { size 8{1} } +I rSub { size 8{2} } +I rSub { size 8{3} } ="14" "." "92"" A"} {}

This is consistent with conservation of charge.

Strategy and Solution for (d)

The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three are known. Let us use P = V 2 R size 12{P= { {V rSup { size 8{2} } } over {R} } } {} , since each resistor gets full voltage. Thus,

P 1 = V 2 R 1 = ( 12 . 0 V ) 2 1 . 00 Ω = 144 W . size 12{P rSub { size 8{1} } = { {V rSup { size 8{2} } } over {R rSub { size 8{1} } } } = { { \( "12" "." 0" V" \) rSup { size 8{2} } } over {1 "." "00 " %OMEGA } } ="144"" W"} {}

Similarly,

P 2 = V 2 R 2 = ( 12 . 0 V ) 2 6 . 00 Ω = 24 . 0 W size 12{P rSub { size 8{2} } = { {V rSup { size 8{2} } } over {R rSub { size 8{2} } } } = { { \( "12" "." 0" V" \) rSup { size 8{2} } } over {6 "." "00 " %OMEGA } } ="24" "." 0" W"} {}

and

P 3 = V 2 R 3 = ( 12 . 0 V ) 2 13 . 0 Ω = 11 . 1 W . size 12{P rSub { size 8{3} } = { {V rSup { size 8{2} } } over {R rSub { size 8{3} } } } = { { \( "12" "." 0" V" \) rSup { size 8{2} } } over {"13" "." "0 " %OMEGA } } ="11" "." 1" W"} {}

Questions & Answers

how do you get the 2/50
Abba Reply
number of sport play by 50 student construct discrete data
Aminu Reply
width of the frangebany leaves on how to write a introduction
Theresa Reply
Solve the mean of variance
Veronica Reply
Step 1: Find the mean. To find the mean, add up all the scores, then divide them by the number of scores. ... Step 2: Find each score's deviation from the mean. ... Step 3: Square each deviation from the mean. ... Step 4: Find the sum of squares. ... Step 5: Divide the sum of squares by n – 1 or N.
kenneth
what is error
Yakuba Reply
Is mistake done to something
Vutshila
Hy
anas
hy
What is the life teble
anas
hy
Jibrin
statistics is the analyzing of data
Tajudeen Reply
what is statics?
Zelalem Reply
how do you calculate mean
Gloria Reply
diveving the sum if all values
Shaynaynay
let A1,A2 and A3 events be independent,show that (A1)^c, (A2)^c and (A3)^c are independent?
Fisaye Reply
what is statistics
Akhisani Reply
data collected all over the world
Shaynaynay
construct a less than and more than table
Imad Reply
The sample of 16 students is taken. The average age in the sample was 22 years with astandard deviation of 6 years. Construct a 95% confidence interval for the age of the population.
Aschalew Reply
Bhartdarshan' is an internet-based travel agency wherein customer can see videos of the cities they plant to visit. The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400 a. what is the probability of getting more than 12,000 hits? b. what is the probability of getting fewer than 9,000 hits?
Akshay Reply
Bhartdarshan'is an internet-based travel agency wherein customer can see videos of the cities they plan to visit. The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400. a. What is the probability of getting more than 12,000 hits
Akshay
1
Bright
Sorry i want to learn more about this question
Bright
Someone help
Bright
a= 0.20233 b=0.3384
Sufiyan
a
Shaynaynay
How do I interpret level of significance?
Mohd Reply
It depends on your business problem or in Machine Learning you could use ROC- AUC cruve to decide the threshold value
Shivam
how skewness and kurtosis are used in statistics
Owen Reply
yes what is it
Taneeya
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 9

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics for ap® courses' conversation and receive update notifications?

Ask