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R s = R 1 + R 2 + R 3 = 1.00 Ω + 6.00 Ω + 13.0 Ω = 20.0 Ω.

Strategy and Solution for (b)

The current is found using Ohm’s law, V = IR size 12{V= ital "IR"} {} . Entering the value of the applied voltage and the total resistance yields the current for the circuit:

I = V R s = 12 . 0 V 20 . 0 Ω = 0 . 600 A . size 12{I= { {V} over {R rSub { size 8{s} } } } = { {"12" "." 0" V"} over {"20" "." "0 " %OMEGA } } =0 "." "600"" A"} {}

Strategy and Solution for (c)

The voltage—or IR size 12{ ital "IR"} {} drop—in a resistor is given by Ohm’s law. Entering the current and the value of the first resistance yields

V 1 = IR 1 = ( 0 . 600 A ) ( 1 . 0 Ω ) = 0 . 600 V . size 12{V rSub { size 8{1} } = ital "IR" rSub { size 8{1} } = \( 0 "." "600"" A" \) \( 1 "." 0 %OMEGA \) =0 "." "600"" V"} {}

Similarly,

V 2 = IR 2 = ( 0 . 600 A ) ( 6 . 0 Ω ) = 3 . 60 V size 12{V rSub { size 8{2} } = ital "IR" rSub { size 8{2} } = \( 0 "." "600"" A" \) \( 6 "." 0 %OMEGA \) =3 "." "60"" V"} {}

and

V 3 = IR 3 = ( 0 . 600 A ) ( 13 . 0 Ω ) = 7 . 80 V . size 12{V rSub { size 8{3} } = ital "IR" rSub { size 8{3} } = \( 0 "." "600"" A" \) \( "13" "." 0 %OMEGA \) =7 "." "80"" V"} {}

Discussion for (c)

The three IR size 12{ ital "IR"} {} drops add to 12 . 0 V size 12{"12" "." 0`V} {} , as predicted:

V 1 + V 2 + V 3 = ( 0 . 600 + 3 . 60 + 7 . 80 ) V = 12 . 0 V . size 12{V rSub { size 8{1} } +V rSub { size 8{2} } +V rSub { size 8{3} } = \( 0 "." "600" +3 "." "60"+7 "." "80" \) " V"="12" "." 0" 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 = IV size 12{P= ital "IV"} {} , where P size 12{P} {} is electric power. In this case, each resistor has the same full current flowing through it. By substituting Ohm’s law V = IR size 12{V= ital "IR"} {} into Joule’s law, we get the power dissipated by the first resistor as

P 1 = I 2 R 1 = ( 0 . 600 A ) 2 ( 1 . 00 Ω ) = 0 . 360 W . size 12{P rSub { size 8{1} } =I rSup { size 8{2} } R rSub { size 8{1} } = \( 0 "." "600"" A" \) rSup { size 8{2} } \( 1 "." "00" %OMEGA \) =0 "." "360"" W"} {}

Similarly,

P 2 = I 2 R 2 = ( 0 . 600 A ) 2 ( 6 . 00 Ω ) = 2 . 16 W size 12{P rSub { size 8{2} } =I rSup { size 8{2} } R rSub { size 8{2} } = \( 0 "." "600"" A" \) rSup { size 8{2} } \( 6 "." "00" %OMEGA \) =2 "." "16"" W"} {}

and

P 3 = I 2 R 3 = ( 0 . 600 A ) 2 ( 13 . 0 Ω ) = 4 . 68 W . size 12{P rSub { size 8{3} } =I rSup { size 8{2} } R rSub { size 8{3} } = \( 0 "." "600"" A" \) rSup { size 8{2} } \( "13" "." 0 %OMEGA \) =4 "." "68"" W"} {}

Discussion for (d)

Power can also be calculated using either P = IV size 12{P= ital "IV"} {} or P = V 2 R size 12{P= { {V rSup { size 8{2} } } over {R} } } {} , where V size 12{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 = IV size 12{P= ital "IV"} {} , where V size 12{V} {} is the source voltage. This gives

P = ( 0 . 600 A ) ( 12 . 0 V ) = 7 . 20 W . size 12{P= \( 0 "." "600"" A" \) \( "12" "." 0" V" \) =7 "." "20"" W"} {}

Discussion for (e)

Note, coincidentally, that the total power dissipated by the resistors is also 7.20 W, the same as the power put out by the source. That is,

P 1 + P 2 + P 3 = ( 0 . 360 + 2 . 16 + 4 . 68 ) W = 7 . 20 W . size 12{P rSub { size 8{1} } +P rSub { size 8{2} } +P rSub { size 8{3} } = \( 0 "." "360"+2 "." "16"+4 "." "68" \) " W"=7 "." "20"" W"} {}

Power is energy per unit time (watts), and so conservation of energy requires the power output of the source to be equal to the total power dissipated by the resistors.

Major features of resistors in series

  1. Series resistances add: R s = R 1 + R 2 + R 3 + . . . . size 12{R rSub { size 8{s} } =R rSub { size 8{1} } +R rSub { size 8{2} } +R rSub { size 8{3} } + "." "." "." "." } {}
  2. The same current flows through each resistor in series.
  3. Individual resistors in series do not get the total source voltage, but divide it.

Resistors in parallel

[link] shows resistors in parallel    , wired to a voltage source. Resistors are in parallel when each resistor is connected directly to the voltage source by connecting wires having negligible resistance. Each resistor thus has the full voltage of the source applied to it.

Each resistor draws the same current it would if it alone were connected to the voltage source (provided the voltage source is not overloaded). For example, an automobile’s headlights, radio, and so on, are wired in parallel, so that they utilize the full voltage of the source and can operate completely independently. The same is true in your house, or any building. (See [link] (b).)

Part a shows two electrical circuits which are compared. The first electrical circuit is arranged with resistors in parallel. The circuit has three paths, with a voltage source V at one end. Just after the voltage source, the circuit has current I. The first path has resistor R sub one and current I sub one after the resistor. The second path has resistor R sub two and current I sub two after the resistor. The third path has resistor R sub three with current I sub three after the resistor. The first circuit is equivalent to the second circuit. The second circuit has a voltage source V and an equivalent parallel resistance R sub p. Part b shows a complicated electrical wiring diagram of a distribution board that supplies electricity to a house.
(a) Three resistors connected in parallel to a battery and the equivalent single or parallel resistance. (b) Electrical power setup in a house. (credit: Dmitry G, Wikimedia Commons)

To find an expression for the equivalent parallel resistance R p size 12{R rSub { size 8{p} } } {} , let us consider the currents that flow and how they are related to resistance. Since each resistor in the circuit has the full voltage, the currents flowing through the individual resistors are I 1 = V R 1 size 12{I rSub { size 8{1} } = { {V} over {R rSub { size 8{1} } } } } {} , I 2 = V R 2 size 12{I rSub { size 8{2} } = { {V} over {R rSub { size 8{2} } } } } {} , and I 3 = V R 3 size 12{I rSub { size 8{3} } = { {V} over {R rSub { size 8{3} } } } } {} . Conservation of charge implies that the total current I size 12{I} {} produced by the source is the sum of these currents:

Questions & Answers

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
what is the solution
bill
how did you arrive at this answer?
bill
-24m+3+3mÁ^2
Susan
i really want to learn
Amira
I only got 42 the rest i don't know how to solve it. Please i need help from anyone to help me improve my solving mathematics please
Amira
Hw did u arrive to this answer.
Aphelele
hi
Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
complete the table of valuesfor each given equatio then graph. 1.x+2y=3
Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
Hi
Enock
given that (7x-5):(2+4x)=8:7find the value of x
Nandala
3x-12y=18
Kelvin
please why isn't that the 0is in ten thousand place
Grace Reply
please why is it that the 0is in the place of ten thousand
Grace
Send the example to me here and let me see
Stephen
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
Marry Reply
how far
Abubakar
cool u
Enock
state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°
Abegail Reply
hello
BenJay
hi
Method
I am eliacin, I need your help in maths
Rood
how can I help
Sir
hmm can we speak here?
Amoon
however, may I ask you some questions about Algarba?
Amoon
hi
Enock
what the last part of the problem mean?
Roger
The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
Munster
difference between rational and irrational numbers
Arundhati Reply
When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
Jakoiya Reply
how to reduced echelon form
Solomon Reply
Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
Zack Reply
d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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