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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

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Kate Reply
what is the change in momentum of a body?
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what is a capacitor?
Raymond Reply
Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
Gautam
A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
Maria Reply
please solve
Sharon
8m/s²
Aishat
What is Thermodynamics
Muordit
velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
Mehmet
A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
Saheed Reply
50 m/s due south east
Someone
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
Ramon Reply
I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
Someone
about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
Someone
definitely of physics
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how many start and codon
Esrael Reply
what is field
Felix Reply
physics, biology and chemistry this is my Field
ALIYU
field is a region of space under the influence of some physical properties
Collete
what is ogarnic chemistry
WISDOM Reply
determine the slope giving that 3y+ 2x-14=0
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Another formula for Acceleration
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a=v/t. a=f/m a
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innocent
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pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?
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how do lnternal energy measures
Esrael
Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.
JALLAH Reply
No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.
Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
Robert
like charges repel while unlike charges atttact
Raymond
What is specific heat capacity
Destiny Reply
Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).
AI-Robot
specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
ROKEEB
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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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