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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:

Practice Key Terms 9

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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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