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These energies must be equal, because there is no other source and no other destination for energy in the circuit. Thus, qV = qV 1 + qV 2 + qV 3 size 12{ ital "qV"= ital "qV" rSub { size 8{1} } + ital "qV" rSub { size 8{2} } + ital "qV" rSub { size 8{3} } } {} . The charge q size 12{q} {} cancels, yielding V = V 1 + V 2 + V 3 size 12{V=V rSub { size 8{1} } +V rSub { size 8{2} } +V rSub { size 8{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 = IR 1 + IR 2 + IR 3 = I ( R 1 + R 2 + R 3 ) . size 12{V= ital "IR" rSub { size 8{1} } + ital "IR" rSub { size 8{2} } + ital "IR" rSub { size 8{3} } =I \( R rSub { size 8{1} } +R rSub { size 8{2} } +R rSub { size 8{3} } \) } {}

Note that for the equivalent single series resistance R s , we have

V = IR s .

This implies that the total or equivalent series resistance R s of three resistors is 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} } } {} .

This logic is valid in general for any number of resistors in series; thus, the total resistance R s of a series connection is

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} } + "." "." "." } {}

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 12 . 0 V size 12{"12" "." 0`V} {} , and the resistances are 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 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:

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"} {}

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Practice Key Terms 9

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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