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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

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hello
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Why is light bright?
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what is radioactive element
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an 8.0 capacitor is connected by to the terminals of 60Hz whoes rms voltage is 150v. a.find the capacity reactance and rms to the circuit
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thanks so much. i undersooth well
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is the study of matter in relation to energy
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what is power?
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power P = Work done per second W/ t. It means the more power, the stronger machine
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e.g. heart Uses 2 W per beat.
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1.75cm
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the same behavior thru the prism out or in water bud abbot
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Jaydie Reply
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Ju
And don't forget to check out YouTube videos on the subject. Videos offer a different visual way to learn easier.
Ju
hope that helps
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I have a exam on 12 february
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what is velocity
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the speed of something in a given direction.
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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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