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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

Determine the total force and the absolute pressure on the bottom of a swimming pool 28.0m by 8.5m whose uniform depth is 1 .8m.
Henny Reply
for the answer to complete, the units need specified why
muqaddas Reply
That's just how the AP grades. Otherwise, you could be talking about m/s when the answer requires m/s^2. They need to know what you are referring to.
Kyle
Suppose a speck of dust in an electrostatic precipitator has 1.0000×1012 protons in it and has a net charge of –5.00 nC (a very large charge for a small speck). How many electrons does it have?
Alexia Reply
how would I work this problem
Alexia
how can you have not an integer number of protons? If, on the other hand it supposed to be 1e12, then 1.6e-19C/proton • 1e12 protons=1.6e-7 C is the charge of the protons in the speck, so the difference between this and 5e-9C is made up by electrons
Igor
what is angular velocity
Obaapa Reply
angular velocity can be defined as the rate of change in radian over seconds.
Fidelis
Why does earth exert only a tiny downward pull?
Mya Reply
hello
Islam
Why is light bright?
Abraham Reply
what is radioactive element
Attah Reply
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
Aisha Reply
thanks so much. i undersooth well
Valdes Reply
what is physics
Nwafor Reply
is the study of matter in relation to energy
Kintu
physics can be defined as the natural science that deals with the study of motion through space,time along with its related concepts which are energy and force
Fidelis
a submersible pump is dropped a borehole and hits the level of water at the bottom of the borehole 5 seconds later.determine the level of water in the borehole
Obrian Reply
what is power?
aron Reply
power P = Work done per second W/ t. It means the more power, the stronger machine
Sphere
e.g. heart Uses 2 W per beat.
Rohit
A spherica, concave shaving mirror has a radius of curvature of 32 cm .what is the magnification of a persons face. when it is 12cm to the left of the vertex of the mirror
Alona Reply
did you solve?
Shii
1.75cm
Ridwan
my name is Abu m.konnek I am a student of a electrical engineer and I want you to help me
Abu
the magnification k = f/(f-d) with focus f = R/2 =16 cm; d =12 cm k = 16/4 =4
Sphere
what do we call velocity
Kings
A weather vane is some sort of directional arrow parallel to the ground that may rotate freely in a horizontal plane. A typical weather vane has a large cross-sectional area perpendicular to the direction the arrow is pointing, like a “One Way” street sign. The purpose of the weather vane is to indicate the direction of the wind. As wind blows pa
Kavita Reply
hi
Godfred
what about the wind vane
Godfred
If a prism is fully imersed in water then the ray of light will normally dispersed or their is any difference?
Anurag Reply
the same behavior thru the prism out or in water bud abbot
Ju
If this will experimented with a hollow(vaccum) prism in water then what will be result ?
Anurag
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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