Chapter 5 Parallel Circuits

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ElectronicCircuits&Electronics Shao-YuLien 1

5.1 Resistors in Parallel

Definition. If there is more than one circuit path (branch) between two points, and if the voltage between those two points also appears across each of the branch, then there is a parallel circuit between those points

5.1 Resistors in Parallel

5.2 Total Parallel Resistance

• As resistors are added, there are more paths for current• There is increased conductance

• Formula for total resistance GT=G1+G2+…+Gn!"#= !

"'+…+ !

RT= !%)%* %)'*⋯* %

Example. Calculate the total parallel resistance between point A and B of the following circuit.

• Find the conductanceG1=1/R1=1/100 Ω=10 mSG2=1/R2=1/47 Ω=21.3 mS

G3=1/R3=1/22 Ω=45.5 mS• The total resistance is

RT= !%)%* %)'* %)-

= !.%*.'*.-

= !!/12*3!.512*67.712

= !89.:12

=13.0 Ω

• The case of two resistors in parallel

RT= !%)%* %)'

= "%"'"%*"'

• The total resistance for two resistors in parallel is equal tothe product of the two resistors divided by the sum of thetwo resistors

Example. Calculate the resistance connected to the voltage source of the following circuit

RT= "%"'"%*"'

=(9:/<)(55/<)9:/<*55/<

= 336,///<'

!,/!/<= 222Ω

• The case of equal-value resistors in parallel: For a parallel circuit consisting of n equal-value resistors connected in parallel, the total resistance is

RT=R/nExample. Find the total resistance between A and B in thefollowing circuit

RT=R/n=(100 Ω)/5=20 Ω

• Notations for parallel resistors: When n resistors are in parallel with each other, the notation can be as follows

R1 || R2 || … || Rn

5.3 Voltage in a Parallel Circuit

• The same voltage appears across each branch in a parallel circuit

Example. Determine the voltage across each resistor in thefollowing circuit

• The five resistors are in parallel, so the voltage across each oneis equal to the source voltage

V1=V2=V3=V4=V5=25 VElectronicCircuits&Electronics Shao-YuLien 15

5.4 Application of Ohm’s Law

Example. Determine the current through each resistor in the following parallel circuit

• The voltage across each resistor (branch) is equal to the source voltage

I1=VS/R1=(20 V)/(1.0 kΩ)=20.0 mAI2=VS/R2=(20 V)/(2.2 kΩ)=9.09 mAI3=VS/R3=(20 V)/(560 Ω)=35.7 mA

Example. Find the voltage across the following parallel circuit.

• The total current into the parallel circuit is 37 mA• The total resistance is

RT= !%)%* %)'* %)-

''@A*%

BC@A*%

%.@DA=136 Ω

• Therefore, the source voltage and the voltage across each branch is

VS=ITRT=(37 mA)(136 Ω)=5.05 V

5.5 Kirchhoff’s Current Law

• Kirchhoff’s current law: The sum of the current into a node (total current in) is equal to the sum of the currents out of that node (total current out)

• A node is any point or junction in a circuit where two or more components are connected

• In a parallel circuit, a node is a point where the parallel branches come together

IT=I1+I2+I3

• By Kirchhoff’s law, the sum of the currents into a node mustequal the sum of the currents out of that node

IIN(1)+IIN(2)+…+IIN(n)- IOUT(1)-IOUT(2)-…-IOUT(m)=0

• The algebraic sum of all of the currents entering and leaving a node is equal to zero

Example. Determine the current through R2 in the following.

IT=I1+I2+I3

I2=IT-I1-I3=100 mA-30 mA-20 mA=50 mA

Example. Use Kerchhoff’s current law to find the current measured by ammeters A3 and A5 in the following

• The total current out of node X is 5 mA• Two currents are into node X: 1.5 mA through resistor R3 and

the current through A3• Kirchhoff’s current law at node X is

5 mA=1.5 mA+IA3

IA3=3.5 mA• The total current out of Y is IA3=3.5 mA• Two currents are into node Y: 1 mA through resistor R2 and

current through A5 and R3

• Kirchhoff’s current law at node Y is 3.5 mA=1 mA+IA5

IA5=2.5 mAElectronicCircuits&Electronics Shao-YuLien 26

5.6 Current Dividers

• A parallel circuit acts as a current divider because the current entering the junction of parallel branches divides up into several individual branch currents

• The total current divides among parallel resistors into currents with values inversely proportional to the resistance values

• Current-divider formula

• Let Ix denote the current through any one of the parallel resistors

• By Ohm’s law, the current any one of the resistors isIx=VS/Rx

• The source voltage VS is equal to the total current times the total parallel resistance

VS=ITRT

• Substituting ITRT for VS results in

𝐼F =𝐼G𝑅G𝑅F

• Rearranging terms yields

𝐼F =𝑅G𝑅F

• The current (Ix) through any branch equals the totalparallel resistance (RT) divided by the resistance (Rx) of thatbranch, and multiplied by the total current (IT) into thejunction of parallel branches

Example. Determine the current through each resistor in the following circuit

• First, find the total parallel resistance

RT= !%

CI@A*%

--@A*%

''@A=110 Ω

• The total current is 10 mA, the current at each branch is

I1= "#"%

𝐼G= !!/<9:/<

10mA=1.63 mA

I2= "#"'

𝐼G= !!/<55/<

10mA=3.36 mA

I3= "#"-

𝐼G= !!/<33/<

10mA=5.05 mA

• Current-divider formulas for two branches

I1= "'"%*"'

I2= "%"%*"'

Example. Find I1 and I2 in the following circuits

I1= "'"%*"'

𝐼G= 68<!68<

100 mA=32.0 mA

I2= "%"%*"'

𝐼G= !//<!68<

100 mA=68.0 mA

5.7 Power in Parallel Circuits

• The formula for total power for any number of resistors inparallel is

PT=P1+P2+…+Pn

• The following formulas are used to calculate the total powerPT=VSIT

PT=IT2RT

𝑃G =𝑉P3

Example. Determine the total amount of power in the followingparallel circuit

• The total resistance is

𝑅G =1

168Ω +

133Ω +

= 11.1Ω

• The total current is 200 mA• The total power is

𝑃G = 𝐼G3𝑅G=(200 mA)2(11.1 Ω)=444 mV

• Let’s calculate the total power by adding individual powerconsumption on each resistor

• The voltage across each branch isVS=ITRT=(200 mA)(11.1 Ω)=2.22 V

• Use P=𝑉P3/R to calculate power for each resistorP1=(2.22 V)2/68 Ω=72.5 mWP2=(2.22 V)2/33 Ω=149 mWP3=(2.22 V)2/22 Ω=224 mW

• The total power isPT=72.5 mA+149 mV+224 mV=446 mV

Chapter 5 Parallel Circuits

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