chapter 5 parallel circuits
TRANSCRIPT
Chapter 5 Parallel Circuits
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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
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5.1 Resistors in Parallel
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5.2 Total Parallel Resistance
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5.2 Total Parallel Resistance
• As resistors are added, there are more paths for current• There is increased conductance
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5.2 Total Parallel Resistance
• Formula for total resistance GT=G1+G2+…+Gn!"#= !
"%+ !
"'+…+ !
"(
RT= !%)%* %)'*⋯* %
)(
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5.2 Total Parallel Resistance
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
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5.2 Total Parallel Resistance
G3=1/R3=1/22 Ω=45.5 mS• The total resistance is
RT= !%)%* %)'* %)-
= !.%*.'*.-
= !!/12*3!.512*67.712
= !89.:12
=13.0 Ω
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5.2 Total Parallel Resistance
• 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
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5.2 Total Parallel Resistance
Example. Calculate the resistance connected to the voltage source of the following circuit
RT= "%"'"%*"'
=(9:/<)(55/<)9:/<*55/<
= 336,///<'
!,/!/<= 222Ω
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5.2 Total Parallel Resistance
• 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
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5.2 Total Parallel Resistance
• Notations for parallel resistors: When n resistors are in parallel with each other, the notation can be as follows
R1 || R2 || … || Rn
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5.3 Voltage in a Parallel Circuit
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5.3 Voltage in a Parallel Circuit
• The same voltage appears across each branch in a parallel circuit
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5.3 Voltage 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
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5.4 Application of Ohm’s Law
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
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5.4 Application of Ohm’s Law
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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
5.4 Application of Ohm’s Law
• Therefore, the source voltage and the voltage across each branch is
VS=ITRT=(37 mA)(136 Ω)=5.05 V
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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
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5.5 Kirchhoff’s Current Law
IT=I1+I2+I3
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5.5 Kirchhoff’s Current Law
• By Kirchhoff’s law, the sum of the currents into a node mustequal the sum of the currents out of that node
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5.5 Kirchhoff’s Current Law
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
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5.5 Kirchhoff’s Current Law
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
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5.5 Kirchhoff’s Current Law
Example. Use Kerchhoff’s current law to find the current measured by ammeters A3 and A5 in the following
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5.5 Kirchhoff’s Current Law
• 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
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5.6 Current Dividers
• 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
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5.6 Current Dividers
• 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
𝐼G
• 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
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5.6 Current Dividers
Example. Determine the current through each resistor in the following circuit
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5.6 Current Dividers
• 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
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5.6 Current Dividers
• Current-divider formulas for two branches
I1= "'"%*"'
𝐼G
I2= "%"%*"'
𝐼G
Example. Find I1 and I2 in the following circuits
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5.6 Current Dividers
I1= "'"%*"'
𝐼G= 68<!68<
100 mA=32.0 mA
I2= "%"%*"'
𝐼G= !//<!68<
100 mA=68.0 mA
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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
𝑅G
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5.7 Power in Parallel Circuits
Example. Determine the total amount of power in the followingparallel circuit
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• The total resistance is
𝑅G =1
168Ω +
133Ω +
122Ω
= 11.1Ω
5.7 Power in Parallel Circuits
• The total current is 200 mA• The total power is
𝑃G = 𝐼G3𝑅G=(200 mA)2(11.1 Ω)=444 mV
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5.7 Power in Parallel Circuits
• 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
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