 # Systematic Circuit Analysis Nodal Analysis Chapter 4 Section 1.

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Systematic Circuit Analysis Nodal Analysis Chapter 4 Section 1

3 Fundamental Principles Ohm’s Law  V = I · R, where I enters at the higher voltage side Kirchoff’s Voltage Law  Algebraic sum of voltages around a loop equal zero Kirchoff’s Current Law  Algebraic sum of currents entering and leaving a node equal zero

Circuit Simplification Series Resistors can be combined into a single equivalent resistance Parallel Resistors can be combined into a single equivalent resistance Current sources in parallel add algebraically Voltage sources in series add algebraically A voltage source is series with a resistor can be replaced by a current source in parallel or vice versa

Shortcuts Voltage Divider – a voltage dividing across a series combination of resistors –Largest voltage is across the largest resistor Current Divider – a current dividing among a parallel combination of resistors –Largest current is through the smallest resistor

Nodal Analysis Employs the 3 fundamental laws May not require any circuit simplification Consists of a straight-forward step-by-step procedure Involves the solution of a set of linear equations simultaneously Used by most circuit analysis computer programs like PSpice

Example Find the voltages and currents

Step One Identify all nodes that have at least 3 components connected to them Choose one of the above nodes as the ground, or reference node, Vo=0 volts Label the other nodes with V 1, V 2, etc.

Voltage Notation Remember that voltage is always measured with respect to some other point So V 1 the voltage at node 1 minus the voltage at node 0, or V 1 -Vo = V 10 = V 1 V 12 would be the voltage at node 1 minus the voltage at node 2, or V 12 = V 1 -V 2

Voltage Notation V 12 could also be obtained by going through any path, say node 3  V 12 = V 132 where V 132 = V 13 + V 32  But V 13 + V 32 = (V 1 -V 3 ) + (V 3 -V 2 )  And (V 1 -V 3 ) + (V 3 -V 2 ) = V 1 -V 2 = V 12

Circuit After Step One V1V1 Vo = 0 volts

Step Two Choose currents for every circuit branch that is connected to any of the unknown voltage nodes –In this circuit, choose currents for the branches connected to V 1 –You can also label voltages at other nodes in the circuit (nodes that have only two components), Va and Vb

Circuit after Step Two I 1 + - I 3 + - + I 2 - V1V1 Vo = 0 volts Va Vb

Step Three Find the value of each current or find an expression for each current in terms of the unknown voltages, V 1, V 2, etc. The idea is to get all the currents to be expressed in terms of the node voltages, so that a KCL equation can be written at each unknown node, giving us as many equations as unknowns

Step Three If you have a resistor between two nodes, then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohm’s Law) For this circuit the 10Ω resistor is between nodes 1 and ground (node 0)  I 2 = (V 1 – 0) / 10Ω = V 1 / 10Ω

Step Three Cont. If you have a resistor and a series voltage source between two key nodes, determine the voltage at the node between the components and then use Ohm’s law There is a 6v source and 14Ω resistor in series between node 1 and ground  Va = Vo+6v. = 6v.  I 1 = (Va - V 1 ) / 14Ω = (6v - V 1 ) / 14Ω

Step Three Cont. There is also 5v source and 10Ω resistor in series between node 1 and ground Determine the voltage at the intermediate node  Vb = Vo+5v. = 5v. Use Ohm’s law to determine the current  I 3 = (V 1 -Vb) / 14Ω = (V 1 -5v) / 10Ω

Circuit after Step Three I 1 =(6-V 1 )/14Ω + - I 3 =(V 1 -5)/10Ω + - + I 2 = V 1 /10Ω - V1V1 Vo = 0 volts Va=6v Vb=5v

Step 4 Write Kirchoff’s Current Law at each node with an unknown voltage  I 1 = I 2 + I 3 Substitute each value or expression into each KCL equation  (6v-V 1 )/14Ω = V 1 /10Ω + (V 1 -5v)/10Ω

Step 5 Solve Solve the set of equations for each unknown voltage  (6v-V 1 )/14Ω = V 1 /10Ω + (V 1 -5v)/10Ω Multiply both sides to clear fractions  70Ω·[(6v-V 1 )/14Ω = V 1 /10Ω + (V 1 -5v)/10Ω]  30v - 5·V 1 = 7·V 1 + 7·V 1 - 35v Group like terms and solve  65v = 19·V 1 or V 1 = 3.42v

Step 5 Solve Cont. Since V 1 = 3.42v Find each current by substituting the voltage values found into each current equation as appropriate  I 1 = (6v-V 1 )/14Ω = (6-3.42)/14 =.184 A  I 2 = V 1 /10Ω = 3.42/10 =.342 A  I 3 = (V 1 -5v)/10Ω = (3.42-5)/10 = -.158 A

Step 6 – Reality Check I 1 =.184A + - I 3 = -.158A + - + I 2 =.342A - V 1 =3.42v Vo = 0 volts Va=6v Vb=5v

Example with a Dependent Source Find the voltages and currents I1I1

Handling the Dependent Source The dependent source will be dependent on some voltage or current in the circuit You will need to express that voltage or current in terms of the unknown node voltages The idea is to only use the node voltages in finding the currents so that there will be as many KCL equations as variables

Step One Identify all nodes that have at least 3 components connected to them Choose one of the above nodes as the ground, or reference node at zero volts Label the other nodes with V 1, V 2, etc.

Circuit after Step 1 V1V1 Vo = 0 volts I1I1

Step Two Choose currents for every circuit branch that are connected to any of the unknown key voltage nodes

Circuit after Step 2 V1V1 Vo = 0 volts - I 3 + + I 2 - I 1 + - Vb

Step Three Find the value of each current or find an expression for each current in terms of the unknown voltages, V 1, V 2, etc.  If you have a resistor between two nodes, then the current is the voltage at the node where the current originates minus the voltage at the other node divided by the resistance (Ohm’s Law)  I 2 = (V 1 – 0) / 4Ω = V 1 / 4Ω

Step Three Cont. If you have a resistor and a series voltage source between two nodes, determine the voltage at the node between the components and then use Ohm’s law  Vb = Vo+3v. = 3v.  I 1 = (V 1 -Vb) / 2Ω = (V 1 -3v) / 2Ω

Step Three Cont. If you have multiple resistors and/or voltage sources between two key nodes, combine them together to get one resistor and one source  If there is a dependent part of the source, express it in terms of the unknown node voltages  Determine the voltage at the node between the equivalent resistor and source  Use Ohm’s law

Circuit for Step 3 V1V1 Vo = 0 volts - I 3 + + I 2 - I 1 + - Vb=3v 5v + 4Ω·I 1 Va Vb

Step Three Cont. The combined dependent source is 5v+4Ω·I 1  So Va = V 1 – (5v+4Ω·I 1 )  But I 1 in the dependent source was determined to be (V 1 -3v)/2Ω  So Va = V 1 –(5v + 4Ω·(V 1 -3v)/2Ω) = -V 1 +1 I 3 = (Vo-Va) / 4Ω = (0-(-V 1 +1)) / 4Ω  I 3 = (V 1 -1v) / 4Ω

Circuit for Step 4 V1V1 Vo = 0 volts - + I 3 =(V 1 -1)/4Ω + I 2 = V 1 /2Ω - I 1 =(V 1 -3)/2Ω + - Vb=3v 5v + 4Ω·I 1 Va= 1v-V 1 Vb

Step 4 Write Kirchoff’s Current Law at each node with an unknown voltage  I 3 = I 2 + I 1 Substitute each value or expression into each KCL equation  (V 1 -1v) / 4Ω = V 1 / 4Ω + (V 1 -3v) / 2Ω

Step 5 Solve Solve the set of equations for each unknown voltage  (V 1 -1v) / 4Ω = V 1 / 4Ω + (V 1 -3v) / 2Ω Multiply both sides to clear fractions  4Ω·[(V 1 -1v) / 4Ω = V 1 / 4Ω + (V 1 -3v) / 2Ω]  V 1 - 1v = V 1 + 2·(V 1 -3v) Group like terms and solve  5v = 2·V 1 or V 1 = 2.5v

Step 5 Solving for Currents Finding each current: I 3 = (V 1 -1v) / 4Ω = (2.5v -1v) / 4Ω =.375 A I 2 = V 1 / 4Ω = 2.5v / 4Ω =.625 A I 1 = (V 1 -3v) / 2Ω = (2.5v-3v) / 2Ω = -.25 A

Step 6 - Checking V 1 =2.5v Vo = 0 volts - + I 3 =.375A + I 2 =.625A - I 1 = -.25A + - Vb=3v 5v+4Ω·I 1 =4v Va= 1v-V 1 = -1.5v Vb

Class Activity Find the current equations at node 1

Class Activity Find I 2 in terms of V 1  For a resistor between nodes, use Ohm’s law  Add polarities if not shown (+ at V 1, - at V 0 )  I 2 = V 1 /60Ω +_+_

Class Activity Find I 3 in terms of its value  For a current source between nodes, the current is fixed by the source  I 3 = 2 A

Class Activity Find I 4 in terms of V 1  If you have series resistors, combine them and then use Ohm’s law  I 4 = V 1 /(30+70)Ω = V 1 /(100Ω) +_+_+_+_

Class Activity Find I 1 in terms of V 1  Combine series resistors, even if not next to each other, then find the voltage between the source and equivalent resistance

Finding I 1 100+80=180Ω Va=100v So I 1 = (Va-V 1 )/180Ω = (100-V 1 )/180Ω Vo=0v V1V1 V1V1 + -

Class Activity Find I 5 in terms of V 1  The voltage on the right side of the 50Ω resistor is known, because of the 150v source  I 5 = (V 1 -150v) / 50Ω + -

Class Activity Write KCL equation at node 1  I 1 + I 3 = I 2 + I 4 + I 5 Substitute all the current equations:  (100-V 1 )/180+2A=V 1 /60+V 1 /100+(V 1 -150v)/50

Class Activity (100-V 1 )/180+2A=V 1 /60+V 1 /100+(V 1 -150v)/50 Multiply both sides by 9000Ω: 5000-50V 1 +18000=150V 1 +90V 1 +180V 1 -27000 Group: 50000 = 470V 1 Solve: V 1 = 106.4 v

Class Activity Checking I 1 =-.036A, I 2 =1.773A, I 3 =2A, I 4 =1.064A, I 5 =-.872A I 1 +I 3 = -.036+2 =1.964 A I 2 +I 4 +I 5 = 1.773+1.064-.872 = 1.965 A – Yes!

Example with 3 Unknown Nodes Find Voltages and Currents

Steps 1 & 2 Find key nodes, assign one to ground, choose currents in branches Vo=0 volts V1V1 V2V2 V3V3 Va I1I1 I5I5 I3I3 I2I2 I4I4 + - +-+- +-+-

Step 3 – Current Equations Currents for resistors between nodes  I 1 =(V 1 -V 3 )/4Ω  I 3 =(V 2 -V 3 )/7Ω  I 4 =V 2 /1Ω  I 5 = V 3 /5Ω Resistor and source between nodes  Va = V 1 + 9v  I 2 = (Va-V 2 )/3Ω = (V 1 +9v-V 2 )/3Ω

Step 4 – KCL Equations KCL Equation at each key node:  At node 1: 0 = I 1 + I 2 + 8A  At node 2: I 2 = I 3 + I 4  At node 3: 25A + I 1 + I 3 = I 5 Substituting for each current  At 1: 0 =(V 1 -V 3 )/4Ω +(V 1 +9v-V 2 )/3Ω +8A  At 2: (V 1 +9v-V 2 )/3Ω =(V 2 -V 3 )/7Ω + V 2 /1Ω  At 3: 25A+(V 1 -V 3 )/4Ω +(V 2 -V 3 )/7Ω = V 3 /5Ω

At node 1 0 =(V 1 -V 3 )/4Ω +(V 1 +9v-V 2 )/3Ω +8A Multiply both sides by 12Ω to clear fractions  12Ω·[0 =(V 1 -V 3 )/4Ω +(V 1 +9v-V 2 )/3Ω +8A]  Or: 0 =3·V 1 -3·V 3 +4·V 1 +36v-4·V 2 +96v Combining terms  7·V 1 -4·V 2 -3·V 3 = -132v

At node 2 (V 1 +9v-V 2 )/3Ω =(V 2 -V 3 )/7Ω + V 2 /1Ω Multiply both sides by 21Ω to clear fractions  21Ω·[(V 1 +9v-V 2 )/3Ω =(V 2 -V 3 )/7Ω + V 2 /1Ω ]  Or: 7·V 1 +63v - 7·V 2 = 3·V 2 - 3·V 3 +21·V 2 Combining terms  7·V 1 -31·V 2 +3·V 3 = -63v

At node 3 25A+(V 1 -V 3 )/4Ω +(V 2 -V 3 )/7Ω = V 3 /5Ω Multiply both sides by 140Ω to clear fractions  140Ω·[25A+(V 1 -V 3 )/4Ω +(V 2 -V 3 )/7Ω =V 3 /5Ω]  Or: 3500v+35·V 1 -35·V 3 +20·V 2 -20·V 3 =28·V 3 Combining terms  35·V 1 +20·V 2 -83·V 3 = -3500v

Set of 3 Simultaneous Eqs. 7·V 1 -4·V 2 -3·V 3 = -132v 7·V 1 -31·V 2 +3·V 3 = -63v 35·V 1 +20·V 2 -83·V 3 = -3500v Solve by hand, calculator, or computer  V 1 = 5.414 v  V 2 = 7.737 v  V 3 = 46.316 v