Experiment 4 Mesh Analysis and Nodal Analysis 1 PDF

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Summary

Lab report...

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Interpretation of Results:

Experiment 4 is all about the mesh analysis and the nodal analysis. The objectives of this experiment are to investigate the effects of mesh analysis on multiple active linear sources in a network, to verify that the linear response at any point in a mesh circuit is similar to Kirchhoff’s voltage law, to investigate the effects of nodal analysis on multiple active linear source in a network and to verify that the linear response at any point in a nodal circuit is similar to Kirchhoff’s current law.

Mesh Analysis is a technique applicable only to those networks which are planar. A planar circuit is a circuit where no branch passes over or under any other branch. In mesh analysis we used the Kirchhoff’s Voltage Law or KVL. The network diagram must strictly be planar or two dimensional. The mesh format requires all sources to be voltage sources and if there is any current source, it should convert to its equivalent voltage. In each mesh, we draw a current loop. All mesh currents must be in the same or uniform in direction. It can be clockwise or counter clockwise direction. We can solve for the unknown mesh current by applying techniques involved in evaluating system of linear equations.

Nodal Analysis is a method that uses Kirchhoff’s Current Law or KCL to obtain a solution of simultaneous equations that when manipulated will provide a means of solving for the voltages on each node and for every branch. The network diagram must strictly be

planar or 2 dimensional. The node format requires all sources to be current sources. If there is any voltage source, convert it to its equivalent current source. In doing the node analysis, 1st we must identify the nodes of the given circuit including the reference node. Solve for the unknown voltages by applying techniques involved in evaluating system of linear equations. Simulated Values

Calculated Values

I1 366.12 mA

Mesh Currents I2 158.47 mA

I1 0.3661 mA

I2 0.1585 A

I3 -360.66 mA I3 0.3607 A

V1 3.66 V

V2 -109.29 mV

V1 3.661 V

V2 0.108 V

Voltages V3 6.23 V

V4 634 V

V5 10.11 V

V3 6 .228V

V4 634 V

V5 10.11 V

In this table, we used the mesh analysis method in order to get the calculated values of mesh currents and voltages. In simulated values, we used tina pro to get the mesh currents. As shown in the table, there is only minimal discrepancy between the values we get in computing the mesh currents and simulated values. This means that the value we computed is right. Simulated Values

Calculated Values

V1 -3.33 V I1 -3.331 V

Node Voltages V2 6.67 V I2 6.67 V

V3 -10 V

I1 -333.33 mA

Currents I2 333.33 mA

I3 -333.33 mA

I3 -10 V

V1 - 0.333 A

V2 0.335 A

V3 -0.3333 A

In this table, we used the nodal analysis method in order to get the calculated values of mesh current. In simulated values, we used tina pro to get the node voltages. As shown in the table, there is only minimal discrepancy between the values we get in computing the

node voltages and simulated values. This means that the value we computed is right. Comparing the data that we have gathered, the values that we obtained in calculating the currents and voltages in both the mesh and nodal analysis is precisely similar or equal.

Answers to Questions and Problems

1. A mesh current is a current that loops around the essential mesh. The sign convention is to have all the mesh currents looping in a clockwise direction. 2. A negative mesh current implies that the mesh current we used is incorrect in direction. It implies that the mesh current flow in opposite direction. 3. a.) If a current source appears on the periphery of only one mesh on a given mesh circuit, the value of the current source in that mesh is equal to the mesh current on that mesh. b.) If a current source is common to two meshes of a given circuit, the technique we should apply is super mesh. 4. The node equation that can be obtained from an N number of nodes present on a given circuit is (N-1). It is because one node serves as a reference node. 5. a.) If a voltage source appears connected to a given node and the reference node in a given nodal circuit, the nodal voltage is equal to the voltage source. b) If a voltage source lies between two given nodes of a given nodal circuit, we apply super node.

6. For mesh analysis, the basic law used was Kirchhoff’s Voltage Law (KVL). For Nodal analysis, the basic law used was Kirchhoff’s Current Law (KCL). 7. Using Mesh Analysis: Converting 2A source to a Voltage source: V s=2 ( 20 ) V s=40 V

At Mesh 1: 24 −40 = ( 5+20 ) I 1 −20 I 2 −16=25 I 1−20 I 2

Where:

I 1 =I x

−16=25 I x −20 I 2

At Mesh 2: 40 −36 =−20 I x +(10 + 20) I 2 4=−20 I x + 30 I 2

Equating Mesh 1 and Mesh 2, we can get: I x =−1.1429 A

8. Using mesh analysis: At Mesh 1:

+2 2¿I ¿ 10−5=¿ 5=4 I 1−2 I 2

Where:

I 2 =I x

At Super mesh 2 and 3: 5−4 I x =−2 I 1 + (2+10 ) I 2 +10 I 3 5=−2 I 1 +16 I 2+10 I 3

Using KCL: 3 V x =I 3−I 2

Where:

V x =2 ( I 1− I 2 )

6 I 1 −6 I 2=I 3−I 2 0=−6 I 1+5 I 2 +I 3

By equating the 3 meshes: I 1 =8 A I 2 =13.5 A I 3 =−19.5 A

9. Using nodal analysis: assuming the reference point is below -8 A and -25A.

At node 1: −8−3=V 1

−11=

( 13 +41)−V ( 13 )−V ( 14 ) 2

3

7 1 1 V − V − V 12 1 3 2 4 3

At node 2: 3=−V 1

3=

( 13 )+V ( 13 +1+ 12 )−V ( 12 ) 2

3

−1 1 11 V 1 + V 2− V 3 2 6 3

At node 3: 25=−V 1

25=

( 14 )−V ( 12 )+V ( 12 + 41 + 51 ) 2

3

−1 19 1 V 1− V 2 + V 3 20 2 4

By equating the 3 node equations, we have: V 1=0.9559 V

V 2=10.5735 V V 3=32.1324 V

10. Using nodal analysis: reference node below -8 A and -25 A

At node 1: −8−3 =V 1

−11=

( 13 +41)−V ( 13 )−V ( 41 ) 2

3

7 1 1 V − V − V 12 1 3 2 4 3

At super node 2 and 3: 25+3=−V 1

28=

( 13 ) +V ( 13 +1) +V ( 51 + 14 ) 2

3

−1 9 4 V 1 + V 2+ V 3 20 3 3

By KVL: V 3 −V 2 =

Ix 2

Where:

I x=

V 1−V 2 4

8 V 3−8 V 2=V 1−V 2 0=V 1+7 V 2 −8 V 3

By equating 3 node equations: V 1=−4.4216 V V 2=15.5029 V

V 3=13.0124 V

Conclusion:  A mesh current is a current that loops around the essential mesh. The sign convention is to have all the mesh currents looping in a clockwise direction.  A negative mesh current implies that the mesh current we used is incorrect in direction. It implies that the mesh current flow in opposite direction.

 If a current source appears on the periphery of only one mesh on a given mesh circuit, the value of the current source in that mesh is equal to the mesh current on that mesh.  The node equation that can be obtained from an N number of nodes present on a given circuit is (N-1). It is because one node serves as a reference node.  If a voltage source appears connected to a given node and the reference node in a given nodal circuit, the nodal voltage is equal to the voltage source.  Mesh analysis is a method that is used to solve planar circuits for the voltage and currents at any place in the circuit.  Mesh analysis uses Kirchhoff’s Voltage Law to solve for the voltages or currents.  Mesh analysis creates a systematic approach for solving the values and reduces the number of equation needed to solve the circuit.  Super mesh occurs when the current source is common to two meshes  For solving the super mesh, we treat the current source absent from the circuit and apply mesh analysis at the meshes that the current source was contained. After obtaining the equation, we use Kirchhoff’s Current Law to equate it to the obtained equations.  Nodal Analysis is the method of determining the voltage between nodes in a circuit in terms of branch currents.

 We apply Kirchhoff’s Current Law in nodal analysis  Super nodes occur when there’s a voltage source between two nodes.  For solving super node, we treat the two nodes that containing the voltage source as one node and have an equation. After obtaining the equation, apply Kirchhoff’s Voltage Law to the two nodes with a voltage source in between. Then, use the equation to equate it to the other obtained equations.  For mesh analysis, the basic law used was Kirchhoff’s Voltage Law (KVL). For Nodal analysis, the basic law used was Kirchhoff’s Current Law (KCL)....