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(c) At first glance- one may assume that when applying mesh analysis t.docx

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(c) At first glance- one may assume that when applying mesh analysis t.docx

(c) At first glance, one may assume that when applying mesh analysis to the circuit must define two separate super-meshes, one for each current source shared However, with some clever rearrangement/redrawing of circuit elements, unknown mesh currents can result and super-meshes can be avoided altogether. Figure 1 i. Re-draw the circuit in Figure 1 to overcome the need for defining two super-meshes. ii. Formulate a minimum number of mesh current equations for the redrawn circuit. Treat all the circuit parameters as variables. iii. Find voltage vx given R1 = R2 = R3 = R4 = 4kOhm, R5 = R6 = 2kOhm, iS1 = 80mA,iS2 =1/2iS1. iv. In Fundamentals of Electric Circuits, Alexander & Sadiku, McGraw Hill, Inc.; 5/e (2013), to which figures between Figure 3.85 and Figure 3.100 can we apply the same technique as applied to the circuit in Figure 1 thereby simplifying the analysis when using the MCM?
Solution
Now applying the technique of mesh current analysis and simplifying we have the loop equations by taking the direction of current clock wise we have
6.4kI 1 -1.6kI 2 -0.8kI 3 =(-80*10 -3 ).........................Loop 1.
Similarily for loop 2 we can write
-1.6kI 1 +6.4KI 2 -0.8KI 3 =(40*10 -3 ).........................Loop 2.
for loop 3 we have
-0.8k I1-0.8kI2+3.6kI3=40*10 -3 ..................................loop3
Solving these equations we have
I1=-0.001mA, i2=0.005mA, I3=0.01mA.
Voltage Vx of the circuit = I3*R4=[0.001*10 -3 *4*10 3 ]= 0.004V.
.

(c) At first glance, one may assume that when applying mesh analysis to the circuit must define two separate super-meshes, one for each current source shared However, with some clever rearrangement/redrawing of circuit elements, unknown mesh currents can result and super-meshes can be avoided altogether. Figure 1 i. Re-draw the circuit in Figure 1 to overcome the need for defining two super-meshes. ii. Formulate a minimum number of mesh current equations for the redrawn circuit. Treat all the circuit parameters as variables. iii. Find voltage vx given R1 = R2 = R3 = R4 = 4kOhm, R5 = R6 = 2kOhm, iS1 = 80mA,iS2 =1/2iS1. iv. In Fundamentals of Electric Circuits, Alexander & Sadiku, McGraw Hill, Inc.; 5/e (2013), to which figures between Figure 3.85 and Figure 3.100 can we apply the same technique as applied to the circuit in Figure 1 thereby simplifying the analysis when using the MCM?
Solution
Now applying the technique of mesh current analysis and simplifying we have the loop equations by taking the direction of current clock wise we have
6.4kI 1 -1.6kI 2 -0.8kI 3 =(-80*10 -3 ).........................Loop 1.
Similarily for loop 2 we can write
-1.6kI 1 +6.4KI 2 -0.8KI 3 =(40*10 -3 ).........................Loop 2.
for loop 3 we have
-0.8k I1-0.8kI2+3.6kI3=40*10 -3 ..................................loop3
Solving these equations we have
I1=-0.001mA, i2=0.005mA, I3=0.01mA.
Voltage Vx of the circuit = I3*R4=[0.001*10 -3 *4*10 3 ]= 0.004V.
.

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(c) At first glance- one may assume that when applying mesh analysis t.docx

1. 1. (c) At first glance, one may assume that when applying mesh analysis to the circuit must define two separate super-meshes, one for each current source shared However, with some clever rearrangement/redrawing of circuit elements, unknown mesh currents can result and super- meshes can be avoided altogether. Figure 1 i. Re-draw the circuit in Figure 1 to overcome the need for defining two super-meshes. ii. Formulate a minimum number of mesh current equations for the redrawn circuit. Treat all the circuit parameters as variables. iii. Find voltage vx given R1 = R2 = R3 = R4 = 4kOhm, R5 = R6 = 2kOhm, iS1 = 80mA,iS2 =1/2iS1. iv. In Fundamentals of Electric Circuits, Alexander & Sadiku, McGraw Hill, Inc.; 5/e (2013), to which figures between Figure 3.85 and Figure 3.100 can we apply the same technique as applied to the circuit in Figure 1 thereby simplifying the analysis when using the MCM? Solution Now applying the technique of mesh current analysis and simplifying we have the loop equations by taking the direction of current clock wise we have 6.4kI 1 -1.6kI 2 -0.8kI 3 =(-80*10 -3 ).........................Loop 1. Similarily for loop 2 we can write -1.6kI 1 +6.4KI 2 -0.8KI 3 =(40*10 -3 ).........................Loop 2. for loop 3 we have -0.8k I1-0.8kI2+3.6kI3=40*10 -3 ..................................loop3 Solving these equations we have I1=-0.001mA, i2=0.005mA, I3=0.01mA. Voltage Vx of the circuit = I3*R4=[0.001*10 -3 *4*10 3 ]= 0.004V.