Network Analysis V Mesh Equations Three Loops
Circuit overview A B V1 12 V R1 R3 C R2 R4 I A I B I C D R6 E F R5 R7 R8 G V2 8 V H Using the method of mesh currents, solve for all the unknown values of voltage and current in the figure shown. To do this, we will complete steps a through p.
Listing of steps a. Identify the components through which the mesh current I A flows. b. Identify the components through which the mesh current I B flows. c. Identify the components through which the mesh current I C flows.
Listing of steps (cont.) d. Which components have opposing mesh currents (if any)? e. Write the mesh equation for mesh A. f. Write the mesh equation for mesh B. g. Write the mesh equation for mesh C.
Listing of steps (cont.) h. Solve for the currents I A, I B and I C using any of the methods for the solution of simultaneous equations. i. Determine the values of currents I 1, I 2, I 3, I 4, I 5, I 6, I 7 and I 8. j. Are the assumed directions of the three mesh (mesh A, mesh B and mesh C) currents correct? How do you know?
Listing of steps (cont.) k. What are the directions of current I 2 and I 5 through the components R 2 and R 5 respectively? l. Solve for the voltage drops V R1, V R2, V R3, V R4, V R5, V R6, V R7 and V R8. m. Using the final solutions for V R1, V R2 and V R3, write a KVL equation for the loop ACDBA going clockwise from point A.
Listing of steps (cont.) n. Using the final solutions for V R2, V R4, V R5 and V R6, write a KVL equation for the loop EFDCE going clockwise from point E. o. Using the final solutions for V R5, V R7 and V R8, write a KVL equation for the loop GHFEG going clockwise from point G. p. Using the final solutions for V R1, V R3, V R4, V R6, V R7 and V R8, write a KVL equation for the loop ACEGHFDBA going clockwise from point A.
Step a solution A B V1 12 V R1 I A R3 C R2 D a. Identify the components through which the mesh current I A flows. I A flows through V 1, R 1, R 2 and R 3 (not necessarily in that order).
Step b solution C R4 E b. Identify the components through which the mesh current I B flows. R2 D I B R6 R5 F I B flows through R 2, R 4, R 5 and R 6 (not necessarily in that order).
Step c solution R5 E F R7 I C R8 G V2 8 V H c. Identify the components through which the mesh current I C flows. I C flows through V 2, R 5, R 7 and R 8 (not necessarily in that order)
Step d solution d. Which components have opposing mesh currents (if any)? A B V1 12 V R1 R3 C R2 R4 R6 R5 R7 I A I B I C D E F R8 G The components with V2 the opposing mesh 8 V currents are R 2 and R 5 (down through R 2 from I A and up through R 2 from H I B, while for R 5 it is down through it from I B and up through it from I C ).
Step e solution A R1 C e. Write the mesh equation for mesh A. V1 12 V I A R2 6I A 2I B 0I C = 12V B R3 D
Step f solution C R4 E f. Write the mesh equation for mesh B. R2 I B R5-2I A 8I B 2I C = 0V R6 D F
Step g solution E R7 G g. Write the mesh equation for mesh C. R5 I C V2 8 V 0I A 2I B 6I C = -8V F R8 H
Step h solution Original equations: Step e: 6I A 2I B 0 = 12V Step f: -2I A 8I B 2I C = 0V Step g: 0 2I B 6I C = -8V h. Solve for the currents I A, I B and I C using any of the methods for the solution of simultaneous equations. Step 1 Since equation from step f is equal to zero, we will rewrite it to solve for I C -2I A 8I B = 2I C (reorder as 2I C = -2I A 8I B )
Step h solution (cont.) Step 2 Divide by 2 to reduce I C = I A 4I B Step 3 Substitute into equation from step g to eliminate third variable (I C ) 0 2I B 6 I A 4I B = 8
Step h solution (cont.) Step 4 Multiply through and combine like terms 0 2I B 6I A 24I B = 8 0 6I A 22I B = 8
Step h solution (cont.) Step 5 Examine remaining expression. Convert how necessary to eliminate one term and then add (Remaining expression: 6I A 2I B = 12; Since this expression contains the value of I A necessary to eliminate I A, nothing more needs to be done except add) 6I A 2I B = 12 6I A 22I B = 8 20I B = 4
Step h solution (cont.) Step 6 Solve for I B by dividing multiplier 20I B 20 = 4 20 I B = 0. 2A
Step h solution (cont.) Step 7 Substitute I B into either equation to solve for I A (we will use one from step e as it is the easiest) 6I A 2(0.2A) = 12 6I A 0.4 = 12
Step h solution (cont.) Step 7a Move all the constants to the right side, which will leave our unknown on the left 6I A = 12.4 Step 7b Divide multiplier to solve for I A 6I A 6 12. 4 = 6 I A = 2. 067A
Step h solution (cont.) Step 8 Substitute I B into equation from step g to solve for I C -2(0.2A) 6I C = -8-0.4 6I C = -8
Step h solution (cont.) Step 8a Move all constants to the right which once again leaves our unknown on the left 6I C = -7.6 Step 8b Divide multiplier to solve for I C 6I C 6 7. 6 = 6 I C = 1. 267A
Step i solution i. Determine the values of currents I 1, I 2, I 3, I 4, A R1 C R4 E R7 I 5, I 6, I 7 and I 8. G I 1 = I 3 = I A = 2.067A V1 12 V R2 R5 V2 8 V I 4 = I 6 = I B = 0.2A I 7 = I 8 = I C = -1.267A I 2 = I A I B = 2.067A 0.2A = 1.867A I 5 = I C I B = -1.267A 0.2A = -1.467A B R3 D R6 F R8 H
Step j solution j. Are the assumed directions of the three mesh (mesh A, mesh B and mesh C) currents correct? How do you know? Following conventional current flow, the meshes of A and B were correct, but the direction of mesh C needs to be reversed (counterclockwise) since it was negative.
Step k solution k. What are the directions of current I 2 and I 5 through the components R 2 and R 5 respectively? Current is flowing down through R 2 (I A and I B are subtractive). The current also flows downward through R 5 (I B and I C are additive).
Step l solution l. Solve for the voltage drops V R1, V R2, V R3, V R4, V R5, V R6, V R7 and V R8. V R1 = I 1 R 1 = 2.067A() = 4.134V V R2 = I 2 R 2 = 1.867A() = 3.734V V R3 = I 3 R 3 = 2.067A() = 4.134V V R4 = I 4 R 4 = 0.2A() = 0.4V
Step l solution (cont.) l. Solving for the voltage drops V R1, V R2, V R3, V R4, V R5, V R6, V R7 and V R8 (cont.). V R5 = I 5 R 5 = -1.467A() = -2.934V V R6 = I 6 R 6 = 0.2A() = 0.4V V R7 = I 7 R 7 = -1.267A() = -2.534V V R8 = I 8 R 8 = -1.267A() = -2.534V
Step m solution A V1 12 V R1 4.134V R2 C 3.734V m. Using the final solutions for V R1, V R2 and V R3, write a KVL equation for the loop ACDBA going clockwise from point A. 4.134V 3.734V 4.134V 12V = 0 R3 B 4.134V D
Step n solution C R2 3.734V R4 0.4V R5 E 2.934V n. Using the final solutions for V R2, V R4, V R5 and V R6, write a KVL equation for the loop EFDCE going clockwise from point E. 2.934V 0.4V 3.734V 0.4V = 0 R6 D 0.4V F
Step o solution E R5 R7 2.534V 2.934V G V2 8 V o. Using the final solutions for V R5, V R7 and V R8, write a KVL equation for the loop GHFEG going clockwise from point G. 8V 2.534V 2.934V 2.534V = 0 F R8 2.534V H
Step p solution p. Using the final solutions for V R1, V R3, V R4, V R6, V R7 and V R8, write a KVL equation for the loop ACEGHFDBA going clockwise from point A. 4.134V 0.4V 2.534V 8V 2.354V 0.4V 4.134V 12V = 0 A R1 C R4 E R7 G 4.134V 0.4V 2.534V V1 12 V V2 8 V B R3 R6 R8 4.134V D 0.4V F 2.534V H
The circuit with voltage measurements according to original current assumption U1-4.133 V U4-0.400 V U7 - -2.533 V R1 R4 R7 V1 12 V R2-3.733 V U2 R5-2.933 V U5 V2 8 V R3 - -4.133 V R6 - -0.400 V R8-2.533 V U3 U6 U8
The circuit with voltage measurements according to corrected current directions U1-4.133 V U4-0.400 V - V U7 2.533 R1 R4 R7 V1 12 V R2-3.733 V U2 R5-2.933 V U5 V2 8 V R3 R6 R8 - V 4.133 - V 0.400-2.533 V U3 U6 U8
The End