Figure 3 Analyze the Circuit Using the Mesh-Current Method The circuit in Figure 3 is difficult to

Figure 3

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Analyze the Circuit Using the Mesh-Current Method The circuit in Figure 3 is difficult to solve, since the series/parallel reduction technique and voltage/current division principle cannot be applied here to reduce the interconnected resistors to a single equivalent resistance. However, we can use the Mesh-Current method to develop a number of simultaneous equations and solve them using the Cramer's Rule. First, we can set the following simultaneous equations based on KCL. Apply KCL to node B and C, we have, 1₁-13-14 0 12+ 13-15 = 0 Since VAD = 5 V, we have, R₁₁ Ral = 5 R₂12 + R515 = 5 Apply KVL to loop ABCA, we have, R₁1₁ + R313 R₂1₂ = 0 So we have 5 equations for 5 unknowns (1₁ to 15), we can use the Cramer's rule to solve the unknowns. Now, follow the steps below to solve the unknown. Refer to your course notes about the details of the Cramer's rule.. (a) Rearrange the above five equations into the following form: a111 + a1212 + a1313 + a1414 + a15/5 = C₁ a2111 + a2212 + a2313 + a2414 + a25l5 = C₂ a3111 + 93212 + 933/3 + a3414 + a3515 = C3 a4111 + a4212 + a4313 + a4414 + a4515 = C4, asılı + a5212 +a53/3 + a5414 + assls = cs Write your equations below: (b) Construct the matrices M, M₁, M2, M3, M4, and Ms based on the above five equations, where M = [911 912 a21 922 ⠀ a52 a51 *** *** a15 a25 I a55. M₁ = [C1 a12 C₂ 922 I a52 I C5 a15] a25 C ⠀ a55 *** *** , M₂ = [911 C₁ a21 C₂ I O a51 C5 **** a15 a25 ass etc. Analyze the Circuit Using the Mesh-Current Method The circuit in Figure 3 is difficult to solve, since the series/parallel reduction technique and voltage/current division principle cannot be applied here to reduce the interconnected resistors to a single equivalent resistance. However, we can use the Mesh-Current method to develop a number of simultaneous equations and solve them using the Cramer's Rule. First, we can set the following simultaneous equations based on KCL. Apply KCL to node B and C, we have, 1₁-13-14 0 12+ 13-15 = 0 Since VAD = 5 V, we have, R₁₁ Ral = 5 R₂12 + R515 = 5 Apply KVL to loop ABCA, we have, R₁1₁ + R313 R₂1₂ = 0 So we have 5 equations for 5 unknowns (1₁ to 15), we can use the Cramer's rule to solve the unknowns. Now, follow the steps below to solve the unknown. Refer to your course notes about the details of the Cramer's rule.. (a) Rearrange the above five equations into the following form: a111 + a1212 + a1313 + a1414 + a15/5 = C₁ a2111 + a2212 + a2313 + a2414 + a25l5 = C₂ a3111 + 93212 + 933/3 + a3414 + a3515 = C3 a4111 + a4212 + a4313 + a4414 + a4515 = C4, asılı + a5212 +a53/3 + a5414 + assls = cs Write your equations below: (b) Construct the matrices M, M₁, M2, M3, M4, and Ms based on the above five equations, where M = [911 912 a21 922 ⠀ a52 a51 *** *** a15 a25 I a55. M₁ = [C1 a12 C₂ 922 I a52 I C5 a15] a25 C ⠀ a55 *** *** , M₂ = [911 C₁ a21 C₂ I O a51 C5 **** a15 a25 ass etc.

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