Rework Prob. 10.5 using mesh analysis.

Name: Problem 19, Chapter 10: Fundamentals of Electric Circuits
Uploaded: 2022-02-06T05:19:14-08:00
Duration: 9 min 49 s
Channel: Ralph Maestre
Description: Rework Prob. 10.5 using mesh analysis.

Key Concepts

Mesh Analysis

Mesh analysis is a systematic method used in circuit analysis that involves applying Kirchhoff's Voltage Law (KVL) to determine the currents circulating around the loops (meshes) in a circuit. This technique simplifies the process of solving complex networks by reducing the number of equations needed compared to node voltage analysis, particularly in planar circuits.

Kirchhoff's Voltage Law (KVL)

Kirchhoff's Voltage Law states that the sum of the electrical potential differences around any closed loop in a circuit is equal to zero. This fundamental principle is applied in mesh analysis to formulate equations for each circuit loop, ensuring that all voltage gains and drops balance out, which lays the groundwork for determining the values of the unknown mesh currents.

Loop Currents

Loop currents are hypothetical currents defined in each independent mesh of a circuit for the purpose of analysis. They simplify the analysis by enabling the use of KVL to write equations for the circuit, which, when solved, yield the actual currents through circuit components after considering the interaction between adjacent loops.

Ohm's Law

Ohm's Law is a basic principle in electrical circuits that relates voltage, current, and resistance through the formula V = IR. In mesh analysis, Ohm's Law is used to express the voltage drop across resistors in terms of the mesh currents and resistance values, making it possible to incorporate the resistive effects into the KVL equations.

System of Equations

Mesh analysis produces a system of simultaneous linear equations based on the application of KVL around each mesh. Solving these equations, often using methods from linear algebra, yields the mesh currents. This systematic approach is key in analyzing and understanding complex circuit behavior.

Transcript

00:01 In this problem, we are asked to rework problem 3 .11 using mesh analysis.

00:07 Now in problem 3 .11, we are asked to find for the current i1 and current i2.

00:16 Now, using mesh analysis, you could get three currents here as ix, iy, iz, where mesh ix gives us an equation of negative 24 plus 10 ix plus 40 times ix minus iy equals 0 so this is just the kvl across mesh ix now we could write a super mesh that goes from the second and the third loops here now, we could label it as color red here for the super mesh.

01:23 Using the kvl at the super mesh, we get 40 times iy minus ix plus 20 oms times iy minus ix plus 20 oms times iy plus 30 times iz.

01:53 Plus 20 times iz is equal to 0.

02:01 30 oms.

02:04 Now, we can simplify these equations as, the first equation being 25 ix minus 20 iy minus 12th is equal to 0.

02:22 That is our first equation.

02:27 For the super mesh, we could simplify it as 40 i .y.

02:32 Minus 40 i x plus 20 iy plus 50 i z equals 0 or 6 iy minus 4 i x plus 5 i z equals 0 now this is our second equation since we have three unknowns i y i z and i x we need three equations to solve for them the third equation can be acquired using junction rule here at this node, let's say node zero.

03:20 So at nodes zero, junction rule gives us i1 or iy is equal i z minus five ampiers.

03:48 Since iy here is to the right and i z is to the right.

03:55 And the 5 ampers is upwards.

04:02 So this is or you can say iy plus 5 ampers is equal to iz.

04:15 Now, rewriting equation, we get iy plus 5 ampers minus iz is equal to 0.

04:25 Now this is our third equation needed to solve for the system of equations...

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