TAILIEUCHUNG - Lecture Electric circuits analysis - Lecture 22: Branch, loop, and node analyses

The branch current method is a circuit analysis method using Kirchhoff's voltage and current laws to find the current in each branch of a circuit by generating simultaneous equations. Once you know the branch currents, you can determine voltages. In this chapter, the following content will be discussed: Branch current method, loop current method, node voltage method. | Previous Lecture 21 Delta-to-Wye (∆-to-Y) and Wye-to-Delta (Y-to-∆) conversions Simultaneous Equations in Circuit Analysis Solutions of Simultaneous Equations Solving by Substitution Solving by Determinants BRANCH, LOOP, AND NODE ANALYSES Branch Current Method Loop Current Method Node Voltage Method Lecture 22 BRANCH CURRENT METHOD The branch current method is a circuit analysis method using Kirchhoff's voltage and current laws to find the current in each branch of a circuit by generating simultaneous equations. Once you know the branch currents, you can determine voltages. Electric Circuit Terminologies Branch A branch is a path that connects two nodes. Loop A closed current path in a circuit. Node A node is a point where two or more components are connected. BRANCH CURRENT METHOD The following are the general steps used in applying the branch current method. Step 1. Assign a current in each circuit branch in an arbitrary direction. Step 2. Show the polarities of the resistor voltages according to the assigned branch current directions. Step 3. Apply Kirchhoff's voltage law around each closed loop (algebraic sum of voltages is equal to zero). Step 4. Apply Kirchhoff's current law at the minimum number of nodes so that all branch currents are included (algebraic sum of currents at a node equals zero). Step 5. Solve the equations resulting from Steps 3 and 4 for the branch current values. Third-order determinants can be evaluated by the expansion method. - 41 = -32, For I1 : Determinant = 179 - 79 = 100 , I1= A 1I1 + 3l2 – 2I3 = 7 0l1 + 4l2 + 1I3 = 8 -5I1 + 1l2 + 6I3 = 9 1I1 + 3l2 – 2I3 = 7 0l1 + 4l2 + 1I3 = 8 -5I1 + 1l2 + 6I3 = 9 Use the branch current method to find each branch current. I1 = mA, I2= mA, I3= mA I1 = mA, I2= mA, I3= mA LOOP CURRENT METHOD In the loop current method (also known as the mesh current method), we will work with loop currents instead of branch currents. An ammeter placed in a given branch .

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