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Power System Analysis and Design (MindTap Course List)
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter6: Power Flows
Section: Chapter Questions
Problem 6.61P
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Problem 1: V. 400Ω ν, 500Ω v. 2002 10V 1002 100mA 10V Node 0 Node-Voltage method: 1. Identify Nodes Here, Vi, Vz, and V3 are the voltage nodes. We calculate currents entering and exiting a particular node. 2. Calculate current equations for each node: For node V1. V1 = 10 V as it is connected to the voltage source and hence no currents are calculated for it. For node V2, considering Vz calculate currents for all branches connected to it using Ohm's law (I = V/R) and KCL. v2 - vi v2 - V3, v2 - 10 = 0 200 400 500 If you notice the equation, the node in question (V2) is always the first node and the term is the voltage difference between two nodes wherever a component (resistor) exists between them. The equation is an application of KCL which states the sum of all currents entering or exiting a particular node is always zero. For node 3, V3 - v2 V3 - 0 500 100+ (-0.1) = 0 The (-0.1) value is the branch current for the branch with the 100 mA source connected to it. If a current source is connected in a branch, it dominates and the current in that branch is equal to that provided by the source. Until now we have considered all the currents exiting the node to be positive (node in question is always first term). The 100 mA or 0.1 A current is entering the node and hence has a negative magnitude to denote the opposite direction of assumed direction. 3. Solve linear equations: You know the value of V1, so there are two unknowns (Vzand Vs) and two equations. Solve the equations linearly to achieve the value of the unknowns. 4. Calculate the required parameter: In the manual, the required parameter is the current across the 200 ohm resistor. v2 - vi v2 - V3 v2 - 10 500 = 0 400 200 In above equation, the highlighted term is the current flowing across the 200 ohm resistor (I200 = Vz00 / R). You have calculated the value of V2 and hence can easily calculate the required current. Mesh Analysis: In mesh analysis, we use KVL to calculate the currents in closed loops to finally calculate the required parameter. 4002 5002 2002 10V iz D31002 100mA 10V Let's assume three currents flowing in the three closed loops in the circuit (i, iz, and is) with the assumed direction. I always assume this clockwise direction as it makes formulating the equations a little bit easier for me. Basically, we are writing the voltages
Transcribed Image Text:Problem 1: V. 400Ω ν, 500Ω v. 2002 10V 1002 100mA 10V Node 0 Node-Voltage method: 1. Identify Nodes Here, Vi, Vz, and V3 are the voltage nodes. We calculate currents entering and exiting a particular node. 2. Calculate current equations for each node: For node V1. V1 = 10 V as it is connected to the voltage source and hence no currents are calculated for it. For node V2, considering Vz calculate currents for all branches connected to it using Ohm's law (I = V/R) and KCL. v2 - vi v2 - V3, v2 - 10 = 0 200 400 500 If you notice the equation, the node in question (V2) is always the first node and the term is the voltage difference between two nodes wherever a component (resistor) exists between them. The equation is an application of KCL which states the sum of all currents entering or exiting a particular node is always zero. For node 3, V3 - v2 V3 - 0 500 100+ (-0.1) = 0 The (-0.1) value is the branch current for the branch with the 100 mA source connected to it. If a current source is connected in a branch, it dominates and the current in that branch is equal to that provided by the source. Until now we have considered all the currents exiting the node to be positive (node in question is always first term). The 100 mA or 0.1 A current is entering the node and hence has a negative magnitude to denote the opposite direction of assumed direction. 3. Solve linear equations: You know the value of V1, so there are two unknowns (Vzand Vs) and two equations. Solve the equations linearly to achieve the value of the unknowns. 4. Calculate the required parameter: In the manual, the required parameter is the current across the 200 ohm resistor. v2 - vi v2 - V3 v2 - 10 500 = 0 400 200 In above equation, the highlighted term is the current flowing across the 200 ohm resistor (I200 = Vz00 / R). You have calculated the value of V2 and hence can easily calculate the required current. Mesh Analysis: In mesh analysis, we use KVL to calculate the currents in closed loops to finally calculate the required parameter. 4002 5002 2002 10V iz D31002 100mA 10V Let's assume three currents flowing in the three closed loops in the circuit (i, iz, and is) with the assumed direction. I always assume this clockwise direction as it makes formulating the equations a little bit easier for me. Basically, we are writing the voltages
4002 5002 2002 10V(*) DS1002 100mA 10V Let's assume three currents flowing in the three closed loops in the circuit (i1, i2, and i3) with the assumed direction. I always assume this clockwise direction as it makes formulating the equations a little bit easier for me. Basically, we are writing the voltages across each element (V = |*R) and using KVL to formulate an equation. KVL states that the sum of voltages in a closed loop is always equal to zero. For first loop with current i1, -10 + 400 i, + 200( i – iz) + 10 = 0 The (-10) is the voltage from the voltage source on the left. The negative sign indicates the flow of current from negative to positive terminal. The (+10) indicates the voltage of the source on the right. The positive sign is for the flow of current from positive to negative terminal. For second loop with current iz, - 10 + 200( iz - i,) + 500 iz + 100( iz – i3) = 0 For the third loop with current i3, the current source dominates and the current iz is equal to the current provided by the current source. iz = - 0.1 A The negative sign indicates that the 100 mA current is opposite in direction to the assumed current i3. Thus, there are 2 unknowns and 2 equations to solve linearly. Find the values of iı and iz. The current across the 200 ohm resistor is i, - iz in the downward direction. If they specify the current flowing from bottom to top, then the subtraction is inversed (i2 – i1).
Transcribed Image Text:4002 5002 2002 10V(*) DS1002 100mA 10V Let's assume three currents flowing in the three closed loops in the circuit (i1, i2, and i3) with the assumed direction. I always assume this clockwise direction as it makes formulating the equations a little bit easier for me. Basically, we are writing the voltages across each element (V = |*R) and using KVL to formulate an equation. KVL states that the sum of voltages in a closed loop is always equal to zero. For first loop with current i1, -10 + 400 i, + 200( i – iz) + 10 = 0 The (-10) is the voltage from the voltage source on the left. The negative sign indicates the flow of current from negative to positive terminal. The (+10) indicates the voltage of the source on the right. The positive sign is for the flow of current from positive to negative terminal. For second loop with current iz, - 10 + 200( iz - i,) + 500 iz + 100( iz – i3) = 0 For the third loop with current i3, the current source dominates and the current iz is equal to the current provided by the current source. iz = - 0.1 A The negative sign indicates that the 100 mA current is opposite in direction to the assumed current i3. Thus, there are 2 unknowns and 2 equations to solve linearly. Find the values of iı and iz. The current across the 200 ohm resistor is i, - iz in the downward direction. If they specify the current flowing from bottom to top, then the subtraction is inversed (i2 – i1).
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