ECE 231 -14

ECE 231 -14 - ECE-231 Circuits and Systems I Fall 2011...

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ECE-231 Circuits and Systems I Fall 2011 Session 14 Dr. Stewart Personick Office: ECEC Room 331 sdp2@verizon.net
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Review
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Review: Solving for the Dependent Currents and Voltages in DC circuits
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The objective is to use Kirchhoff’s Laws, Ohm’s Law, and the information you are given regarding the values of the independent sources, the resistors, etc. to produce a set of equations that can be used to solve for the unknown (dependent) currents and voltages in a circuit
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The Mesh Current Method 500 node e Vah= 50 volts (voltage source) node b - + 1000 500 Controlled voltage source: Vde= 500 V/A x i2 500 node d i1 i2 i3 + - Mesh 2 has only a resistor or a wire (0 Ohm resistor) in every branch Therefore, using Kirchoff’s voltage law, we add the voltages in a closed path around mesh 2, starting at node b: Vbc + Vcf +Vfg +Vgb =0 Volts 0 + 500 (i2 - i3) + 0 + 1000 (i2 - i1) = 0 Volts node c node a node f node g node h Vah
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The Mesh Current Method 500 node e Vah= 50 Volts (voltage source) node b - + 1000 500 Controlled voltage source: Vde= 500 V/A x i2 500 node d i1 i2 i3 + - Mesh 1 has a resistor or a wire (0 Ohm resistor) or a voltage source in every branch Therefore, using Kirchoff’s voltage law, we add the voltages in a closed path around mesh 2, starting at node a: Vab + Vbg + Vgh + Vha =0 Volts 500 (i1) + 1000 (i1 - i2) + 0 -50 Volts = 0 Volts node c node a node f node g node h Vah
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The Mesh Current Method 500 node e Vah= 50 Volts (voltage source) node b - + 1000 500 Controlled voltage source: Vde= 500 V/A x i2 500 node d i1 i2 i3 + - Mesh 3 has a resistor or a wire (0 Ohm resistor) or a voltage source in every branch Therefore, using Kirchoff’s voltage law, we add the voltages in a closed path around mesh 2, starting at node c: Vcd + Vde + Vef + Vfc =0 Volts 500 (i3) + 500 V/A x i2 + 0 + 500 (i3 - i2) = 0 Volts node c node a node f node g node h Vah
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The Mesh Current Method 500 node e Vah= 50 Volts (voltage source) node b - + 1000 500 Controlled current source: i = 5 x i2 500 node d i1 i2 i3 Mesh 3 has a resistor or a wire (0 Ohm resistor) or a current source in every branch The branch current in the branch containing the current source is completely determined by the current source In this example , the branch current in branch e-d is: 5 x i2 This branch current is equal to -1 x the mesh current in mesh 3. i3 = -5 x i2 node c node a node f node g node h Vah
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The Mesh Current Method 500 node e is= 2 Amperes (current source) node b 1000 500 500 node d i1 i2 i3 Mesh 1 has a resistor or a wire (0 Ohm resistor) or a current source in every branch The branch current in the branch containing the current source is completely determined by the current source In this example , the branch current in branch h-a is: 2 Amperes This branch current is equal to the mesh current in mesh 1. i1 = 2 Amperes node c node a node f node g node h is Controlled voltage source: Vde= 500 V/A x i2 + -
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The Node Voltage Method 500 node f Vah= 50 volts (voltage source) node b - + 1000 500 Controlled voltage source: Vcf = 500 V/A x ibg 500 node c + - The voltage at node g is defined as 0 Volts [Node g has been selected
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ECE 231 -14 - ECE-231 Circuits and Systems I Fall 2011...

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