1 / 18

ECE 2006

ECE 2006. Lecture for Chapter 3 S.Norr. Circuit Analysis Methods. Nodal Analysis: Applicable to ANY circuit Uses KCL to determine voltages in cricuit Mesh Analysis: ONLY applicable to planar circuits Uses KVL to determine currents in circuit. Nodal Analysis. Procedure:

tahlia
Télécharger la présentation

ECE 2006

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. ECE 2006 Lecture for Chapter 3 S.Norr

  2. Circuit Analysis Methods • Nodal Analysis: • Applicable to ANYcircuit • Uses KCL to determine voltages in cricuit • Mesh Analysis: • ONLY applicable to planar circuits • Uses KVL to determine currents in circuit

  3. Nodal Analysis • Procedure: • Select One Node as Reference • Assign a Voltage Variable to each remaining Node • Apply KCL at each non-reference Node • Solve the resulting set of simultaneous equations

  4. Grounding • Any SINGLE node in a circuit can be grounded without impact on the performance of the circuit. • Connecting one node of a circuit to ground provides a Zero Voltage reference at that point • Symbols for the Ground Plane:

  5. Example of Nodal Analysis • Assign a Reference Node

  6. Nodal Analysis Example • Assign a Voltage to all Other Nodes:

  7. Nodal Example (Cont.) • Write KCL at One or More Nodes: i1 + i2 + i3 = 0 • Re-Write the Currents using Ohm’s Law: i1= (Va - 5)/2 ; i2 = (Va + 3)/4 ; i3 = (Va – 0)/8 • Substitute: (Va - 5)/2 + (Va + 3)/4 + (Va – 0)/8 = 0 Va = 2 Volts

  8. Nodal Analysis Example… • Use the Resulting Node Voltages to Solve for Currents: Example: i3 = (Va – 0)/8 = (2 – 0)/8 = 1/4 Amps

  9. Example of Nodal Analysiswith a Dependent Source • Establish a Reference Node:

  10. Example of Dependent (Cont.) • Assign a voltage at all other nodes:

  11. Example of Dependent (Cont.) • Applying KCL at Node V2: ix + i + 2i = 0

  12. Example of Dependent (Cont.) Describe Ix using Ohm’s Law: ix = Vx/5 ; Vx = V2 – V1 = V2 – 5 Volts ix = (V2 – 5) /5 Also, Relate ito V2: V2 =i * 10 Result: ix = (i * 10 – 5) /5 = 2i -1 Substitute back into KCL: ix + i + 2i = 0 2i -1 + i + 2i = 0 i = 1/5 Amps ; V2 = 2 Volts

  13. MESH Analysis • ONLYused with PLANAR circuits • Planar meaning the circuit can be drawn on a two-dimensional plane without any branches crossing over another branch • A MESH is a Loop that contains no other Loops within it.

  14. MESH Analysis: • Assign a currentvariable to each MESH in a circuit • Apply KVL to each Mesh, using Ohm’s law to express each Voltage in terms of the assigned currents • Solve the resulting set of simultaneous equations

  15. Mesh Example: • Assign Mesh Currents:

  16. MESH Example… • Write KVL in terms of Mesh Currents: • Mesh 1: -5 + 2i1 + 8(i1 –i2) = 0

  17. Mesh Example…. • Mesh 2 Equation: 8(i2 – i1) + 4i2 – 3 = 0

  18. Mesh Example… • Solve the Set of Simultaneous Equations: 10i1 – 8i2 = 5 -8i1 + 12i2 = 3 14i1 + 0i2 = 21 i1 = 3/2 Amps i2 = 5/4 Amps

More Related