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Circuits with Resistor Combinations (2.6, 7.7)

Circuits with Resistor Combinations (2.6, 7.7). Dr. Holbert September 13, 2001. Solving Circuits with Series and Parallel Combinations. The combination of series and parallel impedances can be used to find voltages and currents in circuits.

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Circuits with Resistor Combinations (2.6, 7.7)

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  1. Circuits with Resistor Combinations (2.6, 7.7) Dr. Holbert September 13, 2001 ECE201 Lect-8

  2. Solving Circuits with Series and Parallel Combinations • The combination of series and parallel impedances can be used to find voltages and currents in circuits. • This process can often yield the fastest solutions to networks. • This process may not apply to complicated networks. ECE201 Lect-8

  3. Series and Parallel Impedances • Impedances are combined to create a simple circuit (usually one source and one impedance), from which a voltage or current can be found • Once the voltage or current is found, KCL and KVL are used to work back through the network to find voltages and currents. ECE201 Lect-8

  4. 1kW 1kW 1kW + + + + 10V V3 V2 V1 1kW - 2kW 2kW - - - Example: Resistor Ladder Find V1, V2, and V3 ECE201 Lect-8

  5. + + + V2 V3 V1 - - - Example: Resistor Ladder 1kW 1kW 1kW Find an equivalent resistance for the network with V1 across it, then find V1 using a voltage divider. + 10V 1kW - 2kW 2kW ECE201 Lect-8

  6. 1kW + + 10V V1 1kW - - Example: Resistor Ladder ECE201 Lect-8

  7. + + + V2 V3 5V - - - Example: Resistor Ladder 1kW 1kW 1kW Find an equivalent resistance for the network with V2 across it, then find V2. + 10V 1kW - 2kW 2kW ECE201 Lect-8

  8. 1kW 1kW + + + 10V 5V V2 1kW - 2kW - - Example: Resistor Ladder ECE201 Lect-8

  9. 1kW 1kW 1kW + + + + 10V 5V 2.5V V3 1kW - 2kW 2kW - - - Example: Resistor Ladder ECE201 Lect-8

  10. Example: Notch Filter 70.4mH 0.1W Find Vout Use w = 1500 100W + + 10V  0 1kW Vout 100mF - - ECE201 Lect-8

  11. j106W 0.1W 100W + + 10V  0 1kW -j6.67W - - Example: Notch Filter Find the equivalent impedance of the resistor, inductor, and capacitor. Vout ECE201 Lect-8

  12. Example: Notch Filter Combine resistor and impedance. 7.12W -89.99 100W + + 10V  0 1kW Vout - - ECE201 Lect-8

  13. 100.3W -4.07 + + 10V  0 1kW Vout - - Example: Notch Filter ECE201 Lect-8

  14. Example: Notch Filter ECE201 Lect-8

  15. Example: Notch Filter 70.4mH 0.1W Find Vout Use w = 377 100W + + 10V  0 1kW Vout 100mF - - ECE201 Lect-8

  16. Example: Notch Filter Vout= 1.23V  -0.17 ECE201 Lect-8

  17. Frequency Response ECE201 Lect-8

  18. Using MATLAB to Solve Circuits ECE201 Lect-8

  19. MATLAB • MATLAB can perform computations with complex numbers. • You can use it as a calculator to compute phasors and impedances for AC SS analysis. • You can also use it to automate computations of frequency responses. ECE201 Lect-8

  20. Using MATLAB • Entering a complex number: >> 1+2j ans = 1.0000 + 2.0000i • Multiplying complex numbers: >> (1+2j)*(3+4j) ans = -5.0000 +10.0000i ECE201 Lect-8

  21. Example: Notch Filter 70.4mH 0.1W Find Vout Use w = 1500 100W + + 10V  0 1kW Vout 100mF - - ECE201 Lect-8

  22. Compute Impedances >> omega = 377 omega = 377 >> xl = j*omega*70.4e-3 xl = 0 +26.5408i >> xc = 1/(j*omega*100e-6) xc = 0 -26.5252i ECE201 Lect-8

  23. Equivalent Capacitor/Inductor Impedance >> zeq = (0.1+xl)*xc /(0.1+xl+xc) zeq = 6.8687e+03 - 1.0981e+03i ECE201 Lect-8

  24. Voltage Divider >> vin = 10 vin = 10 >> vout = vin*1e3 /(100+zeq+1e3) vout = 1.2315 + 0.1697i ECE201 Lect-8

  25. Magnitude and Angle >> abs(vout) ans = 1.2432 >> angle(vout) ans = 0.1369 >> angle(vout)*180/pi ans = 7.8461 ECE201 Lect-8

  26. Class Examples • Learning Extension E2.14 • Learning Extension E2.15 • Learning Extension E7.12 ECE201 Lect-8

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