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ECE201 Lect-71 Circuits with Resistor Combinations (2.6, 8.7) Dr. Holbert February 8, 2006

ECE201 Lect-72 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-73 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-74 1k  2k  1k  2k  1k  +–+– 10V + – V1V1 + – V3V3 + – V2V2 Example: Resistor Ladder Find V 1, V 2, and V 3

ECE201 Lect-75 1k  2k  1k  2k  1k  10V + – V1V1 + – V3V3 + – V2V2 Example: Resistor Ladder Find an equivalent resistance for the network with V 1 across it, then find V 1 using a voltage divider. +–+–

ECE201 Lect-76 1k  10V + – V1V1 Example: Resistor Ladder +–+–

ECE201 Lect-77 1k  2k  1k  2k  1k  10V + – 5V + – V3V3 + – V2V2 Example: Resistor Ladder Find an equivalent resistance for the network with V 2 across it, then find V 2. +–+–

ECE201 Lect-78 Example: Resistor Ladder 1k  2k  1k  10V + – 5V + – V2V2 1k  +–+–

ECE201 Lect-79 1k  2k  1k  2k  1k  10V + – 5V + – V3V3 + – 2.5V Example: Resistor Ladder +–+–

ECE201 Lect-710 Example: Notch Filter Find V out Use  = 1500 1k  0.1  100  10V  0  + – V out 70.4mH 100  F +–+–

ECE201 Lect-711 V out 1k  0.1  100  10V  0  + – j106  –j6.67  Example: Notch Filter Find the equivalent impedance of the resistor, inductor, and capacitor. +–+–

ECE201 Lect-712 V out 1k  100  10V  0  + – 7.12   –89.99  Example: Notch Filter Combine top resistor and impedance. +–+–

ECE201 Lect-713 Example: Notch Filter V out 1k  10V  0  + – 100.3   –4.07  +–+–

ECE201 Lect-714 Example #2: Notch Filter Find V out Use  = 377 1k  0.1  100  10V  0  + – V out 70.4mH 100  F +–+–

ECE201 Lect-715 Example: Notch Filter V out = 1.23V  0.17 

ECE201 Lect-716 Frequency Response

ECE201 Lect-717 Using MATLAB to Solve Circuits MATLAB can perform computations with complex numbers. You can use it as a calculator to compute phasors and impedances for AC steady-state analysis. You can also use it to automate computations of frequency responses.

ECE201 Lect-718 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-719 Example: Notch Filter Find V out Use  = 1500 1k  0.1  100  10V  0  + – V out 70.4mH 100  F +–+–

ECE201 Lect-720 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-721 Equivalent Capacitor/Inductor Impedance >> zeq = (0.1+xl)*xc /(0.1+xl+xc) zeq = 6.8687e+03 - 1.0981e+03i

ECE201 Lect-722 Voltage Divider >> vin = 10 vin = 10 >> vout = vin*1e3 /(100+zeq+1e3) vout = 1.2315 + 0.1697i

ECE201 Lect-723 Magnitude and Angle >> abs(vout) ans = 1.2432 >> angle(vout) ans = 0.1369 >> angle(vout)*180/pi ans = 7.8461

ECE201 Lect-724 Class Examples Learning Extension E2.14 Learning Extension E2.15 Learning Extension E8.12

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