Download presentation

Presentation is loading. Please wait.

1
Lecture 231 Circuits with Series and Parallel Resistors/Impedances

2
Lecture 232 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.

3
Lecture 233 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.

4
Lecture 234 1k 2k 1k 2k 1k + - 10V + - V1V1 + - V3V3 + - V2V2 Example: Resistor Ladder Find V 1, V 2, and V 3

5
Lecture 235 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.

6
Lecture 236 1k + - 10V + - V1V1 Example: Resistor Ladder

7
Lecture 237 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.

8
Lecture 238 Example: Resistor Ladder 1k 2k 1k + - 10V + - 5V + - V2V2 1k

9
Lecture 239 1k 2k 1k 2k 1k + - 10V + - 5V + - V3V3 + - 2.5V Example: Resistor Ladder

10
Lecture 2310 Example: Notch Filter Find V out Use = 1500 1k 0.1 100 + - 10V 0 + - V out 70.4mH 100 F

11
Lecture 2311 V out 1k 0.1 100 + - 10V 0 + - j106 -j6.67 Example: Notch Filter Find the equivalent impedance of the resistor, inductor, and capacitor.

12
Lecture 2312 V out 1k 100 + - 10V 0 + - 7.12 -89.99 Example: Notch Filter Combine resistor and impedance.

13
Lecture 2313 Example: Notch Filter V out 1k + - 10V 0 + - 100.3 -4.07

14
Lecture 2314 Example: Notch Filter

15
Lecture 2315 Example: Notch Filter Find V out Use = 377 1k 0.1 100 + - 10V 0 + - V out 70.4mH 100 F

16
Lecture 2316 Example: Notch Filter V out = 1.23V -0.17

17
Lecture 2317 Frequency Response

18
Lecture 2318 Using MATLAB to Solve Circuits

19
Lecture 2319 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.

20
Lecture 2320 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

21
Lecture 2321 Example: Notch Filter Find V out Use = 1500 1k 0.1 100 + - 10V 0 + - V out 70.4mH 100 F

22
Lecture 2322 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

23
Lecture 2323 Equivalent Capacitor/Inductor Impedance >> zeq = (0.1+xl)*xc /(0.1+xl+xc) zeq = 6.8687e+03 - 1.0981e+03i

24
Lecture 2324 Voltage Divider >> vin = 10 vin = 10 >> vout = vin*1e3 /(100+zeq+1e3) vout = 1.2315 + 0.1697i

25
Lecture 2325 Magnitude and Angle >> abs(vout) ans = 1.2432 >> angle(vout) ans = 0.1369 >> angle(vout)*180/pi ans = 7.8461

Similar presentations

© 2021 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google