1 / 30

Chapter 25: Electric Circuits

Chapter 25: Electric Circuits. Resistors in series. V. V. Resistors in Series and Parallel. Resistors in parallel. V. V. Resistors in Series and Parallel. Example 1:. Resistors in Series and Parallel. Example: (cont’d). I 4. I 2. R 2. R 4. I 3. Resistors in Series and Parallel. I.

zulema
Download Presentation

Chapter 25: Electric Circuits

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. Chapter 25: Electric Circuits • Resistors in series V V Resistors in Series and Parallel

  2. Resistors in parallel V V Resistors in Series and Parallel

  3. Example 1: Resistors in Series and Parallel

  4. Example: (cont’d) I4 I2 R2 R4 I3 Resistors in Series and Parallel I R3 DV

  5. Example: (cont’d) Resistors in Series and Parallel

  6. Loop 2 i i i2 i1 Loop 1 i i i2 • Introduction • Many practical resistor networks cannot be reduced to simple series-parallel • combinations (see an example below). • Terminology: • A junction in a circuit is a point where three or more conductors meet. • A loop is any closed conducting path. junction Kirchhoff’s Rules junction

  7. Kirchhoff’s junction rule • The algebraic sum of the currents into any junction is zero: Kirchhoff’s Rules

  8. Kirchhoff’s loop rule • The algebraic sum of the potential differences in any loop, including • those associated with emfs and those of resistive elements, must equal • zero. Kirchhoff’s Rules

  9. Rules for Kirchhoff’s loop rule Kirchhoff’s Rules

  10. Rules for Kirchhoff’s loop rule (cont’d) Kirchhoff’s Rules

  11. Solving problems using Kirchhoff’s rules Kirchhoff’s Rules

  12. Example 1 Kirchhoff’s Rules

  13. Example 1 (cont’d) Kirchhoff’s Rules

  14. Example 1 (cont’d) Kirchhoff’s Rules

  15. Loop 2 i i i2 i1 Loop 1 i i i2 Find all the currents including directions. • Example 2 Kirchhoff’s Rules Loop 1 Loop 2 multiply by 2 i = i1+ i2

  16. Galvanometer To be discussed in a later class. Electrical Measuring Instruments

  17. Ammeter Electrical Measuring Instruments

  18. Ammeter (cont’d) Electrical Measuring Instruments

  19. Voltmeter Electrical Measuring Instruments

  20. Charging a capacitor R-C Circuits

  21. Charging a capacitor (cont’d) R-C Circuits

  22. Charging a capacitor (cont’d) R-C Circuits

  23. Charging a capacitor (cont’d) R-C Circuits

  24. Charging a capacitor (cont’d) R-C Circuits

  25. Discharging a capacitor R-C Circuits

  26. Discharging a capacitor (cont’d) R-C Circuits

  27. Discharging a capacitor (cont’d) R-C Circuits

  28. Problem 1 Rc=25.0 W • The resistance of a galvanometer coil is 25.0 W, • and the current required for full-scale deflection • is 500 mA. • Show in a diagram how to convert the galvano- • meter to an ammeter reading 20.0 mA full scale, • and compute the shunt resistance. • b) Show how to convert the galvanometer to a • voltmeter reading 500 mV full scale, and compute • the series resistance. 500 mA 20 mA Rs a) ammeter Exercises Rc=25.0 W Solution a) For a 20-mA ammeter, the two resistance are in parallel: Vc=Vs->IcRc=IsRs->(500 x 10-6 A)(25.0 W) = (20 x 10-3 A – 500 x 10-6 A)Rs-> Rs=0.641 W. b) For a 500-mV voltmeter, the resistances are in series: Vab=I(Rc+Rs)->Rs=Vab/I – Rc -> Rs=500 x 10-3 V / 500 x 10-6 A – 25.0 W = 975 W. 500 mA Rs a b Vab=500 mV b) voltmeter

  29. Problem 2 5.00 W 20.0 V + I1 I2 + v 36.0 V 2.00 W 4.00 W v v v + I1-I2 14.0 V Exercises

  30. Problem 3 • What is the potential of point a with respect • to point b when the switch S is open? • Which point, a or b, is at higher potential? • Now the switch S is closed. • What is the final potential of point b? • How much charge flows through switch S • when it is closed? V=18.0 V 6.00 W 6.00 mF Solution b a S • With an open switch: • Also, there is a current in the left branch: • So • Point b is at the higher potential. • If the switch is closed: • New charges are: 3.00 mF Exercises 3.00 W The total charge flowing through the switch is 5.40 x 10-5 C.

More Related