1 / 29

Kirchhoff’s Laws

I. I. R. r. I. a. I. V. R. e. b. C. e. R. Kirchhoff’s Laws. Yesterday. Ohm’s Law V=IR Ohm’s law isn’t a true law but a good approximation for typical electrical circuit materials Resistivity =1/ (Conductivity) : Property of the material

gerard
Download Presentation

Kirchhoff’s Laws

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. I I R r I a I V R e b C e R Kirchhoff’s Laws

  2. Yesterday • Ohm’s LawV=IR • Ohm’s law isn’t a true law but a good approximation for typical electrical circuit materials • Resistivity =1/ (Conductivity): Property of the material • Resistance proportional to resistivity and length, inversely proportional to area

  3. Question 1 Two cylindrical resistors are made from the same material, and they are equal in length. The first resistor has diameter d, and the second resistor has diameter 2d. Compare the resistance of the two cylinders. a) R1 > R2 b) R1 = R2 c) R1 < R2

  4. 0 of 300 0 Question 1 • a • b • c

  5. Question 1 Two cylindrical resistors are made from the same material, and they are equal in length. The first resistor has diameter d, and the second resistor has diameter 2d. Compare the resistance of the two cylinders. a) R1 > R2 b) R1 = R2 c) R1 < R2 • Resistance is proportional to Length/Area

  6. Question 2 Two cylindrical resistors are made from the same material, and they are equal in length. The first resistor has diameter d, and the second resistor has diameter 2d. If the same current flows through both resistors, compare the average velocities of the electrons in the two resistors: a) v1 > v2 b) v1 = v2 c) v1 < v2

  7. 0 of 300 0 Question 2 • a • b • c

  8. Question 2 Two cylindrical resistors are made from the same material, and they are equal in length. The first resistor has diameter d, and the second resistor has diameter 2d. If the same current flows through both resistors, compare the average velocities of the electrons in the two resistors: a) v1 > v2 b) v1 = v2 c) v1 < v2 Current  Area  Current Density Current Density  average velocity of electrons I is the same A1<A2  v1>v2

  9. a I R1 b R2 a c Reffective c Resistors in Series • What is the same effective single resistance to two resistances in series? • Whenever devices are in SERIES, the current is the same through both. • By Ohm’s law, the Voltage difference across resistance R1 is • Across R2 is • Total voltage difference • the effective single resistance is

  10. R1 L1 V L2 R2 Another (intuitive) way… • Consider two cylindrical resistors with lengths L1 andL2 • Put them together, end to end to make a longer one...

  11. R1 V0 V R2 By varying R2 we can controllably adjust the output voltage! The World’s Simplest (and most useful) circuit:Voltage Divider

  12. Question 3 Two resistors are connected in series to a battery with emf E. The resistances are such that R1 = 2R2. The currents through the resistors are I1 and I2 and the potential differences across the resistors V1 and V2. Are: • I1>I2 and V2=E • I1=I2 and V2= E • I1=I2 and V2=1/3E • I1<I2 and V2=1/2E • I1<I2 and V2=1/3E

  13. I a V R I d Resistors in Parallel • Very generally, devices in parallel have the same voltage drop I a I1 I2 • Current through R1 is I1. • Current through R2 is I2. V R1 R2 I d • But current is conserved Þ Þ

  14. A2 A1 V R1 R2 Another (intuitive) way… Consider two cylindrical resistors with cross-sectional areas A1 and A2 Put them together, side by side … to make one “fatter”one, Þ

  15. KVL: e1 R1 e2 R2 I -e2 +e1 =0 -IR1 -IR2 Kirchhoff’s First Rule“Loop Rule” or “Kirchhoff’s Voltage Law (KVL)” "When any closed circuit loop is traversed, the algebraic sum of the changes in potential must equal zero." • This is just a restatement of what you already know: that thepotential difference is independent of path!

  16. e1 R1 e2 R2 I Rules of the Road • Our convention: • Voltage gains enter with a + sign, and voltage drops enter with a - sign. • We choose a direction for the current and move around the circuit in that direction. • When a battery is traversed from the negative terminal to the positive terminal, the voltage increases, and hence the battery voltage enters KVL with a + sign. • When moving across a resistor, the voltage drops, and hence enters KVL with a - sign. +e1 -e2 =0 -IR1 -IR2

  17. R1 R4 e1 b f a I I d c e R2 e2 R3 Þ KVL: Þ Current in a Loop Start at point a (could be anywhere) and assume current is in direction shown (could be either)

  18. 12V R I a 12V 12V R b Question 3 • Consider the circuit shown. • The switch is initially open and the current flowing through the bottom resistor isI0. • Just after the switch is closed, the current flowing through the bottom resistor isI1. • What is the relation betweenI0andI1? (c) I1 > I0 (b) I1 = I0 (a) I1 < I0

  19. 0 of 300 0 Question 3 • a • b • c

  20. 12V R I a 12V 12V R b Question 3 • Consider the circuit shown. • The switch is initially open and the current flowing through the bottom resistor isI0. • Just after the switch is closed, the current flowing through the bottom resistor isI1. • What is the relation betweenI0andI1? (c) I1 > I0 (b) I1 = I0 (a) I1 < I0 • From symmetry the potential (Va-Vb) before the switch is closed is Va-Vb = +12V. • Therefore, when the switch is closed, potential stays the same and NO additional current will flow! • Therefore, the current before the switch is closed is equal to the current after the switch is closed.

  21. (c) I1 >I0 (b)I1 = I0 (a) I1 < I0 Question 3 • Consider the circuit shown. • The switch is initially open and the current flowing through the bottom resistor isI0. • After the switch is closed, the currentflowing through the bottom resistor isI1. • What is the relation betweenI0andI1? 12V R a I 12V 12V R b • Write a loop law for original loop: 12V +12V -I0R-I0R = 0 I0 = 12V/R • Write a loop law for the new loop: 12V -I1R = 0 I1 = 12V/R

  22. Kirchhoff’s Second Rule“Junction Rule” or“Kirchhoff’s Current Law (KCL)” • In deriving the formula for the equivalent resistance of 2 resistors in parallel, we applied Kirchhoff's Second Rule (the junction rule). • "At any junction point in a circuit where the current can divide (also called a node), the sum of the currents into the node must equal the sum of the currents out of the node." • This is just a statement of the conservation of charge at any given node. • The currents entering and leaving circuit nodes are known as “branch currents”. • Each distinct branch must have a current, Iiassigned to it

  23. R1 I3 e 2 I2 I1 R2 R3 e 1 How to use Kirchhoff’s Laws A two loop example: • Assume currents in each section of the circuit, identify all circuit nodes and use KCL. • (1)I1 = I2 + I3 • Identify all independent loops and use KVL. • e1-I1R1-I2R2 = 0 • I2R2-e2-I3R3 = 0 • e1 - I1R1 - e2 - I3R3 = 0

  24. R1 I3 I2 I1 e 2 R2 R3 e 1 From eqn. (2) From eqn. (3) Þ How to use Kirchoff’s Laws • Solve the equations for I1, I2, and I3: • Firstfind I2 and I3 in terms of I1: Now solve for I1using eqn. (1):

  25. R1 I3 I2 I1 e 2 R2 R3 e 1 Let’s plug in some numbers e1 = 24 V e 2 = 12 V R1= 5W R2=3W R3=4W Then,and I1=2.809 A I2= 3.319 A, I3= -0.511 A

  26. Outside loop: e1 R Top loop: I1 I2 e2 R Junction: I3 e3 R Junction Demo

  27. Summary • Kirchhoff’s Laws • KCL: Junction Rule (Charge is conserved) • Review KVL (V is independent of path) • Non-ideal Batteries & Power • Discharging of capacitor through a Resistor: Reading Assignment: Chapter 26.6 Examples: 26.17,18 and 19

  28. Two identical light bulbs are represented by the resistors R2 and R3 (R2 = R3 ). The switch S is initially open. 2) If switch S is closed, what happens to the brightness of the bulb R2? a) It increases b) It decreases c) It doesn’t change 3) What happens to the current I, after the switch is closed ? a) Iafter = 1/2 Ibefore b) Iafter = Ibefore c) Iafter = 2 Ibefore

  29. I R2 R1 Four identicalresistors are connected to a battery as shown in the figure. R4 R3 E 5) How does the current through the battery change after the switch is closed ? Before:Rtot = 3R Ibefore = 1/3 E/R After:R23 = 2R R423 = 2/3 R Rtot = 5/3 R Iafter = 3/5 E/R a) Iafter > Ibefore b) Iafter = Ibefore c) Iafter < Ibefore

More Related