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Kirchhoff’s Laws

Physics 102: Lecture 06. Kirchhoff’s Laws. Today’s lecture will cover Textbook Sections 18.5, 7. Conflict Exam sign up available in grade book. Be Careful with round off errors in homework 3!. Solved Circuits. Today. What about this one?. Last Time. Resistors in series:.

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Kirchhoff’s Laws

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  1. Physics 102:Lecture06 Kirchhoff’s Laws • Today’s lecture will cover Textbook Sections 18.5, 7 Conflict Exam sign up available in grade book Be Careful with round off errors in homework 3! Physics 102: Lecture 6, Slide 1

  2. Solved Circuits Today • What about this one? Last Time • Resistors in series: Current thru is same; Voltage drop across is IRi Last Lecture • Resistors in parallel: Voltage drop across is same; Current thru is V/Ri Physics 102: Lecture 6, Slide 2

  3. Kirchhoff’s Rules • Kirchhoff’s Voltage Rule (KVR): • Sum of voltage drops around a loop is zero. • Kirchhoff’s Current Rule (KCR): • Current going in equals current coming out. Physics 102: Lecture 6, Slide 3

  4. R1 I1 A - + + + R2 E3 - - + B I2 E1 I3 I4 - - - + R3 E2 - + + R5 - + Kirchhoff’s Laws • (1) Label all currents • Choose any direction (2) Label +/- for all elements Current goes +  - (for resistors) • Choose loop and direction • Must start on wire, not element. R4 • Write down voltage drops • First sign you hit is sign to use. Physics 102: Lecture 6, Slide 4

  5. - + + - - + + - B R1=5 W I - + + e1= 50V - - + A + - R2=15 W Example KVR Practice B R1=5 W I Find I: e1= 50V A R2=15 W e2= 10V What if only went from A to B? VA - VB= VA – VB = e2= 10V Physics 102: Lecture 6, Slide 5 22

  6. - + + - - + + - B R1=5 W I - + + e1= 50V - - + A + - R2=15 W Example KVR Practice Label currents Label elements +/- Choose loop Write KVR B R1=5 W I Find I: –e1+IR1 +e2+IR2 = 0 -50 + 5 I + 10 +15 I = 0 I = +2 Amps e1= 50V A R2=15 W e2= 10V What if only went from A to B? VA - VB= –E1+ IR1 = -50 + 25 = -40 Volts VA - VB = -IR2 – E2 = -215 - 10 = -40 Volts e2= 10V Physics 102: Lecture 6, Slide 6

  7. ACT: KVR I1 R1=10 W Resistors R1 and R2 are: 1) in parallel 2) in series 3) neither E2 = 5 V I2 R2=10 W IB - + E1 = 10 V Physics 102: Lecture 6, Slide 7

  8. ACT: KVR I1 R1=10 W Resistors R1 and R2 are 1) in parallel 2) in series 3) neither E2 = 5 V I2 R2=10 W Definition of parallel: Two elements are in parallel if (and only if) you can make a loop that contains only those two elements. IB - + E1 = 10 V Definition of series: Two elements are in series if (and only if) every loop that Contains R1 also contains R2 Physics 102: Lecture 6, Slide 8

  9. Preflight 6.1 Calculate the current through resistor 1. I1 R1=10 W + - • I1 = 0.5 A • I1 = 1.0 A • I1 = 1.5 A E2 = 5 V I2 R2=10 W IB - + E1 = 10 V ACT: Voltage Law How would I1 change if the switch was closed? 1) Increase 2) No change 3) Decrease Physics 102: Lecture 6, Slide 9

  10. Preflight 6.1 Calculate the current through resistor 1. I1 R1=10 W + - E2 = 5 V I2 R2=10 W 1) I1 = 0.5 A 2) I1 = 1.0 A 3) I1 = 1.5 A -E1+ I1R1 = 0  I1 = E1/R1 = 1A IB - + E1 = 10 V ACT: Voltage Law How would I1 change if the switch was opened? 1) Increase 2) No change 3) Decrease Physics 102: Lecture 6, Slide 10

  11. Preflight 6.2 Calculate the current through resistor 2. I1 R1=10 W • I2 = 0.5 A • I2 = 1.0 A • I2 = 1.5 A E2 = 5 V I2 R2=10 W - + IB - + E1 = 10 V Physics 102: Lecture 6, Slide 11

  12. Preflight 6.2 Calculate the current through resistor 2. I1 R1=10 W 1) I2 = 0.5 A 2) I2 = 1.0 A 3) I2 = 1.5 A E2 = 5 V I2 R2=10 W - + -E1 +E2+ I2R2 = 0  I2 = 0.5A IB - + E1 = 10 V Physics 102: Lecture 6, Slide 12

  13. Preflight 6.3 I1 R=10 W E = 5 V I2 R=10 W IB - + E1 = 10 V Kirchhoff’s Junction Rule Current Entering = Current Leaving I1 I1 = I2 + I3 I2 I3 • IB = 0.5 A • IB = 1.0 A • IB = 1.5 A Physics 102: Lecture 6, Slide 13

  14. Preflight 6.3 I1 R=10 W E = 5 V I2 R=10 W IB - + E1 = 10 V Kirchhoff’s Junction Rule Current Entering = Current Leaving I1 I1 = I2 + I3 I2 I3 1) IB = 0.5 A 2) IB = 1.0 A 3) IB = 1.5 A IB = I1 + I2 = 1.5 A Physics 102: Lecture 6, Slide 14

  15. Example You try it! In the circuit below you are given e1, e2, R1, R2 and R3. Find I1, I2 and I3. • Label all currents 2. Label +/- for all elements 3. Choose loop and direction • Write down voltage drops Loop 1: R1 I3 I1 Loop 2: - + I2 + - + 5. Write down node equation e1 R2 R3 - - + Node: - + e2 Physics 102: Lecture 6, Slide 15

  16. R1 I1 A R2 E3 B I2 E1 I3 I4 R3 R4 E2 R5 Kirchhoff’s Laws • (1) Label all currents • Choose any direction (2) Label +/- for all elements Current goes +  - (for resistors) • Choose loop and direction • Your choice! • Write down voltage drops • Follow any loops • Write down node equation • Iin = Iout Physics 102: Lecture 6, Slide 16 36

  17. Example You try it! In the circuit below you are given e1, e2, R1, R2 and R3. Find I1, I2 and I3.  • Label all currents(Choose any direction)  2. Label +/- for all elements(Current goes +  - for resistor)  3. Choose loop and direction(Your choice!)  • Write down voltage drops(First sign you hit is sign to use!) Loop 1: –e1+I1R1 –I2R2 = 0 +I2R2 +I3R3 +e2= 0 R1 I3 I1 Loop 2: - + I2 +  - + 5. Write down node equation e1 R2 Loop 1 R3 - - Loop 2 + Node: I1 +I2 = I3 - + e2 3 Equations, 3 unknowns the rest is math! Physics 102: Lecture 6, Slide 17

  18. See you next lecture! • Read Sections 18.10,11 Physics 102: Lecture 6, Slide 18

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