1 / 27

Kirchhoff’s Laws

Physics 102: Lecture 06. Kirchhoff’s Laws. Solved Circuits. Today. What about this one?. Last Time. Resistors in series:. Current thru is same; Voltage drop across is IR i. Last Lecture. Resistors in parallel:. Voltage drop across is same; Current thru is V/R i. Kirchhoff’s Rules.

clementec
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. Physics 102:Lecture06 Kirchhoff’s Laws

  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

  3. Kirchhoff’s Rules • Kirchhoff’s Junction Rule (KJR): • Current going in equals current coming out. • Kirchhoff’s Loop Rule (KLR): • Sum of voltage drops around a loop is zero.

  4. R1 I1 A - + + + R2 E3 - - + B I2 E1 I3 I4 - - - + R3 E2 - + + R5 - + Using Kirchhoff’s Rules • (1) Label all currents • Choose any direction (2) Label +/- for all elements Current goes +  - (for resistors) • Choose loop and direction R4 • Write down voltage drops • Be careful about signs

  5. Example Loop Rule Practice B R1=5 W I Find I: e1= 50V A R2=15 W e2= 10V

  6. - + + - - + + - Example Loop Rule Practice B R1=5 W I Find I: e1= 50V Label currents Label elements +/- Choose loop Write KLR A R2=15 W e2= 10V –e1+IR1 +e2+IR2 = 0 -50 + 5 I + 10 +15 I = 0 I = +2 Amps

  7. ACT: KLR 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 25

  8. Preflight 6.1 Calculate the current through resistor 1. I1 R=10 W + - E2 = 5 V I2 R=10 W 1) I1 = 0.5 A 2) I1 = 1.0 A 3) I1 = 1.5 A  I1 = E1 /R = 1A -E1+ I1R = 0 IB - + E1 = 10 V 27

  9. Preflight 6.1 Calculate the current through resistor 1. I1 R=10 W + - E2 = 5 V I2 R=10 W 1) I1 = 0.5 A 2) I1 = 1.0 A 3) I1 = 1.5 A -E1+ I1R = 0  I1 = E1/R = 1A IB - + E1 = 10 V ACT: Voltage Law How would I1 change if the switch was opened? 1) Increase 2) No change 3) Decrease 32

  10. Preflight 6.2 Calculate the current through resistor 2. I1 R=10 W 1) I2 = 0.5 A 2) I2 = 1.0 A 3) I2 = 1.5 A E2 = 5 V I2 R=10 W - + -E1 +E2+ I2R = 0  I2 = 0.5A IB - + E1 = 10 V 35

  11. Preflight 6.2 How do I know the direction of I2? It doesn’t matter. Choose whatever direction you like. Then solve the equations to find I2. If the result is positive, then your initial guess was correct. If result is negative, then actual direction is opposite to your initial guess. I1 R=10 W E2 = 5 V I2 R=10 W + - Work through preflight with opposite sign for I2? IB - + E1 = 10 V -E1 +E2- I2R = 0 Note the sign change from last slide  I2 = -0.5A Answer has same magnitude as before but opposite sign. That means current goes to the left, as we found before. 35

  12. 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 “The first two can be calculated using V=IR because the voltage and resistance is given, and the current through E1 can be calculated with the help of Kirchhoff's Junction rule, that states whatever current flows into the junction must flow out. So I1 and I2 are added together.” 38

  13. 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 junction equation • Iin = Iout 36 39

  14. R1 I3 I1 - + I2 - + e1 R2 R3 - + - + e2 Example You try it! In the circuit below you are given e1, e2, R1, R2 and R3. Find I1, I2 and I3. + - 45

  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(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 junction equation e1 R2 Loop 1 R3 - - Loop 2 + Node: I1 +I2 = I3 - + e2 3 Equations, 3 unknowns the rest is math! 45

  16. 5 I3 I1 - + I2 + - + 10 20 10 - - + - + 2 Example Let’s put in real numbers In the circuit below you are given e1, e2, R1, R2 and R3. Find I1, I2 and I3. • 1. left loop: -20+5I1-10I2 = 0 • outer loop: -20 +5I1+10I3+2=0 • 3. junction: I3=I1+I2 solution: substitute Eq.3 for I3 in Eq. 2: -20 + 5I1 + 10(I1+I2) + 2 = 0 rearrange: 15I1+10I2 = 18 rearrange Eq. 1: 5I1-10I2 = 20 Now we have 2 eq., 2 unknowns. Continue on next slide 45

  17. 15I1+10I2 = 18 5I1 - 10I2 = 20 Now we have 2 eq., 2 unknowns. Add the equations together: 20I1=38 I1=1.90 A Plug into bottom equation: 5(1.90)-10I2 = 20 I2=-1.05 A note that this means direction of I2 is opposite to that shown on the previous slide Use junction equation (eq. 3 from previous page) I3=I1+I2 = 1.90-1.05 I3 = 0.85 A We are done!

More Related