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# Lect06 handout

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• 1 Lecture timing is fine. No need to rush
• Have students label I5, since it isn’t shown in their drawing
• Can ask if R1 and R5 are in series, parallel. Also make sure they understand potential at B is higher than at A.
• Note that nothing is in series or in parallel!
• Note that nothing is in series or in parallel!
• Have them go back to this slide and fill in (5)
• 1) This will be given on the exam. Comment on 1 equation 2 unknowns. 2 eqs 3 unknowns finally 3 and 3
• 1) This will be given on the exam. Comment on 1 equation 2 unknowns. 2 eqs 3 unknowns finally 3 and 3
• 1) This will be given on the exam. Comment on 1 equation 2 unknowns. 2 eqs 3 unknowns finally 3 and 3
• 1) This will be given on the exam. Comment on 1 equation 2 unknowns. 2 eqs 3 unknowns finally 3 and 3
• ### Lect06 handout

1. 1. Kirchhoff’s Laws Physics 102: Lecture 06
2. 2. Last Time <ul><li>Resistors in series: </li></ul><ul><li>Resistors in parallel: </li></ul>Current thru is same; Voltage drop across is IR i Voltage drop across is same; Current thru is V/R i Last Lecture <ul><li>Solved Circuits </li></ul><ul><li>What about this one? </li></ul>Today
3. 3. Kirchhoff’s Rules <ul><li>Kirchhoff’s Junction Rule (KJR): </li></ul><ul><ul><li>Current going in equals current coming out. </li></ul></ul><ul><li>Kirchhoff’s Loop Rule (KLR): </li></ul><ul><ul><li>Sum of voltage drops around a loop is zero. </li></ul></ul>
4. 4. <ul><li>Conceptual basis: conservation of charge </li></ul><ul><li>At any junction in a circuit, the current that enters the junction equals the current that leaves the junction </li></ul><ul><li>Example: </li></ul>Kirchhoff’s Junction Rule (KJR) At junction: I 1 + I 2 = I 3 R 1 R 2 R 3 I 1 I 3 I 2  1  2
5. 5. <ul><li>Conceptual basis: conservation of energy </li></ul><ul><li>Going around any complete loop in a circuit, the sum total of all the potential differences is zero </li></ul><ul><li>Example: </li></ul>Kirchhoff’s Loop Rule (KLR) Around the right loop:  2 + I 3 R 3 + I 2 R 2 = 0 R 1 R 2 R 3 I 1 I 3 I 2  1  2
6. 6. Using Kirchhoff’s Rules <ul><li>(1) Label all currents </li></ul><ul><ul><li>Choose any direction </li></ul></ul>(2) Label +/- for all elements Current goes +  – (for resistors) <ul><li>Choose loop and direction </li></ul><ul><li>Write down voltage drops </li></ul><ul><ul><li> Be careful about signs </li></ul></ul>R 4 + - + + + + - - - - + + + - - - R 1 ε 1 R 2 R 3 ε 2 ε 3 R 5 A B I 1 I 3 I 2 I 4 I 5
7. 7. Loop Rule Practice R 1 =5  I +  1 - IR 1 -  2 - IR 2 = 0 +50 - 5 I - 10 - 15 I = 0 I = +2 Amps  1 = 50V R 2 =15   2 = 10V A B Find I: Example Label currents Label elements +/- Choose loop Write KLR + - + - + - + -
8. 8. ACT: KLR R 1 =10  E 1 = 10 V I B I 1 E 2 = 5 V R 2 =10  I 2 Resistors R 1 and R 2 are 1) in parallel 2) in series 3) neither + -
9. 9. Preflight 6.1 R=10  E 1 = 10 V I B I 1 E 2 = 5 V R=10  I 2 1) I 1 = 0.5 A 2) I 1 = 1.0 A 3) I 1 = 1.5 A + - + - E 1 - I 1 R = 0 Calculate the current through resistor 1.  I 1 = E 1 /R = 1A
10. 10. Preflight 6.1 R=10  E 1 = 10 V I B I 1 R=10  I 2 1) I 1 = 0.5 A 2) I 1 = 1.0 A 3) I 1 = 1.5 A + - + - E 1 - I 1 R = 0  I 1 = E 1 /R = 1A How would I 1 change if the switch was opened? E 2 = 5 V 1) Increase 2) No change 3) Decrease Calculate the current through resistor 1. ACT: Voltage Law
11. 11. Preflight 6.2 R=10  E 1 = 10 V I B I 1 E 2 = 5 V R=10  I 2 1) I 2 = 0.5 A 2) I 2 = 1.0 A 3) I 2 = 1.5 A + - + - E 1 - E 2 - I 2 R = 0  I 2 = 0.5A Calculate the current through resistor 2.
12. 12. Preflight 6.2 R=10  E 1 = 10 V I B I 1 E 2 = 5 V R=10  I 2 - + + - + E 1 - E 2 + I 2 R = 0 Note the sign change from last slide  I 2 = -0.5A Answer has same magnitude as before but opposite sign. That means current goes to the right, as we found before. How do I know the direction of I 2 ? It doesn’t matter. Choose whatever direction you like. Then solve the equations to find I 2. If the result is positive, then your initial guess was correct. If result is negative, then actual direction is opposite to your initial guess. Work through preflight with opposite sign for I 2 ?
13. 13. Kirchhoff’s Junction Rule Current Entering = Current Leaving I 1 I 2 I 3 I 1 = I 2 + I 3 1) I B = 0.5 A 2) I B = 1.0 A 3) I B = 1.5 A I B = I 1 + I 2 = 1.5 A Calculate the current through battery. R=10  E 1 = 10 V I B I 1 E = 5 V R=10  I 2 + - Preflight 6.3
14. 14. Kirchhoff’s Laws <ul><li>(1) Label all currents </li></ul><ul><ul><li>Choose any direction </li></ul></ul>(2) Label +/- for all elements Current goes +  - (for resistors) <ul><li>Choose loop and direction </li></ul><ul><ul><li>Your choice! </li></ul></ul><ul><li>Write down voltage drops </li></ul><ul><li>Follow any loops </li></ul><ul><li>Write down junction equation </li></ul><ul><li>I in = I out </li></ul>R 4 R 1 ε 1 R 2 R 3 ε 2 ε 3 I 1 I 3 I 2 I 4 R 5 A B
15. 15. You try it! In the circuit below you are given  1 ,  2 , R 1 , R 2 and R 3 . Find I 1 , I 2 and I 3 . Example R 1 R 2 R 3 I 1 I 3 I 2  1  2
16. 16. You try it! R 1 R 2 R 3 I 1 I 3 I 2 + - + + + Loop 1: +  1 - I 1 R 1 + I 2 R 2 = 0 <ul><li>Label all currents (Choose any direction) </li></ul>2. Label +/- for all elements (Current goes +  - for resistor) 3. Choose loop and direction (Your choice!) <ul><li>Write down voltage drops (Potential increases or decreases?) </li></ul>- - - Loop 2:  1 5. Write down junction equation Node: I 1 + I 2 = I 3  2 3 Equations, 3 unknowns the rest is math! In the circuit below you are given  1 ,  2 , R 1 , R 2 and R 3 . Find I 1 , I 2 and I 3 . Loop 1  + -   +  1 - I 1 R 1 - I 3 R 3 -  2 = 0   Example Loop 2
17. 17. ACT: Kirchhoff loop rule What is the correct expression for “Loop 3” in the circuit below? Loop 3 R 2 R 3  1  2 R 1 1) +  2 – I 3 R 3 – I 2 R 2 = 0 2) +  2 – I 3 R 3 + I 2 R 2 = 0 3) +  2 + I 3 R 3 + I 2 R 2 = 0 I 1 I 3 I 2 + - + + + - - - + -
18. 18. Let’s put in real numbers In the circuit below you are given  1 ,  2 , R 1 , R 2 and R 3 . Find I 1 , I 2 and I 3 . Example 1. Loop 1: 20 -5I 1 +10I 2 = 0 2. Loop 2: 20 -5I 1 -10I 3 -2=0 3. Junction: I 3 =I 1 +I 2 solution: substitute Eq.3 for I 3 in Eq. 2: 20 - 5I 1 - 10(I 1 +I 2 ) - 2 = 0 rearrange: 15I 1 +10I 2 = 18 rearrange Eq. 1: 5I 1 -10I 2 = 20 Now we have 2 eq., 2 unknowns. Continue on next slide Loop 1 Loop 2  5 10 10 I 1 I 3 I 2 + - + + + - - -  + -
19. 19. 15I 1 +10I 2 = 18 5I 1 - 10I 2 = 20 Now we have 2 eq., 2 unknowns. Add the equations together: 20I 1 =38 I 1 =1.90 A Plug into bottom equation: 5(1.90)-10I 2 = 20 I 2 =-1.05 A note that this means direction of I 2 is opposite to that shown on the previous slide Use junction equation (eq. 3 from previous page) I 3 =I 1 +I 2 = 1.90-1.05 I 3 = 0.85 A We are done!
20. 20. See you next time…