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. 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. 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. 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. 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. 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. 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
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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
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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. 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. 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!
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