phy1161_10_kirchhoffcurrent/voltagelaw loop rule and junction rule-1.ppt
- 1. Physics 1161: Lecture 10 Kirchhoff’s Laws
- 2. Kirchhoff’s Rules • Kirchhoff’s Junction Rule: –Current going in equals current coming out. • Kirchhoff’s Loop Rule: –Sum of voltage changes around a loop is zero.
- 5. Using Kirchhoff’s Rules (1) Label all currents (3)Choose loop and direction • Choose any direction • You will need one less loop than unknown currents (4) Write down voltage changes Be careful about signs • For batteries – voltage change is positive when summing from negative to positive • For resistors – voltage change is negative when summing in the direction of the current R4 I1 I3 I2 I4 R1 1 R2 R3 2 3 R5 A B (2) Write down junction equation Iin = Iout I5
- 6. Loop Rule Practice R1=5 W I 1= 50V R2=15 W 2= 10V A B Find I:
- 7. Loop Rule Practice R1=5 W I +1 - IR1 - 2 - IR2 = 0 +50 - 5 I - 10 - 15 I = 0 I = +2 Amps 1= 50V R2=15 W 2= 10V A B Find I: Label currents Choose loop Write KLR
- 8. Resistors R1 and R2 are 1 2 3 0% 0% 0% 1. In parallel 2. In series 3. neither I1 R1=10 W R2=10 W E1 = 10 V IB E2 = 5 V I2 + -
- 9. Resistors R1 and R2 are 1 2 3 0% 0% 0% 1. In parallel 2. In series 3. neither I1 R1=10 W R2=10 W E1 = 10 V IB E2 = 5 V I2 + - Definition of parallel: Two elements are in parallel if (and only if) you can make a loop that contains only those two elements. Upper loop contains R1 and R2 but also E2.
- 10. Preflight 10.1 R=10 W E1 = 10 V IB I1 E2 = 5 V R=10 W I2 1) I1 = 0.5 A 2) I1 = 1.0 A 3) I1 = 1.5 A E1 - I1R = 0 24% 62% 24% Calculate the current through resistor 1. 27 I1 = E1 /R = 1A
- 11. How would I1 change if the switch was opened? 1 2 3 0% 0% 0% E1 = 10 V IB R=10 W I1 R=10 W I2 E2 = 5 V 1. Increase 2. No change 3. Decrease
- 12. How would I1 change if the switch was opened? 1 2 3 0% 0% 0% E1 = 10 V IB R=10 W I1 R=10 W I2 E2 = 5 V 1. Increase 2. No change 3. Decrease
- 13. Preflight 10.2 R=10 W E1 = 10 V IB I1 E2 = 5 V R=10 W I2 1) I2 = 0.5 A 2) I2 = 1.0 A 3) I2 = 1.5 A E1 - E2 - I2R = 0 43% I2 = 0.5A Calculate the current through resistor 2. 35 28% 28%
- 14. Preflight 10.2 R=10 W E1 = 10 V IB I1 E2 = 5 V R=10 W I2 - + + - +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. 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. Work through preflight with opposite sign for I2?
- 15. Kirchhoff’s Junction Rule Current Entering = Current Leaving I1 I2 I3 I1 = I2 + I3 1) IB = 0.5 A 2) IB = 1.0 A 3) IB = 1.5 A IB = I1 + I2 = 1.5 A 7% 37% 57% R=10 W E1 = 10 V IB I1 E = 5 V R=10 W I2 + - Preflight 8.3 “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.”
- 16. Kirchhoff’s Laws (1) Label all currents Choose any direction (2) Write down the junction equation Iin = Iout (3) Choose loop and direction Your choice! (4) Write down voltage changes Follow any loops (5) Solve the equations by substitution or combination . R4 R1 E1 R2 R3 E2 E3 I1 I3 I2 I4 R5 A B
- 17. You try it! In the circuit below you are given 1, 2, R1, R2 and R3. Find I1, I2 and I3. R1 R2 R3 I1 I3 I2 + - 1 2 + -
- 18. You try it! R1 R2 R3 I1 I3 I2 + - Loop 1: +1- I1R1 + I2R2 = 0 1. Label all currents (Choose any direction) 3. Choose loop and direction (Your choice!) 4. Write down voltage changes Loop 2: 1 Node: I1 + I2 = I3 2 3 Equations, 3 unknowns the rest is math! In the circuit below you are given 1, 2, R1, R2 and R3. Find I1, I2 and I3. Loop 1 Loop 2 + - - I2R2 - I3R3 - 2 = 0 2. Write down junction equation
- 19. Let’s put in actual numbers In the circuit below you are given 1, 2, R1, R2 and R3. Find I1, I2 and I3. 2 5 10 10 I1 I3 I2 + - 20 + - 1. junction: I3=I1+I2 2. left loop: 20 - 5I1+10I2 = 0 3. right loop: -2 - 10I2 - 10I3 = 0 solution: substitute Eq.1 for I3 in Eq. 3: rearrange: -10I1 - 20I2 = 2 rearrange Eq. 2: 5I1-10I2 = 20 Now we have 2 eq., 2 unknowns. Continue on next slide
- 20. -10I1-20I2 = 2 2*(5I1 - 10I2 = 20) = 10I1 – 20I2 = 40 Now we have 2 eq., 2 unknowns. Add the equations together: -40I2 = 42 I2 = -1.05 A note that this means direction of I2 is opposite to that shown on the previous slide Plug into left loop equation: 5I1 -10*(-1.05) = 20 I1=1.90 A Use junction equation (eq. 1 from previous page) I3=I1+I2 = 1.90-1.05 I3 = 0.85 A