Physics 102: Lecture 06 Kirchhoff s Laws Physics 102: Lecture 6, Slide 1
Today Last Lecture Last Time Resistors in series: R eq = R 1 R 2 R 3 Current through each is same; Voltage drop is IR i Resistors in parallel: 1 R eq = 1 R 1 1 R 2 1 R 3 Voltage drop across each is the same; Current through each is V/R i Solved Circuits What about this one? Physics 102: Lecture 6, Slide 2
When would you ever find a circuit like this?!?! ε1 and ε2 arise from surface electrochemistry so what does the EKG monitor really measure? Physics 102: Lecture 6, Slide 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. Physics 102: Lecture 6, Slide 4
Kirchhoff s Junction Rule (KJR) Conceptual basis: conservation of charge At any junction in a circuit, the current that enters the junction equals the current that leaves the junction Example: R 1 I 3 At junction: e 1 R 2 R 3 = I 3 e 2 Physics 102: Lecture 6, Slide 5
Kirchhoff s Loop Rule (KLR) Conceptual basis: conservation of energy Going around any complete loop in a circuit, the sum total of all the potential differences is zero Example: R 1 I 3 Around the right loop: e 1 R 2 R 3 e 2 I 3 R 3 R 2 = 0 e 2 Physics 102: Lecture 6, Slide 6
Using Kirchhoff s Rules (1) Label all currents Choose any direction (2) Label / for all elements Current goes (for resistors) (3) Choose loop and direction (4) Write down voltage drops Be careful about signs A ε 1 R 1 ε 2 B R 2 I 5 ε 3 R 3 R 4 R 5 I 3 I 4 Physics 102: Lecture 6, Slide 7
Loop Rule Practice Find I: R 1 =5 W I B 1. Label currents 2. Label elements / 3. Choose loop 4. Write KLR e 1 IR 1 e 2 IR 2 = 0 50 5 I 10 15 I = 0 I = 2 Amps e 1 = 50V A R 2 =15 W e 2 = 10V What is the electric potential at V B (assume V A = 0): V A e 1 IR 1 = V B 0 50 2 5 = 40V = V B Physics 102: Lecture 6, Slide 8
Checkpoint 1 Calculate the current through R 1. R 1 =10 W 1) = 0.5 A 2) = 1.0 A 3) = 1.5 A e 2 = 5 V R 2 =10 W I B e 1 = 10 V Physics 102: Lecture 6, Slide 9
Checkpoint 2 Calculate the current through R 1. R 1 =10 W 1) = 0.5 A 2) = 1.0 A 3) = 1.5 A e 2 = 5 V R 2 =10 W I B e 1 = 10 V ACT: Voltage Law How would change if the switch was opened? 1) Increase 2) No change 3) Decrease Physics 102: Lecture 6, Slide 10
ACT/Checkpoint 2 Calculate the current through R 2. R 1 =10 W A) = 0.5 A B) = 1.0 A C) = 1.5 A e 2 = 5 V R 2 =10 W I B e 1 = 10 V Physics 102: Lecture 6, Slide 11
Checkpoint 2 How do I know the direction of? It doesn t matter. Choose whatever direction you like. Then solve the equations to find. If the result is positive, then your initial guess was correct. If result is negative, then actual direction is opposite to your initial guess. e 2 = 5 V R=10 W R=10 W Work through Checkpoint with opposite sign for : I B e 1 = 10 V Physics 102: Lecture 6, Slide 12
Kirchhoff s Junction Rule Current Entering = Current Leaving = I 3 I 3 ACT/Checkpoint 3 Calculate the current through battery. R=10 W 1) I B = 0.5 A 2) I B = 1.0 A 3) I B = 1.5 A e 2 = 5 V R=10 W Physics 102: Lecture 6, Slide 13 I B e 1 = 10 V
Kirchhoff s Laws (1) Label all currents Choose any direction (2) Label / for all elements A R 1 Current goes (for resistors) R 2 ε 3 (3) Choose loop and direction Your choice! (4) Write down voltage drops Follow any loops ε 1 ε 2 B R 5 R 3 I 3 I 4 R 4 (5) Write down junction equation I in = I out Physics 102: Lecture 6, Slide 14
You try it! In the circuit below you are given e 1, e 2, R 1, R 2 and R 3. Find, and I 3. R 1 I 3 e 1 R 2 R 3 e 2 Physics 102: Lecture 6, Slide 15
You try it! In the circuit below you are given e 1, e 2, R 1, R 2 and R 3. Find, and I 3. 1. Label all currents (Choose any direction) 2. Label / for all elements (Current goes for resistor) 3. Choose loop and direction (Your choice!) 4. Write down voltage drops (Potential increases or decreases?) Loop 1: e 1 R 1 R 2 = 0 Loop 2 Loop 2: e 1 R 1 I 3 R 3 e 2 = 0 R 1 I 3 5. Write down junction equation Node: = I 3 e 1 Loop 1 R 2 R 3 3 Equations, 3 unknowns the rest is math! Physics 102: Lecture 6, Slide 16 e 2
ACT: Kirchhoff loop rule What is the correct expression for Loop 3 in the circuit below? e 1 R 1 I 3 R 2 R Loop 3 3 1) e 2 I 3 R 3 R 2 = 0 2) e 2 I 3 R 3 R 2 = 0 3) e 2 I 3 R 3 R 2 = 0 e 2 Physics 102: Lecture 6, Slide 17
Let s put in real numbers 20 In the circuit below you are given e 1, e 2, R 1, R 2 and R 3. Find, and I 3. 5 Loop 1 I 3 10 10 Loop 2 1. Loop 1: 20 5 10 = 0 2. Loop 2: 20 5 10I 3 2=0 3. Junction: I 3 = 2 solution: substitute Eq.3 for I 3 in Eq. 2: 20 5 10( ) 2 = 0 rearrange: 15 10 = 18 rearrange Eq. 1: 5 10 = 20 Now we have 2 eq., 2 unknowns. Continue on next slide Physics 102: Lecture 6, Slide 18
15 10 = 18 5 10 = 20 Now we have 2 eq., 2 unknowns. Add the equations together: 20 =38 =1.90 A Plug into bottom equation: 5(1.90)10 = 20 =1.05 A note that this means direction of is opposite to that shown on the previous slide Physics 102: Lecture 6, Slide 19 Use junction equation (eq. 3 from previous page) I 3 = = 1.901.05 I 3 = 0.85 A We are done!
Back to where we started V A I V B I V EKG = V A V B V A IR c1 ε1 IR t1 V sig IR t2 ε2 IR c2 = V B V A V B = IR c1 ε1 IR t1 V sig IR t2 ε2 IR c2 Physics 102: Lecture 6, Slide 20 Surface potentials tend to cancel!