Important Announcements Makeup lecture 5 Video of lecture

Important Announcements • Makeup lecture #5 – – – Video of lecture available online (see course website) Homework #3 now due Thursday Feb. 10 My (and M. Lundsgaard’s) office hours moved to Wednesday Lectures resume as normal this week (#6 and #7) – • Discussions & Labs – – Discussions & Labs resume as normal this week Missed discussion or lab? See course website Physics 102: Lecture 6, Slide 1

Exam I two weeks from today! • How do you study for a Phys 102 exam? – – – Start studying now! (cramming DOES NOT work) Emphasize understanding concepts & problem solving, NOT memorization Review lecture notes, problem solver summary Understand formula sheet (i. e. when to use and when NOT to use an equation) & know what each symbol means Do practice exam problems (time yourself!) Go to office hours & review session Physics 102: Lecture 6, Slide 2

Physics 102: Lecture 06 Kirchhoff’s Laws Physics 102: Lecture 6, Slide 3

Last Time • Resistors in series: Last Lecture Current thru is same; Voltage drop across is IRi • Resistors Voltage drop across is same; Current thru is V/Ri • Solved Today in parallel: • What Circuits about this one? Physics 102: Lecture 6, Slide 4

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 5

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 R 1 I 1 • Example: At junction: I 1 + I 2 = I 3 Physics 102: Lecture 6, Slide 6 e 1 I 3 I 2 R 3 R 2 e 2

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 1 e 1 Around the right loop: e 2 + I 3 R 3 + I 2 R 2 = 0 Physics 102: Lecture 6, Slide 7 I 3 I 2 R 3 R 2 e 2

Using Kirchhoff’s Rules (1) Label all currents Choose any direction (2) Label +/– for all elements R 1 A + I 1 - + R 2 Current goes + – (for resistors) (3) Choose loop and direction + B ε 1 - ε 2 (4) Write down voltage drops Be careful about signs Physics 102: Lecture 6, Slide 8 I 2 ε 3 I 3 - + R 3 I 5 R 5 + + - + I 4 - R 4 +

Loop Rule Practice R 1=5 W Find I: Label currents Label elements +/Choose loop Write KLR - + + e 1= 50 V - A - + R 2=15 W +e 1 - IR 1 - e 2 - IR 2 = 0 +50 - 5 I - 10 - 15 I = 0 I = +2 Amps What is the electric potential at VB (assume VA = 0): VA+e 1 - IR 1 = VB 0 + 50 – 2 x 5 = 40 V = VB Physics 102: Lecture 6, Slide 9 B I - + e 2= 10 V

ACT: KLR Resistors R 1 and R 2 are 1) in parallel 2) in series 3) neither 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 R 1 and R 2 but also ε 2. Physics 102: Lecture 6, Slide 10 R 1=10 W I 1 E 2 = 5 V I 2 R 2=10 W IB + E 1 = 10 V

Preflight 6. 1 Calculate the current through resistor 1. 28% 60% 1) I 1 = 0. 5 A 2) I 1 = 1. 0 A 12% 3) I 1 = 1. 5 A E 1 - I 1 R = 0 I 1 = E 1 /R = 1 A I 1 R 1=10 W + - E 2 = 5 V I 2 R 2=10 W IB + E 1 = 10 V Physics 102: Lecture 6, Slide 11

Preflight 6. 1 Calculate the current through resistor 1. 28% 60% 1) I 1 = 0. 5 A 2) I 1 = 1. 0 A 12% 3) I 1 = 1. 5 A I 1 R 1=10 W + - E 2 = 5 V I 2 E 1 - I 1 R = 0 I 1 = E 1 /R = 1 A R 2=10 W IB + E 1 = 10 V ACT: Voltage Law How would I 1 change if the switch was opened? 1) Increase Physics 102: Lecture 6, Slide 12 2) No change 3) Decrease

Preflight 6. 2 Calculate the current through resistor 2. 57% 30% 1) I 2 = 0. 5 A 2) I 2 = 1. 0 A 13% 3) I 2 = 1. 5 A R 1=10 W I 1 E 2 = 5 V E 1 - E 2 - I 2 R = 0 I 2 = 0. 5 A Physics 102: Lecture 6, Slide 13 I 2 R 2=10 W + IB + E 1 = 10 V

Preflight 6. 2 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: R=10 W I 1 E 2 = 5 V I 2 R=10 W + - IB + E 1 = 10 V +E 1 - E 2 + I 2 R = 0 Note the sign change from last slide I 2 = -0. 5 A Answer has same magnitude as before but opposite sign. That means current goes to the right, as we found before. Physics 102: Lecture 6, Slide 14

Kirchhoff’s Junction Rule Current Entering = Current Leaving I 1 = I 2 + I 3 I 2 I 3 Preflight 6. 3 R=10 W I 1 Calculate the current through battery. 12% 32% 57% E=5 V I 2 R=10 W 1) IB = 0. 5 A 2) IB = 1. 0 A 3) IB = 1. 5 A IB = I 1 + I 2 = 1. 5 A IB + E 1 = 10 V Physics 102: Lecture 6, Slide 15

Kirchhoff’s Laws (1) Label all currents Choose any direction (2) Label +/– for all elements R 1 I 1 A R 2 Current goes + – (for resistors) (3) Choose loop and direction Your choice! (4) Write down voltage drops Follow any loops (5) Write down junction equation Iin = Iout Physics 102: Lecture 6, Slide 16 B ε 1 ε 3 I 2 I 3 R 3 ε 2 R 5 I 4 R 4

You try it! In the circuit below you are given e 1, e 2, R 1, R 2 and R 3. Find I 1, I 2 and I 3. R 1 I 3 I 1 I 2 e 1 R 2 R 3 e 2 Physics 102: Lecture 6, Slide 17

You try it! In the circuit below you are given e 1, e 2, R 1, R 2 and R 3. Find I 1, I 2 and I 3. 1. 2. 3. 4. Label all currents Label +/- for all elements (Choose any direction) (Current goes + - for resistor) Choose loop and direction (Your choice!) Write down voltage drops (Potential increases or decreases? ) Loop 1: +e 1 - I 1 R 1 + I 2 R 2 = 0 Loop 2: +e 1 - I 1 R 1 - I 3 R 3 - e 2 = 0 5. + R 1 - I 1 I 3 I 2 Write down junction equation Node: I 1 + I 2 = I 3 + e 1 - 3 Equations, 3 unknowns the rest is math! Physics 102: Lecture 6, Slide 18 Loop 1 R 2 + - R 3 - + - e 2 +

ACT: Kirchhoff loop rule What is the correct expression for “Loop 3” in the circuit below? + R 1 - I 1 I 3 I 2 e 1 + - R 2 Loop 3 - + - Physics 102: Lecture 6, Slide 19 1) +e 2 – I 3 R 3 – I 2 R 2 = 0 + e 2 + R 3 2) +e 2 – I 3 R 3 + I 2 R 2 = 0 3) +e 2 + I 3 R 3 + I 2 R 2 = 0

Let’s put in real numbers In the circuit below you are given e 1, e 2, R 1, R 2 and R 3. Find I 1, I 2 and I 3. Loop 2 + 5 - I 1 I 3 I 2 + 20 - Loop 1 10 + - 10 - + - 2 1. Loop 1: 20 -5 I 1+10 I 2 = 0 2. Loop 2: 20 -5 I 1 -10 I 3 -2=0 3. Junction: I 3=I 1+I 2 + solution: substitute Eq. 3 for I 3 in Eq. 2: 20 - 5 I 1 - 10(I 1+I 2) - 2 = 0 rearrange: 15 I 1+10 I 2 = 18 rearrange Eq. 1: 5 I 1 -10 I 2 = 20 Now we have 2 eq. , 2 unknowns. Continue on next slide Physics 102: Lecture 6, Slide 20

15 I 1+10 I 2 = 18 5 I 1 - 10 I 2 = 20 Now we have 2 eq. , 2 unknowns. Add the equations together: 20 I 1=38 I 1=1. 90 A Plug into bottom equation: 5(1. 90)-10 I 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! Physics 102: Lecture 6, Slide 21

See you next time… Physics 102: Lecture 6, Slide 22