RC Cola. I was really confused about the calculus that lead to the current equation. Could you derive this in class? - We ll do that today

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1 ola was really confused about the calculus that lead to the current euation. ould you derive this in class? - We ll do that today Phys 122 Lecture 15 G. ybka

2 Business Exam Thursday! (no circuits on this exam) Practice exam solutions posted on Wed. ll you your seats Wed will be a review, with a few problems and a review of the key points of this unit Lecture Material on Exam: Lectures 7-14;; From Electrical Potential Energy through urrent, Voltage and esistance, as listed on the syllabus feel as if 'm starting to get this material, but with the midterm coming up, nothing new is going to stick.

Lecture Material on Exam: Lectures 7-14;; From Electrical Potential Energy through urrent, Voltage and

3 Kirchoff ules 1) Label all currents hoose any direction 2) Label +/ for all elements urrent goes + to (for resistors) or remember to track direction of loop vs direction of current 3) hoose loop and direction Must start on wire, not element. 4) Write down voltage drops First sign you hit is sign to use. 5) Write down node euation in = out A + + E E 2 B E

and direction Must start on wire, not element.

4 E 3 + E 1 E

5 apacitors: Okay, a brief summary We learned how they add in series, parallel and how a dielectric increases the capacitance esistors: We learned how they add in series, parallel and how the material they are made of affects the resistance Networks Voltage Loop: Drops (+);; Gains (-);; Label currents;; Label the +/- in dir of ;; Sum of Drops/Gains around loop = urrent rule: Sum of currents in junction = sum out Power From battery: P = V n resistors, P = V = 2 = V 2 / n general, look at sum of V s for total

Voltage Loop: Drops (+);; Gains (-);; Label currents;; Label the +/- in dir of ;; Sum of Drops/Gains around loop = urrent rule:

6 Batteries (non-ideal) Parameterized with "internal resistance" V = ε r ε + r = V r ε = ε + r Þ V = ε + r

7 Power Batteries & esistors Energy expended chemical to electrical to heat ate is: What s happening? energy = time power harges per time Assert: P = V Energy drop per charge 2 For esistors: P = ( ) = Units okay? Joule oulomb oulomb second = J s = Watt

energy = time power harges per time Assert: P = V Energy drop

8 More complex now... Add a apacitor to circuit ecall voltage drop on is V = + - ε Write KVL: + ε = onsider that = d dt and substitute. Now en has only d + ε = dt Differential Euation!

9 licker 1A At t = the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. What is the value of the current just after the switch is thrown? a ε b (a) = (b) = ε/2 (c) = 2ε/ Just after the switch is thrown, the capacitor still has no charge à therefore the voltage drop across the capacitor = à it acts like a wire in the circuit Applying KVL to the loop at t =, à ε = Þ = ε /2 an you go over why charge flows through the capacitor right when the switch is flipped? Essentially, why does it act as a wire immediately after the switch is thrown? Thanks.

a ε b (a) = (b) = ε/2 (c) = 2ε/ Just after the switch is thrown, the capacitor still has no charge à therefore the voltage drop across the capacitor = à

10 licker a b ε What is the value of the current after a very 1B long time? (a) = (b) = ε/2 (c) > 2ε/ As the current continues to flow, the charge on the capacitor continues to grow. As the charge on the capacitor continues to grow, the voltage across the capacitor will increase. The voltage across the capacitor is limited to ε;; the current goes to.

11 ircuit Time Dependence: harging Loop euation ε = d + dt Guess solution: a ε b ( / ) = ε 1 e t heck that it is a solution: Þ d dt d dt t = εe / 1 t/ e ( t/ + = ε + ε 1 e ) = ε! Note that this guess incorporates the boundary conditions: t = = t = = ε

12 ircuit Time Dependence: harging harge capacitor: a ( / ) = ε 1 e t b urrent is found from differentiation: ε d = = dt ε / e t Þ onclusion: apacitor reaches its final charge (Q=ε) exponentially with time constant τ =. urrent decays from max ( max = ε/) with same time constant.

onclusion: apacitor reaches its final charge (Q=ε) exponentially

13 harging apacitor harge on ε 1 2 Max = ε ( / ) = ε 1 e t Q f( x).5 63% Max at t= t urrent d = = dt Max = ε/ ε / e t ε/ 1 f( x) 1.5 x t/ 37% Max at t= t x 4

5 63% Max at t= 1 2 3 4 t urrent d = = dt

14 licker At t = the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. At time t = t 1 = τ, the charge Q 1 on the capacitor is (1-1/e) of its asymptotic charge Q f =ε. What is the relation between Q 1 and Q 2, the charge on the capacitor at time t = t 2 = 2τ? Hint: think graphically! a b ε (a) Q 2 < 2 Q 1 (b) Q 2 = 2 Q 1 (c) Q 2 > 2 Q 1 The charge on the capacitor increases with time as: = ( e t /2 ε 1 ) The uestion is: how does this charge increase differ from a linear increase? See graph: the charge increase is not as fast as linear. The rate of increase is proportional to the current (d/dt) which decreases with time. Therefore, Q 2 < 2Q 1. Q f( x) 1.5 Q 1 2Q 1 Q 2 τ 2τ

What is the relation between Q 1 and Q 2, the charge on the capacitor at time t = t 2 = 2τ? Hint: think graphically!

15 ircuit Time Dependence: Discharging initially charged with Q = ε onnect switch to b at t =. alculate current and charge as function of time. a ε b Loop theorem Þ + = onvert to differential euation for : d = Þ d dt dt + =

16 ircuit Time Dependence: Discharging Discharge capacitor: d dt + = Guess solution: a ε b = εe -t/ heck that it is a solution: d dt t = εe / 1 t/ t/ Þ + = εe + εe =! d dt Note that this guess incorporates the boundary conditions: t = = ε t = =

17 ircuit Time Dependence: Discharging Discharge capacitor: = εe -t/ urrent is found from differentiation: a ε b d ε dt e = = t / Þ Does current really "flow backwards" when the capacitor is discharged? think it would help me have a better understanding of what's going on if you explained this process in terms of the flow of electrons. onclusion: apacitor discharges exponentially with time constant τ = urrent decays from initial max value (= -ε/) with same time constant

think it would help me have a better understanding of what's going on if you explained this process in terms of the flow

18 Discharging apacitor harge on = εe -t/ Max = ε 37% Max at t= urrent d ε dt e = = t / Max = -ε/ 37% Max at t= 1 f( x) Q f( x) ε ε/ x x t t

19 licker At t = the switch is connected to position a in the circuit shown: The capacitor is initially uncharged. ε 1 At t = t, the switch is thrown from position a to position b. Which of the following graphs best represents the time dependence of the charge on? ε 1 (a) (b) (c) ε ε 1 a b 2 f( x).5 Q f( x).5 Q f ( x ).5 t x t/ t t t t For < t < t, the capacitor is charging with time constant τ = For t > t, the capacitor is discharging with time constant τ = 2 (a) has eual charging and discharging time constants (b) has a larger discharging τ than a charging τ (c) has a smaller discharging τ than a charging τ x t/ x t/ t

ε 1 (a) (b) (c) ε ε 1 a b 2 f( x).5 Q f( x).5 Q f ( x ).

20 harging Discharging ε ε 1 2 Q f( x).5 ( / ) = ε 1 e t f( x).5 = εe -t/ ε/ x t/ t x 4 t f( x).5 d = = dt ε / e t Q f( x).5 d = = dt ε / e t t -ε/ t x 4 x

21 This is the tough one heckpoint 8 A B After both switches have been closed for a long time The current through the capacitor is zero The current through = current through 2 V capacitor = V 2 V 2 = 2/3 V

22 heckpoint 1 & 12 Same charge initially Which circuit has the largest time constant? A) ircuit 1 B) ircuit 2 ) Same τ = Q elationship of Q 1 to Q 2 at any time after t =? A) Q 1 > Q 2 B) Q 1 > Q 2 ) Q 1 = Q 2 D) Something wrong E) Something else wrong = 1 =

23 heckpoints 2 & 4 Not involved for these uestions lose S1, V 1 = voltage across immediately after V 2 = voltage across a long time after mmediately after the switch S 1 is closed: A) V 1 = V V 2 = V B) V 1 = V 2 = V ) V 1 = V 2 = D) V 1 = V V 2 = After the switch S 1 has been closed for a long time Q = V = Q/ V 1 = = V = V 2 = V

24 lose S 1 at t = (leave S 2 open) V 2 S1 S2 V V = V = Q/ = At t = V = V At t = big

25 heckpoint 6 + A B D V 2 V

26 V 2 Open S1 at t = big and close S2 S2 S1 V V 2 = V/2

Your Comments. This was a very confusing prelecture. Do you think you could go over thoroughly how the LC circuits work qualitatively?

Your Comments. This was a very confusing prelecture. Do you think you could go over thoroughly how the LC circuits work qualitatively? Your omments I am not feeling great about this mierm...some of this stuff is really confusing still and I don't know if I can shove everything into my brain in time, especially after spring break. an you