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I a I I R R b C e r V e R I Lecture 10, ACT 1 • Consider the circuit shown: 1A 50W a – What is the relation between Va -Vd and Va -Vc ? b 12V (a) (Va -Vd) < (Va -Vc) (b) (Va -Vd) = (Va -Vc) (c) (Va -Vd) > (Va -Vc) 1B 1B (b) I1 = I2 20W 80W d – What is the relation between I1 and I2? (a) I1 < I2 I2 I1 (c) I1 > I2 c Lecture 10, ACT 1 • Consider the circuit shown: 1A 50W a – What is the relation between Va -Vd and Va -Vc ? (a) (Va -Vd) < (Va -Vc) (b) (Va -Vd) = (Va -Vc) (c) (Va -Vd) > (Va -Vc) b I2 I1 12V 20W 80W d c • Do you remember that thing about potential being independent of path? Well, that’s what’s going on here !!! (Va -Vd) = (Va -Vc) Point d and c are the same, electrically Lecture 10, ACT 1 • Consider the circuit shown: 1A 50W a – What is the relation between Va -Vd and Va -Vc ? b 12V (a) (Va -Vd) < (Va -Vc) (b) (Va -Vd) = (Va -Vc) (c) (Va -Vd) > (Va -Vc) 1B 1B 20W • Note that: Vb -Vd • Therefore, (b) I1 = I2 (c) I1 > I2 = Vb -Vc I1 (20W) I 2 (80W) 80W d – What is the relation between I1 and I2? (a) I1 < I2 I2 I1 I1 4I 2 c Summary of Simple Circuits • Resistors in series: Reffective R1 R2 R3 ... Current thru is same; • Resistors 1 in parallel: R effective Voltage drop across is IRi 1 1 1 ... R1 R2 R3 Voltage drop across is same; Current thru is V/Ri Kirchhoff’s laws: (Tipler problems on Kirchhoff’’s rules) V n 0 loop I in I out Batteries (“Nonideal” = cannot output arbitrary current) I I • Parameterized with R "internal resistance" V e Ir r V e e Ir IR 0 e I Rr R V e Rr Internal Resistance Demo As # bulbs increases, what happens to “R”?? I r V R e How big is “r”? Power Batteries & Resistors Energy expended chemical to electrical to heat Rate is: energy power sJ time What’s happening? Assert: Charges per time P VI Potential difference per charge or you can write it as For Resistors: Units okay? P IR I I 2 R P V V R V 2 R Joule Coulomb J Watt Coulomb second s More complex now… I Let’s try to add a Capacitor to our simple circuit I R Recall voltage “drop” on C? V C e Q C Q Write KVL: e IR 0 C What’s wrong here? Consider that I dQ dt and substitute. Now eqn has only “Q”. KVL gives Differential Equation ! e R dQ Q 0 dt C The Big Idea • Previously: – Analysis of multi-loop circuits with batteries and resistors. – Main Feature: Currents are attained instantaneously and do not vary with time!! • Now: – Just added a capacitor to the circuit. – What changes?? • KVL yields a differential equation with a term proportional to Q and a term proportional to I = dQ/dt. dQ Q eR 0 dt C The Big Idea dQ Q eR 0 dt C • Physically, what’s happening is that the final charge cannot be placed on a capacitor instantly. • Initially, the voltage drop across an uncharged capacitor = 0 because the charge on it is zero ! (V=Q/C) • As current starts to flow, charge builds up on the capacitor, the voltage drop is proportional to this charge and increases; it then becomes more difficult to add more charge so the current slows 2 Lecture 10, ACT 2 2A • At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. a 2B (b) I0+ = e /2R b (c) I0+ = 2e /R long time? (b) I = e /2R C e – What is the value of the current I after a very (a) I = 0 I R – What is the value of the current I0+ just after the switch is thrown? (a) I0+ = 0 I (c) I > 2e /R R Lecture 10, ACT 2 2A • At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. a (b) I0+ = e /2R I R b C e – What is the value of the current I0+ just after the switch is thrown? (a) I0+ = 0 I R (c) I0+ = 2e /R • Just after the switch is thrown, the capacitor still has no charge, therefore the voltage drop across the capacitor = 0! • Applying KVL to the loop at t=0+, e IR 0 IR = 0 I = e /2R Lecture 10, ACT 2 a 2A • At t=0 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 I0+ just after the switch is thrown? I I R b C e R (a) I0+ = 0 2B (b) I0+ = e /2R (c) I0+ = 2e /R – What is the value of the current I after a very long time? (a) I = 0 (b) I = e /2R (c) I > 2e /R • The key here is to realize that 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 e ; the current goes to 0. Behavior of Capacitors • Charging – Initially, the capacitor behaves like a wire. – After a long time, the capacitor behaves like an open switch. • Discharging – Initially, the capacitor behaves like a battery. – After a long time, the capacitor behaves like a wire. Discharging Capacitor • The capacitor is initially fully charged, Q = Q0. At t = 0 the switch is thrown from position a to position b in the circuit shown. • From KVL: R dQ dt dQ dt Q 0 C 1 RC I a b Q(t ) e Therefore, e Q dQ ae at aQ dt where From initial condition, Q(0) = Q0, we get: Q(t ) Q0e t RC R + + C Q(t) must be an exponential function of the form: at I 1 a RC - - Summary • Kirchhoff’s Laws – KCL: Junction Rule (Charge is conserved) – Review KVL (V is independent of path) • Non-ideal Batteries & Power • Discharging of capacitor through a Resistor: Q (t ) Q0 e t RC Reading Assignment: Chapter 26.6 Examples: 26.17,18 and 19 Two identical light bulbs are represented by the resistors R2 and R3 (R2 = R3 ). The switch S is initially open. 2) If switch S is closed, what happens to the brightness of the bulb R2? a) It increases b) It decreases c) It doesn’t change 3) What happens to the current I, after the switch is closed ? a) Iafter = 1/2 Ibefore b) Iafter = Ibefore c) Iafter = 2 Ibefore I Four identical resistors are connected to a battery as shown in the figure. R2 R1 R4 E R3 5) How does the current through the battery change after the switch is closed ? a) Iafter > Ibefore b) Iafter = Ibefore c) Iafter < Ibefore Before: Rtot = 3R Ibefore = 1/3 E/R After: R23 = 2R R423 = 2/3 R Rtot = 5/3 R Iafter = 3/5 E/R Appendix: A three-loop KVL example • Identify all circuit nodes these are where KCL eqn’s are found I3 I1 I2 – determine which KCL equations are algebraically independent (not all are in this circuit!) – – – – I1=I2+I3 I4=I2+I3 I4=I5+I6 I1=I5+I6 I1=I4 I1=I2+I3 I4+I5+I6 • Analyze circuit and identify all independent loops where S DVi = 0 < KVL I6 I4 I5 A three-loop KVL example • Here are the node equations from applying Kirchoff’s current law: I1=I4 I1=I2+I3 I4+I5+I6 • Now, for Kirchoff’s voltage law: (first, name the resistors) I6R6+I5R5=0 I2R2b+I2R2a- I3R3 =0 VB-I1R1-I2R2a- I2R2b-I4R4-I5R5 = 0 Six equations, six unknowns…. I3 R1 I1 I2 R3 R2b I6 R6 R5 • There are simpler ways of analyzing this circuit, but this illustrates Kirchoff’s laws R2a I5 R4 I4