* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Lecture08: Multi-Loop and RC Circuits
Wien bridge oscillator wikipedia , lookup
Schmitt trigger wikipedia , lookup
Valve RF amplifier wikipedia , lookup
Josephson voltage standard wikipedia , lookup
Regenerative circuit wikipedia , lookup
Power electronics wikipedia , lookup
Topology (electrical circuits) wikipedia , lookup
Wilson current mirror wikipedia , lookup
Two-port network wikipedia , lookup
Surge protector wikipedia , lookup
Operational amplifier wikipedia , lookup
Opto-isolator wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Power MOSFET wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Current source wikipedia , lookup
RLC circuit wikipedia , lookup
Current mirror wikipedia , lookup
Physics 121 - Electricity and Magnetism Lecture 08 - Multi-Loop and RC Circuits Y&F Chapter 26 Sect. 2 - 5 • • • • • • • • • • • Definitions of Circuit Terms Kirchhoff Rules Problem solving using Kirchhoff’s Rules Multi-Loop Circuit Examples RC Circuits – Charging a Capacitor – Discharging a Capacitor Discharging Solution of the RC Circuit Differential Equation The Time Constant Examples Charging Solution of the RC Circuit Differential Equation Features of the Solution Examples Copyright R. Janow Fall 2015 Definitions of Circuit Terms Basic circuit elements have two terminals. Generic symbol: ECE 2 1 Node: A point where 2 or more circuit elements are joined. i1 Essential Node (Junction): A node where at least 3 circuit elements are joined. i2 Path: A route (or trace) through adjacent basic elements with no element included more than once. May pass through essential and/or non-essential nodes. Branch: A path that connects two nodes (essential or not). Includes 1 or more elements Essential Branch: • A path that connects two essential nodes without passing through another essential node. • A circuit section between two essential nodes with 1 or more elements in series. • Non-essential (trivial) nodes connect circuit elements in “series”. • There is exactly one current per essential node. Loop: A closed path whose last node is the same as the starting node. Mesh: A loop that does not enclose any other loops. Planar Circuit: A circuit whose diagram can be drawn on a plane with no crossing branches. Copyright R. Janow Fall 2015 i Example: Circuit Terms b a + v1 + f i1 R1 - c v2 Source: Nilsson & Reidel Electric Circuits i2 d R2 R4 i3 i5 R3 i4 R5 i7 e R7 i6 R6 g Ideal, Independent Current Sources: i6 Ideal, Independent Voltage Sources (ideal EMFs) : v1, v2 Nodes: a, b, c, d, e, f, g Essential Nodes (Junctions): b, c, e, g Branches: v1, v2, R1, R2, R3, R4, R5, R6, R7, i6 Paths: Many, traversing 1, 2, …..n branches Essential Branches (7): v1-R1, R2-R3, v2-R4, R5, R6, R7, i6 Currents: 1 per essential branch, total of 7: i1 . . . i7 Meshes (subset of loops): v1-R1-R5-R3-R2, v2-R2-R3-R6-R4, R5-R7-R6, R7-i6 Loops: Meshes + V1-R1-R5-R6-R4-V2, V1-R1-R7-R4-V2, v1-R1-i R6Janow -R5-i6 Fall 2015 6-R4-V2 ,R. Copyright Solving Circuit Problems using Kirchhoff’s Rules i CIRCUITS CONSIST OF: JUNCTIONS (ESSENTIAL NODES) ... and … ESSENTIAL BRANCHES (elements connected in series, one current/branch) 1 i i2 i The current through all series elements in an essential branch is the same; i.e., the number of currents = the number of essential branches Often, all element capacities are specified but the currents are unknowns. Analysis strategy: 1) Use Kirchhoff Rules to generate N independent equations in N unknowns • Junction Rule or Current rule or Node Law (charge iin iout conservation): At any junction the algebraic sum of the currents at any (essential) node equals zero • Loop Rule (energy conservation): The algebraic sum V 0 of all potential changes is zero for every closed path around a circuit. Corollary: voltage difference is the same for all paths connecting a pair of nodes. 2) Solve the resulting set of simultaneous equations (you need a strategy). • Linear, algebraic for resistances and EMFs only Copyright R. Janow Fall 2015 Applying Kirchhoff’s Rules: We use “active” sign convention: voltage drops considered negative “passive” sign convention: voltage drops considered positive For brevity: “branch” means “essential branch “junction” means “essential node” Procedure for Generating Circuit Equations • • • • • • • • • Name the currents or other unknowns In each branch, arbitrarily assign direction to current - a negative result opposite flow Apply Junction Rule. Apply Loop Rule to write down terms in equations: - choose direction for traversing a closed loop. possibly traverse branches with or against assumed current directions. When crossing resistances: - voltage drop (V = - iR) negative when following assumed current - positive voltage change V = +iR for crossing opposite to assumed current - when crossing EMFs from – to +, V = +E. Otherwise V= -E Keep generating equations until you have N independent ones After solving, calculate power or other quantities as needed Dot product i.E determines whether EMFs supply or dissipate power For Later: when crossing C with current write –VC= -Q/C. Copyright R. Janow Fall 2015 when crossing inductance write VL=-Ldi/dt Example: Equivalent resistance for resistors in series Junction Rule: The current through all of the resistances in series (a single branch) is identical. No information from Junction/Current/Node Rule i i1 i2 i3 Loop Rule: The sum of the potential differences around a closed loop equals zero. Only one loop path exists: iR 1 iR 2 iR 3 0 i R1 R 2 R 3 The equivalent circuit replaces the series resistors with a single equivalent resistance: same E, same i as above iR eq 0 i R eq CONCLUSION: The equivalent resistance for a series combination is the sum of the individual resistances and is always greater than any one of them. R eq R 1 R 2 R 3 R eq n Ri i 1 Copyright R. Janow Fall 2015 inverse of series capacitance rule Example: Equivalent resistance for resistors in parallel Loop Rule: The potential differences across each of the 4 parallel branches are the same. Four unknown currents. Apply loop rule to 3 paths. E i1R1 0 i1 E i2R 2 0 E E E , i2 , i3 R1 R2 R3 E i3R 3 0 i not in these equations Junction Rule: The sum of the currents flowing in equals the sum of the currents flowing out. Combine equations for all the upper junctions at “a” (same at “b”). 1 1 1 i i1 i2 i3 E R R R 2 3 1 The equivalent circuit replaces the series resistors with a single equivalent resistance: same E, same i as above. i iR eq 0 R eq CONCLUSION: The reciprocal of the equivalent resistance for a parallel combination is the sum of the individual reciprocal resistances and is always smaller than any one of them. 1 1 1 1 R eq R1 R 2 R 3 1 R eq n 1 R i 1 i inverse of parallel capacitance rule R eq R 1R 2 R1 R 2 Copyright R. Janow Fall 2015 EXAMPLE: MULTIPLE BATTERIES SINGLE LOOP i + - R1= 10 W E1 = 8 V + - i E2 = 3 V + R2= 15 W A battery (EMF) absorbs power (charges up) when I is opposite to E Pemf E is opposite to Vdrop Ei E i -3.0x0.2 Copyright R. Janow Fall 2015 Example: Multi-loop circuit with 2 EMFs + R1 i1 E2 A - E1 - D R3 i3 + Given all resistances and EMFs in circuit: • Find currents (i1, i2, i3), then potential drops and power dissipated by resistors • 3 unknowns (currents) imply 3 independent equations needed E R2 i2 Apply Procedure: C B • Identify essential branches (3) & junctions (2). • Name all currents (3) and other variables. • Same current flows through all elements in any series branch. • Assume arbitrary current directions; negative result means opposite direction. • At junctions, write Current Rule (Junction Rule) equations. iin iout • Same equation at junctions A and B (not independent). • Junction Rule yields only 1 of 3 equations needed • Are points C, D, E, F junctions? (not essential nodes) i2 i1 i3 (1) Copyright R. Janow Fall 2015 F Procedure, continued: V E1 - + equations that include all the unknowns (3). • Traversal direction is arbitrary. • When following the assumed current direction voltage change = - iR. When going against assumed current voltage steps up by +IR • EMF’s count positive when traversed from – to + side • EMF’s count negative when traversed from + to - sides D R1 i1 A E2 R3 i3 C + - • Apply Loop Rule as often as needed to find B 0 Loop equations for the example circuit: ADCBA - CCW ADCBFEA - CCW ABFEA - CCW E1 i1 R1 i3R 3 0 E1 i1 R1 i2R2 E2 0 i3 R 3 i2R 2 E2 0 Solution: (after a lot of algebra) Define: R1R 2 R 2R 3 R1R 3 • Only 2 of these three are independent • Now have 3 equations in 3 unknowns i1 E1R 2 E1R 3 E2R 3 i2 E1R 3 E2R 3 E2R1 E2R1 E1R 2 i3Copyright R. Janow Fall 2015 E R2 i2 F Example: find currents, voltages, power 6 BRANCHES 6 CURRENTS. •JUNCTION RULE: Junctions C & E are the same point, as are D & F -> 4 currents left. Remaining 2 junction equations are dependent -> 1 junction equation i i1 i2 i3 LOOP RULE: ABCDA - CW CEFDC - CW E i1 R1 0 E i1 R1 i1 E/R1 12 / 3 4.0 A. i2R2 i1R1 0 i2 i1R1/R2 4x3 / 8 1.5 A. i3R 3 i2R 2 0 i3 i2R 2/R 3 1.5x8 / 6 2.0 A. R eq 1.6 W CHECK: i i1 i2 i3 4.0 1.5 2.0 7.5 A. E should VR1 i1R1 4.0x3.0 12.0 Volts EGHFE - CW POWER: PR1 i12R1 48.0 Watts PR2 i22R 2 18.0 Watts PR3 i23R 3 24.0 Watts PE E i 90.0 Watts PR1 PR2 PR3 Copyright R. Janow Fall 2015 Multiple EMF Example: find currents, voltages, power R2 = 4 W E2 = 6 V R1 = 2 W E1 = 3 V MULTIPLE EMF CIRCUIT JUNCTION RULE at A & B: USE THE SAME RULES i3 i1 i2 LOOP ACDBA: i1R1 E1 i1R1 E2 i2R 2 0 i2 i1 3/4 LOOP BFEAB: i3R1 E2 i3R1 i2R 2 E2 0 USE JUNCTION EQUATION: i3 i2 i1 i2 EVALUATE i 1/2, i2 - 1/4, NUMERICALLY: 1 i3 i2 i1 2 i2 i3 1/4 For power use: V iiRi Pi ii2Ri Copyright R. Janow Fall 2015 RC Circuits: Time dependance 27.8 i a Can constant current flow through R + + E - b C - i a capacitor indefinitely? Vc • Given Capacitance + Resistance + EMF • Use Loop Rule + Junction Rule • New term Vc= Q/C in Loop Rule • Find Q, i, Vc, U for capacitor as functions of time First charge up C (switch to “a”) then discharge (switch to “b”) Charging: “Step Response” Switch to “a” then watch. Loop equation: E iR Vc 0 Discharging: Switch to “b”. no EMF, Loop equation: iR Vc 0 • Assume clockwise current i through R • Vcap E as t infinity, Q Qinf = CE • Expect zero current as t infinity • Expect largest current at t = 0, • Energy is stored in C, some is dissipated in R • Energy stored in C is now dissipated in R • Arbitrarily assume current is still CW • Vcap= E at t =0, but it must die away • Q0= full charge = CE • Result: i through R is actually CCW Copyright R. Janow Fall 2015 RC Circuit: solution for discharging Loop Equation is : Substitute : iR Vc 0 dQ i(t ) dt Circuit Equation: dQ Q(t ) dt RC Q( t ) V c (t ) C First order differential equation, form is Q’ = -kQ Exponential solution Charge decays exponentially: • t/RC is dimensionless Q(t ) Q 0e t / RC RC = t = the TIME CONSTANT Q falls to 1/e of original value Voltage across C also decays exponentially: Current also decays exponentially: E Q0 / C Vc (t) Q(t) / C dQ i(t ) i0e t / RC dt Q0 e 1 0.37 Q t 2t 3t t Vc (t) E e t / RC E Q0 i0 R RC Copyright R. Janow Fall 2015 Solving for discharging phase by direct integration dQ Q( t ) dt RC dQ dt Q RC RC is constant Initial conditions (“boundary conditions”) At t 0 : Q(t) Q0 CE dQ' 1 t dt' Q0 Q' 0 RC exponentiate both sides of above right ln ( e Q ) Q0 Q t ln ( ) Q0 RC Q Q e Q0 t RC eln (x) x Q( t ) Q 0 e t / RC exponential decay RC = time constant = time for Q to fall to 1/e of its initial value RC t e 1 1 1 .37 e 2.71828 Time t 2t 3t Value e-1 e-2 e-3 % left 36.8 13.5 5.0 4t 5t e-4 e-5 1.8 0.67 After 3-5 time constants the action is over Copyright R. Janow Fall 2015 Units for RC 8-1: We defined t = RC, which of the choices best captures the physical units for the decay constant t ? [t] = [RC] =[(V/i)(Q/V)]=[Q/Q/t]=[t] A. B. C. D. E. WF C/A WC/V VF/A s (ohmfarad) (coulomb per ampere) (ohmcoulomb per volt) (voltfarad per ampere) (second) Copyright R. Janow Fall 2015 Examples: discharging capacitor C through resistor R a) When has the charge fallen to half of it’s initial value Q0? 1 e t / t 2 1 2 set: Q(t ) Q 0 Q 0e t / t take log: ln( 12 ) t / t ln(1) 0 ln( 2) - t/t (solve for t - depends only on t) ln(a/b) ln(a) - ln(b) ln(2) 0.69 t 0.69 t b) When has the stored energy fallen to half of its original value? Q2 recall: U(t ) 2C and Q(t ) Q 0e t / RC at any time t: U(t) U0 e 2t / RC evaluate for: U(t) U0 U0 e 2 t / t 2 take log: ln( 1 ) 2t / t 2 where Q20 U0 U(t 0) 2C t 0.69 t /2 0.35 t c) How does the power delivered by C vary with time? power: P dU U0 d [ e 2t / t ] U0 [ 2 ]e 2t / t 2 Q 0 Q 0 e 2 t / t dt dt t 2 C RC Q0 C supplies rather than absorbs power recall: Q 0 E i0 Drop minus sign C RC power supplied by C: P i0e t / t Ee t / t i(t) V(t) P0e 2 t / t P(t ) i(t) V(t) Copyright R. Janow Fall 2015 RC Circuit: solution for charging Loop Equation is : Substitute : E - iR Vc 0 dQ i(t ) dt Circuit Equation: dQ Q( t ) E dt RC R Q( t ) V c (t ) C • First order differential equation again: form is Q’ = - kQ + constant • Same equation as for discharge, but with i0 = E / R added on right side • At t = 0: Q = 0 & i = i0. Large current flows (C acts like a plain wire) • As t infinity: Current 0 (C acts like a broken wire) Q Qinf = CE = limiting charge Solution: Charge starts from zero, grows as a saturating exponential. Q(t) Qinf 1 e t / RC Qinf Q • RC = t = TIME CONSTANT describes time dependance again • Q(t) 0 as t 0 • Q(t) Qinf as t infinity 1 e 1 0.63 t 2t t Copyright R. Janow Fall 2015 3t RC Circuit: solution for charging, continued Voltage across C while charging: Vc (t) Q(t) / C and Vc (t ) E (1 - e t / RC ) E Qinf / C Voltage across C also starts from zero and saturates exponentially Current in the charging circuit: dQ(t ) d i(t ) Qinf 1 e t / RC dt dt 1 t / RC Qinf e RC i(t ) i0 e t / RC i0 E R Qinf RC Current decays exponentially just as in discharging case Growing potential Vc on C blocks current completely at t = infinity At t=0 C acts like a wire. At t=infinity C acts like a broken wire Voltage drop VR across the resistor: VR (t) E e t / RC VR (t ) i(t )R i0 Re t / RC Voltage across R decays exponentially, reaches 0 as t infinity Form factor: 1 – exp( - t / t ) Factor .63 .865 .95 .982 .993 Time t 2t 3t 4t 5t After 3-5 time constants the Copyright R. Janow Fall 2015 6t action is over .998 RC circuit – multiple resistors at t = 0 8-2: Consider the circuit shown, The battery has no internal resistance. The capacitor has zero charge. Just after the switch is closed, what is the current through the battery? A. 0. C B. /2R. C. 2/R. D. /R. E. impossible to determine R R Copyright R. Janow Fall 2015 RC circuit – multiple resistors at t = infinity 8-3: Consider the circuit shown. The battery has no internal resistance. After the switch has been closed for a very long time, what is the current through the battery? A. 0. B. /2R. C C. 2/R. D. /R. R R E. impossible to determine Copyright R. Janow Fall 2015 Discharging Example: A 2 mF capacitor is charged and then connected in series with a resistance R. The original potential across it drops to ¼ of it’s starting value in 2 seconds. What is the value of the resistance? Use: Vc (t ) V0 e t / RC Vc (t ) 1 e t / RC V0 4 Set: Take natural log of both sides, t = 2 s: 2 ln(1) ln(4) ln[ e 2 / RC ] RC ln(4) 1.39 1.39 RC 2 ln(1) 0 R ln[ ex ] x 2 1 1.39 2x10 6 R 0.72 MW Define: 1 MW = 106 W Copyright R. Janow Fall 2015 Example: Discharging C = 500 mF R = 10 KW E = 12 V Capacitor C is charged for a long time to E, then discharged. E a) Find current at t = 0 i(t ) dQ i0e t / RC dt i0 E R Q0 RC i(t 0) E 0 12 e 1.2 mA 4 R 10 b) When does VCap (voltage on C) reach 1 Volt? Vcap (t ) E e t / RC Vcap 1 et / 5 E 12 RC 104 x 5 x 102 x 106 5 sec ln(12) t / 5 V0 E 12 Volts t 5 ln(12) 12.4 sec c) Find the current in the resistor at that time dQ i(t ) i0e t / RC dt i(t 12.4 sec) 1.2 mA x e12.4 / 5 i(t 12.4 sec) 0.1 mA Copyright R. Janow Fall 2015 Charging Example: How many time constants does it take for an initially uncharged capacitor in an RC circuit to become 99% charged? Use: Require: Q(t) Q 1 e t / t Q( t ) 0.99 1 e t / t Q Take natural log of both sides: ln (0.01) - 4.61 - t/t t RC time constant 0.01 e t / t t/t 4.61 # of time constants Did not need specific values of RC Copyright R. Janow Fall 2015 Example: Charging a 100 mF R capacitor in series with a 10,000 W resistor, using EMF E = 5 V. E C a) How long after voltage is applied does Vcap(t) reach 4 volts? Vc (t ) E (1 - e t / RC ) Vc (t ) RC 104 x 100 x 10-6 1.0 sec 4 0.8 1 - e t / RC E 5 Take natural log of both sides: ln( 0.2 ) 1.61 ln [e - t/RC ] e t / RC 0.2 t t RC 1.0 t 1.61 sec b) What’s the current through R at t = 2 sec? i(t ) i0 e t / RC i0 i(t 2) i0e 2.0 / 1.0 E R E 2.0 / 1.0 5 e 4 (0.37)2 6.77 x 10- 5 R 10 i(t 2) 6.8 mA. Copyright R. Janow Fall 2015 Example: Multiple loops and EMFs • Open switch S for a long time. • Capacitor C charges to potential of battery 2 • Then close S for a long time What is the CHANGE in charge on C? First: E2 charges C to have: Vc E2 3 volts with current i1 0 Q 0 final charge for first phase CE2 3.0 x 10-5 Q0 inital charge for second phase 30 mC Second: Close switch for a long time At equilibrium, current i3 though capacitor zero Find outer loop current i = i1 = 12 using loop rule E2 iR 2 iR 1 E1 0 3 i(0.4 0.2) 1 0 i 2.0/0.6 3.33 A. Now find Voltage across C, same as voltage across right hand branch Vb Va E2 iR 2 3 - 3.33x0.4 1.67 V Final charge on C: Q final C (Vb Va ) 10x10-6 x 1.67 Q final 16.7 mC Copyright Q final Q0 - 13.3 mC R. Janow Fall 2015