Physics 202, Lecture 10. Basic Circuit Components

Transcription

1 Physics 202, Lecture 10 Tody s Topics DC Circuits (Chpter 26) Circuit components Kirchhoff s Rules RC Circuits Bsic Circuit Components Component del ttery, emf Resistor Relistic Bttery (del) wire Cpcitor nductor (del) Switch Trnsformer Diodes, Trnsistors, Symol Behvior in circuit ΔV=V + -V - =ε ΔV= -R ε r ΔV=0 ( R=0, C=0) ΔV=V - - V + = - q/c We ll see this lter C=0, R=0 (on), R= (off) Future Topics 1

2 emf: Electromotive Force Bttery: source of Electromotive force (emf) Chemicls inside the ttery mintins chrge distriution which provides persistent potentil difference emf. Emf lso produced y chnging mgnetic flux (lter in semester). emf is potentil difference, it is not force! del tteries: emf E= V (voltge provided y ttery) Non-idel tteries: "internl resistnce r V =! " r V ++++ e Resistor R e - Direct Current (DC) Circuit: Circuit driven y ε ~ constnt Circuit Anlysis s We ve seen simple circuits (series/prllel).. Generl cse: more complicted topology my contin more thn one emf Resistor/cpcitor comintions my not e s simple s in series or in prllel Circuits my contin multiple s nd junctions. junctions 2

3 Circuit Anlysis Kirchoff s Lws #1 Conservtion of electric chrge All the chrge tht flows into junction of conductors per unit time, the sme mount must leve in the sme time intervl. #2 Conservtion of energy A complete trip round the circuit (the end point is the sme s the eginning poin must result in zero net energy chnge. Kirchhoff s Rules: Junction Rule Rule #1: Junction Rule The net current entering ny junction equls the net current leving tht junction. Σ in =Σ out 1 = Determined y ssigned for ech current: in : current with ssigned towrds junction out : current with ssigned off junction 3

4 (Very) Quick Quiz: Junction Rule Wht is the junction rule for the current ssignment shown? = 3 1 = = Although eqution 2 nd 3 re equivlent, eqution 3 does not follow templte form in = out Quick Quiz: Junction Rule Wht is the junction rule for the current ssignment shown? = =0 Neither While the ctul currents cn not ll goes into junction, the ssigned currents cn. 4

5 Kirchhoff s Rules: Loop Rule Loop Rule (Energy Conservtion): The sum of potentil drops cross components long ny closed circuit must e zero. "!V = 0 The exct expression of the potentil drop is determined y the type of component nd the ssigned current (see next slides) Determine Potentil Difference(1) Choose for the current, nd move round the in tht. ΔV=V -V = - ε ΔV=V -V = +ε When ttery is trversed from the positive terminl to the negtive terminl, the voltge drops (- sign). When ttery is trversed from the negtive terminl to the positive terminl, the voltge increses (+ sign). 5

6 Determine Potentil Differences (2) Choose for the current, nd move round the in tht. ΔV=V -V = -R ΔV=V -V = +R When moving cross resistor in the of the ssigned current, the voltge drops (- sign). When moving cross resistor n opposite to the ssigned current, the voltge increses (+ sign). Determine Potentil Differences (3) Choose for the current, nd move round the in tht. q q ΔV=V -V = - q/c =dq/dt ΔV=V -V = + q/c =dq/dt When moving cross cpcitor from the positive plte to the negtive plte, the voltge drops (- sign). When moving cross cpcitor from the negtive plte to the positive plte, the voltge increses (+ sign). 6

7 Steps to Apply Kirchhoff s Rules 1. Assign l currents for ech rnch of the circuit. 2. Set up junction rules t certin (ny) junctions. Normlly: # of junctions = # of currents Select numer of closed s to pply rule. For ech, follow clockwise (or counterclockwise), find ΔV cross ech component, pply rule. # of s determined y # of unknowns. 4. Solve for unknowns. (A negtive current indictes its is opposite to the ssigned.) Exmple 1: Resistors n Series Show R eq =R 1 + R 2. Applying Kirchhoff Rules: Junction : 1 = Junction c: 2 = Loop: ΔV - 1 R 1 2 R 2 =0 1 2 Questions: Wht re the voltge drops cross R 1 nd R 2? Wht is the power delivered to R 1 nd R 2? 7

8 Exmple 2: Resistors n Prllel Show 1/R eq =1/R 1 + 1/R 2. Applying Kirchhoff Rules: Junction: = Loop 1: ΔV - 1 R 1 =0 Loop 2: ΔV - 2 R 2 = Questions: Wht re the voltge drops cross R 1 nd R 2? Wht is the power delivered to R 1 nd R 2? Question: find 1, 2, 3 Kirchhoff s Rules: Exmple 3: Multi-Loop ε 2 Junction c: = 3 R 2 Loop 1: ε 1-1 R 1-3 R 3 =0 ε 1 R 1 Loop 2: -ε R 1 ε 1-2 R 2 =0 Solve: 1 = 2.0A, 2 = -3.0A, 3 = -1.0A, (Wht do the signs men?) R 3 8

9 Exmple 3: nterprettion of Results 3.0A 2.0A 1.0A 1 = 2.0A, 2 = -3.0A, 3 = -1.0A, Actul sitution Exmple 3 Agin: Different nitil Directions Different initil s for 1, 2 Apply Kirchhoff s Rules: ε 2 Junction c: 0= Loop 1: ε R 1-3 R 3 =0 Loop 2: -ε 2-1 R 1 -ε R 2 =0 R 2 ε 1 R 1 Solve: 1 = -2.0A, 2 = +3.0A, 3 = -1.0A, R 3 Sme result s previous slide Text exmples: 26.19, 26.36,

10 RC Circuits First exmple: time-dependent currents r Light Bul Chrging A Cpcitor in RC Circuit Current nd chrge when cpcitor is eing chrged in RC circuit: " dq(! q( / C! R = 0 dt! t / RC! t / RC q( = " C(1! e ) ( = " e R Note: τ RC is the time constnt Chrging 10

11 Dischrging A Cpcitor in RC Circuit Current nd chrge when cpcitor is eing dischrged in RC circuit: q( = Qe! t / RC ( =! Q RC e! t / RC dq(! q( / C + R = 0 dt Note the time constnt τ=rc Text: dischrging 11