PHY2049: Chapter 2731CircuitsÎThe light bulbs in the circuits below are identical. Which configuration produces more light? (a) circuit I (b) circuit II (c) both the sameCircuit II has ½ current of each branchof circuit I, so each bulb is ¼ as bright.The total power in circuit I is thus 4x thatof circuit II.PHY2049: Chapter 2732CircuitsÎThe three light bulbs in the circuit are identical. What is the brightness of bulb B compared to bulb A? a) 4 times as much b) twice as much c) the same d) half as much e) 1/4 as muchUse P = I2R. Thus 2x current in Ameans it is 4x brighter.PHY2049: Chapter 2733Circuit Problem (1)ÎThe light bulbs in the circuit are identical. What happens when the switch is closed? a) both bulbs go out b) the intensity of both bulbs increases c) the intensity of both bulbs decreases d) nothing changesBefore switch closed: Va= 12V because ofbattery. Vb=12 because equal resistancedivides 24V in half.After switch closed: Nothing changessince (a) and (b) are still at same potential.(a)(b)PHY2049: Chapter 2734Circuit Problem (2)ÎThe light bulbs in the circuit shown below are identical. When the switch is closed, what happens to the intensity of the light bulbs? a) bulb A increases b) bulb A decreases c) bulb B increases d) bulb B decreases e) nothing changes(b)(a)Before switch closed: Va= 12V because ofbattery. Vb=12 because equal resistancedivides 24V in half.After switch closed: Nothing changessince (a) and (b) are still at same potential.PHY2049: Chapter 2735(a)(b)Circuit Problem (3)ÎThe bulbs A and B have the same R. What happens when the switch is closed? a) nothing happens b) A gets brighter, B dimmer c) B gets brighter, A dimmer d) both go outBefore: Va= 24, Vb= 18. Bulb A and bulb B both have 18V across them.After: Va= 24, Vb= 24 (forced bythe batteries). Bulb A has 12V across it and bulb B has 24V across it.24VPHY2049: Chapter 2736Kirchhoff’s RulesÎJunction rule (conservation of charge) Current into junction = sum of currents out of itÎLoop rule (conservation of energy) Algebraic sum of voltages around a closed loop is 0II2I1I331 22 2 31 20IRIREIRE−−−−+=123IIII=++1211 1 11 2 220IREIREIR−−− ++ =12PHY2049: Chapter 2737Problem Solving Using Kirchhoff’s RulesÎLabel the current in each branch of the circuit Choice of direction is arbitrary Signs will work out in the end (if you are careful!!) Apply the junction rule at each junction Keep track of sign of currents entering and leavingÎApply loop rule to each loop (follow in one direction only) Resistors: if loop direction matches current direction, voltage drop Batteries: if loop direction goes through battery in “normal”direction, voltage gainÎSolve equations simultaneously You need as many equations as you have unknownsPHY2049: Chapter 2738Kirchhoff’s rules ÎDetermine the magnitudes and directions of the currents through the two resistors in the figure below. Take two loops, 1 and 2, as shownUse I1= I2+ I3323615 022 9 15 0III+−=−++ =321236/15 0.4015/22 0.681.08IIIII=====+=1212PHY2049: Chapter 2739CircuitsÎWhich of the equations is valid for the circuit shown below? a) 2 − I1− 2I2= 0 b) 2 − 2I1− 2I2− 4I3= 0 c) 2 − I1− 4 − 2I2= 0 d) I3− 2I2− 4I3= 0 e) 2 − 2I1− 2I2− 4I3= 012V4V6VI1I2I3PHY2049: Chapter 2740Wheatstone BridgeÎAn ammeter A is connected between points a and b in the circuit below. What is the current through the ammeter? a) I / 2 b) I / 4 c) zero d) need more informationBefore ammeter is added:• The top branch divides the voltageevenly, so Va= V/2.• The bottom branch also dividesthe voltage evenly, so Vb= V/2.•Thus Va= Vband current is 0across ammeter.VPHY2049: Chapter 2741Wheatstone BridgeÎAn ammeter A is connected between points a and b in the circuit below. What is the current through the ammeter? a) I / 2 b) I / 4 c) zero d) need more information2R2RSame analysis. Before ammeter is added:• The top branch divides the voltageevenly, so Va= V/2.• The bottom branch also dividesthe voltage evenly, so Vb= V/2.•Thus Va= Vband current is 0across ammeter.PHY2049: Chapter 2742Res-Monster Maze (p. 725)All batteries are 4VAll resistors are 4ΩFind current in RHint: follow the batteriesPHY2049: Chapter 2743Res-Monster Maze (p. 725)All batteries are 4VAll resistors are 4ΩFind current in RHint: follow the batteries0V4V8V12V8V0VIR= 2PHY2049: Chapter 2744Problem solving ÎFind the value of R that maximizes power emitted by R. IT12ΩI1I218VR6Ω121218 12124612TRRRRIRRR=++==+++&222PIR=122111212 1212 4RTRIIIRR==⇒=++ +()22221444RPIRR==+Maximize249WRP=Ω=PHY2049: Chapter 2745Light Bulbs ÎA three-way light bulb contains two filaments that can be connected to the 120 V either individually or in parallel. A three-way light bulb can produce 50 W, 100 W or 150 W, at the usual household voltage of 120 V. What are the resistances of the filaments that can give the three wattages quoted above?Use P = V2/R¾R1= 1202/50 = 288Ω (50W)¾R2= 1202/100 = 144Ω (100W)PHY2049: Chapter 2746ProblemÎWhat is the maximum number of 100 W light bulbs you can connect in parallel in a 100 V circuit without tripping a 20 A circuit breaker? (a) 1 (b) 5 (c) 10 (d) 20 (e) 100Each bulb draws a current of 1A. Thus only 20 bulbs are allowed before the circuit breaker is tripped.PHY2049: Chapter 2747ERC CircuitsÎCharging a capacitor takes time in a real circuit Resistance allows only a certain amount of current to flow Current takes time to charge a capacitorÎAssume uncharged capacitor initially Close switch at t = 0 Initial current is (no charge on capacitor)ÎCurrent flows, charging capacitor Generates capacitor potential of q/CÎCurrent decreases continuously as capacitor charges! Goes to 0 when fully charged/iER=/EqCiR−=PHY2049: Chapter 2748Analysis of RC CircuitsÎCurrent and charge are relatedÎSo can recast previous equation as “differential equation“ÎGeneral solution is (Check and see!) K = −EC (necessary to make q = 0 at t = 0)ÎSolve for charge q and current i/idqdt=dq q Edt RC R+=/tRCqECKe−=+()//1tRC tRCdq EqEC e i edt R−−=− ==/EqCiR−=PHY2049: Chapter 2749Charge and Current vs Time(For Initially Uncharged Capacitor)()()/01tRCqt q e−=−()/0tRCit ie−=PHY2049: Chapter 2750Exponential BehaviorÎt = RC is the
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