PHY2054: Chapter 1834Wheatstone BridgeÎAn ammeter A is connected between points a and b in the circuit below, in which the four resistors are identical. What is the current through the ammeter? a) I / 2 b) I / 4 c) zero d) need more informationThe parallel branches have the sameresistance, so equal currents flow ineach branch. Thus (a) and (b) are atthe same potential and there is nocurrent flow across the ammeter.PHY2054: Chapter 1835Circuit 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) A gets brighter and B gets dimmer (e) nothing changesBefore: Potential at (a) is 24V, but sois potential at (b) because equalresistance divides 48V in half. When theswitch is closed, nothing will changesince (a) and (b) are still at same potential.PHY2054: Chapter 1836Circuit Problem (2)ÎBulbs A and B are identical. What happens when the switch is closed? (a) nothing happens (b) A gets brighter, B dimmer (c) A gets dimmer, B brighter (d) Both A and B get brighter (e) Both A and B get dimmerBefore: Bulb A and bulb B both have 9V across them.After: Bulb A has 12V across it andbulb B has 6V across it (these potentials are forced by the batteries).PHY2054: Chapter 1837Calculate Currents (Not Simplest Way)2132112 3 063 0iiiii=++− =+− =1221342422iii===−=PHY2054: Chapter 1838Res-Monster MazeAll batteries are 4VAll resistors are 4ΩFind current in R. (Hint: Find voltagesalong path not connected by resistors)PHY2054: Chapter 1839Res-Monster MazeAll batteries are 4VAll resistors are 4ΩFind current in R. (Hint: Find voltagesalong path not connected by resistors)2PHY2054: Chapter 1840Res-Monster Maze (Part 2)All batteries are 4VAll resistors are 4ΩFind currents acrossthese resistorsPHY2054: Chapter 1841212011310111011111First find currents in resistors usingpotentials established by batteries000PHY2054: Chapter 18422120113101110111111411238 4677 1010008Then use junction rules to find other currentsPHY2054: Chapter 18432120113101110111111411238 4677919810012111 10 9 99000Continue using junction rules until finished132PHY2054: Chapter 1844Light 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)PHY2054: Chapter 1845ProblemÎWhat 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.PHY2054: Chapter 1846Find Currents Through All Resistors60Fig. 18-8PHY2054: Chapter 1847Find Currents (2)ÎFind total current (need equivalent R) 10.00Ω + 5.00Ω (parallel) Î 3.33Ω 3.33Ω + 4.00Ω (series) Î 7.33Ω 7.33Ω + 3.00Ω (parallel) Î 2.13Ω 2.13Ω + 3.00Ω (series) Î 5.13Ω I = 60 / 5.13 = 11.70 AÎCurrent through outside3.00Ω resistor = 11.70 A.60IPHY2054: Chapter 1848Find Currents (3)ÎTotal current I = 11.70, find current I2One approach, find VA–VB VB= 0, VA= 60 – 3.00*11.70 = 24.90 I2= 24.9 / 3.00 = 8.30 A60I1II2ABCPHY2054: Chapter 1849Find Currents (4)ÎFind I1I1= I – I2= 3.40ÎFind remaining currents Find ΔV = VA-VC VC= 0 + 3.40*4.00 = 13.60 ΔV = 24.90 – 13.60 = 11.30 I10= 11.30 / 10 = 1.13 I5= 11.30 / 5 = 2.2660I1II2ABCPHY2054: Chapter 1850Find Currents (5)ÎFind remaining currents Use I1= 3.40  Find ΔV = VA-VC VC= 0 + 3.40*4.00 = 13.60 ΔV = 24.90 – 13.60 = 11.30 I10= 11.30 / 10 = 1.13 I5= 11.30 / 5 = 2.2660I1II2ABCPHY2054: Chapter 1851VRC 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/iVR=resistor/VVqCRiΔ=−=iPHY2054: Chapter 1852Analysis of RC CircuitsÎCurrent and charge are relatedÎSo can recast previous equation as “differential equation“ÎGeneral solution is (Check and see!) K = −CV (necessary to make q = 0 at t = 0)ÎSolve for charge q and current i/idqdt=dq q Vdt RC R+=/tRCqCVKe−=+()//1tRC tRCdq VqCV e i edt R−−=− ==/VqCRi−=PHY2054: Chapter 1853Charge and Current vs Time(For Initially Uncharged Capacitor)()()/01tRCqt q e−=−()/0tRCit ie−=PHY2054: Chapter 1854RC ExampleÎLet V = 12 volts, C = 6μF, R = 100Ω τ = RC = 600μsÎCharge vs time Using units of μC and μsecÎCurrent vs time Using units of amps and μsecÎThis circuit will fully charge in a few millisec()()/ /6001721tRC tqCV e e−−=− =−/ /6000.12tRC tVie eR−−==PHY2054: Chapter 1855Exponential BehaviorÎt = RC is the “characteristic time” of any RC circuit Only t / RC is meaningfulÎt = RC Current falls to 37% of maximum value Charge rises to 63% of maximum valueÎt =2RC Current falls to 13.5% of maximum value Charge rises to 86.5% of maximum valueÎt =3RC Current falls to 5% of maximum value Charge rises to 95% of maximum valueÎt =5RC Current falls to 0.7% of maximum value Charge rises to 99.3% of maximum valuePHY2054: Chapter 1856Discharging a CapacitorÎConnect fully charged capacitor to a resistor at t = 0ÎSolution isÎSolve for current iR0qiRC−−=0dq qdt RC+=//0tRC tRCqqe CVe−−==//0tRC tRCdq Vieiedt R−−== =CSq, i fall exponentiallyPHY2054: Chapter 1857Charge and Current vs Time(For Initially Charged Capacitor)()/0tRCqt qe−=()/0tRCit ie−=PHY2054: Chapter 1858Same RC ExampleÎC = 6μF, R = 100Ω τ = RC = 600μsÎCharge capacitor to 12 volts, then pull out battery and let the RC circuit dischargeÎCharge vs time Using units of μC and μsecÎCurrent vs time Using units of amps and μsecÎThis circuit will fully discharge in a few millisec/ /60072tRC tqCVe e−−==/