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# week 8 222 Phys 222

U of L

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This 4 page Class Notes was uploaded by Mary-elizabeth Notetaker on Sunday October 16, 2016. The Class Notes belongs to Phys 222 at University of Louisville taught by Dr. Ming Yu in Fall 2016. Since its upload, it has received 3 views. For similar materials see Fundamentals of Physics II in Physics at University of Louisville.

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Date Created: 10/16/16

Week 8 Tuesday, October 11, 2016 10:52 AM Superconducting magnets-: ○ Meissner effect- small permanent mag floats freely above N cooled ceramic superconductor bc has 0 resistance & expels any mag field from inside super creating mirror image of mag pole Classify superconductors: ○ Conventional- explained by BCS theory BCS- e- form cooper pairs to move around… phonon from vibration(causes pair formation), attract instead of repel here.. e- move easily w/out collisions ○ Unconventional- if not ○ Critical temp: High T… Tc>77K □ Cool w liquid N Low T… if need other techniques to be cooled ○ Resp to mag field Type 1: single critical field, all superconductivity lost above it Type 2: 2 critical fields, btwn them allows penetration of mag field Direct-current electric circuits- control flow of electricity and E assoc w it ○ Current varies in time if circuit has source, resistor, and capacitors Source-produces power for circuit ○ Combo of resistors: Kirchoff's Rules- simplifyanalysis of complex circuits ○ RC circuits (has resistor and capacitor) Emf- electromotive force- produces pot diff across 2 open circuit terminals ○ Does external work expended per unit of charge ○ Drives current flow in closed circuit = E in notes from now on ○ Ex) battery, generator ○ Measured in volts ○ Real battery- has internal resistance(r) If neglected, Vab=E ○ Terminal voltage(Vab)- pot diff btwn a & b (Does not equal E) Vab= E-Ir=IR ○ Load resistance(R) ○ Relations: For closed circuit: E=IR + Ir Current depends on resistance external to battery and internal □ --> I= E/(R+r) □ If not current(open circuit): V=E-Ir=E …E=Vab when I=0.."open current voltage" R>>>r, r can be ignored □ E=IR+Ir --->E=IR Power relationship: □ IE=I R+I r □ When R>>r, 2ower delivered to battery is transferred to load resistor IE=>I R Resistors in series- when 2+ resistors connected end-to-head ○ Features in circuit: I1 & I2? Same in all resistors bc any charge flowing thru 1 flows through the other □ Charge conserved □ Ohm's law for each resistor: I1= V1/R1 & I2= V2/R2 V1 DNE V2 ---> ( V1/R1)=( V2/R2) V1/ V2= R1/R2 □ Larger R= larger V & larger P □ Sum of pot diff across resistors(Vac) Vac= V= V1+ V2 Req? Ohm;s law for each resistor: ◊ V1=IR1 & V2=IR2 ◊ V= V1+ V2=I(R1+R2)=Ireq □ Conservation of energy □ Req=R1+R2 □ Eq resistance in series has effect on circuit is sum of resistors Always greater than any one of the indiv resistors If one element in series fails, whole circuit not complete so none work □ Ohm's law for eq resistors: I= Vac/Req □ Ex) 4 resistors in series w eq resistancer Phys 222 Page 1 I= Vac/Req □ Ex) 4 resistors in series w eq resistancer □ Ex) Open switch on circuit, ammeter reading shows decrease bc current will then go thru resistor above switch so circuit still closed. Req>R1 (I= V/R) □ Ex) 2 resistors in circuit…closed->Pc delivered to R1. open, what happens with Po delivered to R1? P=I R1 Po<Pc Resistors in parallel ○ Pot diff across each resistor is the same Larger Ri has smaller Ii and smaller power Pi (=Ii V) ○ Short circuit- more charge travels thru path w less resistance\ ○ Req? Conservation of charge: current I that enters point must be equal to the total current leaving that point □ I=I1+I2 □ Currents generally not the same Equivalent resistance for the 2 parallel resistors □ Ohm's law: I= V/Req □ Eq resistance replaces the 2 original resistances ○ Inverse of the equivalent resistance in parallel is the algebraic sum of the inverses of the indiv resistance ○ Eq is always less than the smallest resistor in the group Req<Rmin) ○ Advantages: Running several devices using the full V of the supply □ Varying the current to the need of each device □ If one fails, other still run □ If one shorts, others get no V so no overload damage ○ Household circuits are in parallel Circuit breakers used in series w other circuit elements for saftey purposes ○ Ex 1) 3 resistors in parallel, calc eq resistance for this system: Total current: Power doesn’t change based on switch being open/closed Complicated circuit- several resistors and batteries can often be reduced to simplecircuit w only one resistor Eq resistance btwn a & c? Currents on each resistor? V on each "? ○ Strategy 1 to solve: Combine all resistors in series ..carry same current (I1=I2) Pot diffs across them not same ( V1 DNE V2) Phys 222 Page 2 ○ Strategy 1 to solve: Combine all resistors in series ..carry same current (I1=I2) Pot diffs across them not same ( V1 DNE V2) Eq resitance of series (Req=R1+R2) Draw simplified circuit diagram ○ Strategy 2: Combine all in parallel Pot diff across is same Currents thru them are not same Eq resistance of parallel combo: (1/Req)=(1/R1)+(1/R2) □ Must invert answer after summing reciprocals Draw simplified circuit diagram ○ Strategy 3: Repeat first 2 steps as necessary Replace any resistors in series or In parallel using steps 1 or 2 Sketch new circuit after changes have been made Continue to replace any series or parallel combos Continue until one eq resistance is found ○ Strategy 4: Use ohms law… V=Ireq to det current in eq resistor Start w final circuit found in step 3 and gradually work back thru circuits Applying useful facts from steps 1 and 2 to find the voltage and currents in other resistors □ Each device: V=(Ii)(Ri) □ Series: I1=I2 □ Parallel: V1= V2 Ex2) 4 resistors are connected ○ Eq resistance btwn a & c, Rac> ○ Current ihn each resistor if a 42V battery is connected btwn a & c ( I I1 and I2) V in each resistor if a 42 V battery is connected btwn a & c? ○ Then combine resulting resistors into series Rac=12+2=14 Final eq resistance: Req=Rac=14 V drop in parallel (Vbc) Vbc=(Ibc)(Rbc) Rab and Rbc are in series Iab=Ibc=I To find current in each resistor of parallel circuit: I1= Vbc/R1= 6/6=1 I2= Vbc/R2=6/3=2 Apply ohms law to find V in each resistor of series circuit: V3=IR3= 3 x 8=24 V4=IR4= 3 x 4=12 Applying physics(lightbulb combos) ○ Compare brightness of the 4 identical light bulbs (Ra=Rb=Rc=Rd) A & B in series: Ia=Ib, Va=IaRa, Vb=IbRb ○ Eq resitance is [arallel to C: Vc= Vab ○ Vc>Va, Vc>Vb, Pc>Pa=Pb Phys 222 Page 3 ○ Compare brightness of the 4 identical light bulbs (Ra=Rb=Rc=Rd) A & B in series: Ia=Ib, Va=IaRa, Vb=IbRb ○ Eq resitance is [arallel to C: Vc= Vab ○ Vc>Va, Vc>Vb, Pc>Pa=Pb ○ D has wire c connected acrossd it(short circuit)… if Vd=0, D deoes not glow ○ So… A fails, b goes out, c stays lit C fails, no affect on a & b D fails, no effect on any ○ 3 way light bulbs: Bulb has 2 filaments and can provide 3 lvls of lightintensity Filaments in parallel bc if was in series and one failed, there would be no current in bulb and it wouldn’t glow no matter what position the switch was in □ In parallel, bulb still works if one fails ○ Ex) 75 fil offers 1 R value 100W offers 2nd R value 3rd R is obtained by combining the 2 filaments in parallel □ S1 closed and S2 open, only 75W filament carries current □ S1 open, and s2 closed, only 100W " " " □ Both closed, both carry current and total illumination =175W How to analyze complex dc circuit? ○ Resistors connected so that the circuits formed cannot be reduced to a single eq resistor ○ Current and pot diff in each resistor? Kirchoffs rules analyze I and pot diff in each Junction rule- sum of currents entering any junction must equal the sum of the currents leaving that junction □ Iin=Iout (conservation of charge) □ Current I1 entering the junction must equal the sum of the currents I2 and I3 leaving the junction □ Net volume flow rate in must eq net volume and flow rate out □ Number times junction rule can be used is one fewer than the number of junction points in the circuit Phys 222 Page 4

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