Lecture 15Chapter 28CircuitsReview• emf device label terminal at higher V as + and lower V as –•Draw emf, Е, arrow from – to + terminal• + charge carriers move against E field in emfdevice from lower (-) to higher (+) VReview• Kirchhoff’s loop rule – in traversing a circuit loop the sum of the changes in V is zero, ∆V =0• Resistance rule – Move through resistor in direction of current V =-iR, in opposite direction V =+iR•Emfrule– Move through emf device in direction of emf arrow V =+E, in opposite direction V =-EReview•Put real battery in circuit • Using Kirchhoff’s loop rule and starting at point a gives• For ideal battery, r = 0 and we get same as before 0=−− iRirE)( Rri +=EiR=ECircuits (25)• Arbitrarily label currents, using different subscript for each branch• Using conservation of charge can write• At point d • At point b231iii =+231iii =+• At point a• At point c11ii =22ii =outinii =Circuits (26)• Kirchhoff’s junction rule (or current law) –– From conservation of charge– Sum of currents entering a junction is equal to sum of currents leaving that junction • Kirchhoff’s loop rule (or voltage law) –– From conservation of energy – Sum of changes in potential going around a complete circuit loop equals zeroCircuits (27)• Resistors in series• Have identical currents, i, through them• Use Kirchhoff’s loop rule 0321=−−− iRiRiRE321RRREi++=Circuits (28)• Resistors in series• Replace 3 resistors with equivalent resistor Req321RRREi++=321RRRReq++=Circuits (29)• Resistors in series –• Resistors have identical currents, i• Sum of V s across resistors = applied V• Reqis sum of all resistors∑==njjeqRR1Circuits (30)• Checkpoint #2 – If R1>R2>R3, rank greatest first• A) current through resistorsi is same for all, tie• B) V across themR1, R2, R3iRV =Circuits (31)• Resistors in parallel• Have same V across them• Arbitrarily choose direction for currents in each branch• Write down current relation for each resistor11RVi =22RVi =33RVi =Circuits (32)• Resistors in parallel • Apply Kirchhoff’s junction rule at point a• Substitute current values321iiii ++=++=321111RRRViCircuits (33)• Resistors in parallel • Replace 3 resistors with equivalent resistor, Req++=3211111RRRReqCircuits (34)• Resistors •Series• Parallel• Capacitors•Series• Parallel∑==njjeqRR1∑==njjeqCC1∑==njjeqRR111∑==njjeqCC111Circuits (35)• Checkpoint #4 – Battery with potential V supplies current i to 2 identical resistors• What is V across and i through either of the resistors if they are connected in• A) Series – What is constant?i is same, V1= V /2 • B) Parallel – What is constant?V is same, i1= i /2iRV=Circuits (36)• What is i of the circuit?• Use Kirchhoff’s loop rule• Clockwise from point a gives• Counterclockwise from point a gives01122=+−−−− EE iriRir02211=++++− EE iriRirCircuits (37)• Solve for i2121rrR ++−=EEi02211=++++− EE iriRirV4.41=EV1.22=EΩ= 3.21r Ω= 8.12rΩ= 5.5RmAA 2402396.0 ≈=iCircuits (38)• What is V across battery 1’s teminals?• Sum potential differences from point b to point a– Clockwise– CounterclockwiseabVirV =+−11EabViriRV =+++22E()22E++=− rRiVVba11irVVba−=− EV.VVba843=−Circuits (39)• Checkpoint #3 – A real battery has Е =12V and r = 2Ω. Is the V across the terminals greater than, less than or equal to 12V if the current in the battery is • A) from – to + terminal• LESS THAN• B) from + to -• GREATER THAN•C) i = 0• EQUAL TO 12VbaVirV =−+ EibaVirV =++
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