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Exam 1 Result Physics 202, Lecture 10 Today’s Topics Phy202 Exam1 Scores Direct Current Circuits (Ch. 28) 60 Basic circuit components( ε, R, …) 50 Kirchhoff’s Rules 40 Circuits Analysis (For circuits of R’s and ε’s) Frequen ncy Median: 69 A: [83-97] AB,B: [69-82] BC,C: [45-68] D: [<45] 30 20 Preview Requirements: emf, junction rule, loop rule, …. 10 0 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 Bin Next Tuesday: RC circuit (ch. 28.4), Magnetic Field(ch. 29.1-3) Basic Circuit Components Component Symbol Behavior in circuit ΔV=V+-V- =ε Ideal battery, emf Resistor Realistic Battery (Ideal) wire Capacitor Æ ΔV= -IR ε r ΔV=0 (ÆR=0, L=0, C=0) ΔV=V- - V+ = - q/C, dq/dt =I Inductor ΔV= - LdI/dt (Ideal) Switch L=0, C=0, R=0 (on), R=∞ (off) Note: Letter boundaries are for reference only. ( Based on nominal A=15%, B=50% etc. and subject to final curve at the end of semester) Simple Circuit 1: Resistors In Series Exercise: show Req=R1+ R2. I1=I2=I ΔV =VR1+VR2 = IR1+IR2 Æ ΔV =I(R1+R2) i Req=R i.e. R1+R2 R1 R2 I1 I2 I R1 +R2 Transformer Diodes, Transistors,… Future Topics In general: Req=R1+ R2+ R3+ …. 1 Quiz/Exercise: Equivalent Resistance of a Combined Parallel and Serial Circuit Simple Circuit 2: Resistors In Parallel VR1=I1R1 Show 1/Req=1/R1+ 1/R2. VR1=VR2 =ΔV I1 + I2= I ÆI1R1 /R1 + I2R2 /R2=I ÆΔV (1/R1+1/R2)=I ÆΔV=I 1/ ((1/R1+1/R2) i.e. What is the Req for the combination shown? R1= R2=1Ω, R3=2Ω, R4=4Ω. I1 VR2=I2R2 I 2 I Req R1 1. 2. 3 Î 3. 4. 8Ω 6Ω 5Ω None of above R1 R2 R4 R3 1/(1/R1 +1/R2) R2 In general: 1/Req=1/R1+ 1/R2+ 1/R3+ …. A Complicated Circuit loops Kirchhoff’s Rules: Junction Rule Junction Rule (Charge conservation): I1=I2+I3 The sum of currents entering any junction equals the sum of currents leaving that junction. A complicated circuit : May contain more than one emf May not be simplified as “in series” or “in parallel” May contain multi loops and junctions. ΣIin =ΣIout In practice, the classifications of “in” and “out” are determined by assigned direction for each current. The assignment of current directions can be arbitrary. They may not be the same as actual directions, which are not known a prior. “in” : current with assigned direction towards junction “out” : current with assigned direction off junction in in junctions out 2 (Very) Quick Quiz: Junction Rule Quick Quiz: Junction Rule What is the junction rule for the current assignment shown? 1. I1+I2=I3 I1 2. I1=I2+I3 Í I2 I3 3. I1-I2=I3 What is the junction rule for the current assignment shown? 1. I1+I2=I3 I1 2. I1+I2+I3 =0 Í I2 I3 3. Neither Although equation 2 and 3 are equivalent, equation 3 does not follow template form Iin=Iout While the actual currents can not all goes into a junction, the assigned currents can. Kirchhoff’s Rules: Loop Rule Loop Rule (Energy Conservation): The sum of potential drops across components along any closed circuit loop must be zero. Σ ΔV =0 ¾ The potential “drop” across a component is always defined as Vdown_stream_end –V up_sream_end where, the stream direction is the same as loop direction ¾ The exact expression of the potential drop is determined by the type of component and the assigned current direction. (See next slides) Determine Potential Difference b a b a loop direction loop direction ΔV=Vb-Va= - ε ΔV=Vb-Va= +ε I I b a loop direction ΔV=Vb-Va= -IR b a loop direction ΔV=Vb-Va= +IR 3 Steps to Apply Kirchhoff’s Rules Example 1: Multi-Loop 1. Assign a directional current for each branch (segment) of a circuit. The assigned direction for each current can be arbitrarily chosen but, once assigned, need to be observed. 2. Set up junction rules at certain (any) junctions. Normally, # of junctions = # of branches -1. 3 Select a number of closed loops to apply loop rule. 3. rule For each closed loop, assign a loop direction (clockwise or counter clockwise). Follow that assigned direction, find ΔV drop across each component, and apply loop rule. # of loops determined by # of unknowns. 4. Solve for unknowns. 5. If a current is found to be negative, it means its actual direction is opposite to the originally chosen one. The magnitude is always correct. 3.0A 2.0A 1.0A Actual situation ε2 ε1 R1 R3 Example 1 Again: Different Initial Directions Example 1: Interpretation of Results I1= 2.0A, I2= -3.0A, I3= -1.0A, (Text example 28.9) Find out I1, I2, I3 Kirchhoff’s Rules: Junction c: R2 I1+I2=I3 L Loop abcda: b d ε1 - I1R1 - I3R3 =0 Loop befcb: -ε2 + I1R1 –ε1 - I2R2 =0 Solving three equations: I1= 2.0A, I2= -3.0A, I3= -1.0A, What does the – sign mean? Different initial direction for I1, I2 Apply Kirchhoff’s Rules: Junction c: 0=I3+I1+I2 R2 Loop abcda: ε1 ε1 + I1R1 - I3R3 =0 0 Loop befcb: -ε2 - I1R1 -ε1 + I2R2 =0 Solving three equations: I1= -2.0A, I2= +3.0A, I3= -1.0A, ε2 R1 R3 Same effective result as in previous slide 4