Lect10 - Physics 212 Lecture 10 Today's Concept:...

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Unformatted text preview: Physics 212 Lecture 10 Today's Concept: Kirchhoff’s Rules Circuits with resistors & batteries 35 30 25 20 15 10 5 0 Confused Avg = 3.3 Confident Physics 212 Lecture 10, Slide 1 Music 1 Who is the Artist? A) B) C) D) E) Marvin Gaye Aaron Neville Michael Bolton Al Green Ray Charles 1. 2. 2. 3. 3. 4. 5. 5. 6. 7. 7. 8. 8. 9. 9. 10. 10. Pretty Woman Unchained Melody Funny How Time Slips Away Funny Ain't No Mountain High Enough How Can you Mend A Broken Heart Take Me To The River I’m So Lonesome I Could Cry I Can't Get Next To You My Girl Light My Fire Something upbeat for Something the day after the exam the Different.. Different.. Mostly covers.. Very good… Very Roy Orbison Roy Righteous Brothers Righteous Willie Nelson Willie Marvin Gaye Marvin Bee Gees Al Green Al Hank Williams Hank Temptations Temptations Temptations Temptations Doors Physics 212 Lecture 10, Slide 2 Physics 212 Lecture 10 Today's Concept: Kirchhoff’s Rules Circuits with resistors & batteries 35 30 25 20 15 10 5 0 Confused Avg = 3.3 Confident Physics 212 Lecture 10, Slide 3 Key Concepts: 1) Understanding Kirchoff’s Rules Understanding Kirchoff 2) Using Kirchoff’s Rules Using Kirchoff Today’s Plan: • Summary of Kirchoff’s rules Summary Kirchoff • Example problem • Review Preflights Review Preflights Physics 212 Lecture 10, Slide 4 Last Time Resistors in series: Current through is same. Voltage drop across is IRi Reffective = R1 + R 2 + R3 + ... Resistors in parallel: Voltage drop across is same. 1 Current through is V/Ri Reffective 1 1 1 =+ + + ... R1 R2 R3 Solved Circuits R2 R1 V = R3 V I1234 R1234 R4 Physics 212 Lecture 10, Slide 5 5 New Circuit R1 R3 V1 V2 R2 How Can We Solve This One? R1 R3 V1 V2 R2 = V I1234 R12 THE ANSWER: Kirchhoff’s Rules Physics 212 Lecture 10, Slide 6 5 Kirchoff’s Voltage Rule ∑ ∆V i =0 Kirchoff's Voltage Rule states that the sum of the voltage changes caused by any elements (like wires, batteries, and resistors) around a circuit must be zero. If we model voltage as height above the ground floor, see if you can come up with the analogy to Kirchoff's Voltage Rule in terms of someone walking around in the hallways and stairways and elevators of a high-rise building. If we start at the ground floor with a potential of 0, and walk around up and down If stairs, take the elevator a few flights, go to the roof, parachute off, and end up stairs, te back on the ground floor, our potential is still 0. Therefore, the potential back he difference is 0. We may have increased and decreased our potential as we traveled difference al through the building, but we still start and end at a potential of 0. OR: OR: The potential difference between a point and itself is zero ! The Physics 212 Lecture 10, Slide 7 Kirchoff’s Current Rule ∑I in = ∑ I out Kirchoff's Current Rule states that the sum of all Current currents entering any given point in a circuit must equal the sum of all currents leaving the same point. If we model electrical current as water, see if you can come up with an analogy to Kirchoff's Current Rule in terms of household plumbing. If you have a main water line coming into your house, it will split off to lit service all utilities, such as sink, toilet, shower, etc. The water in all of those service ter lines must equal the amount of water coming out of the main line, and when lines and all the household water drains out of the house into a main line again, all the all again, smaller lines must include as much water as the drain carries out. smaller t. OR: OR: Electric Charge is Conserved Electric Physics 212 Lecture 10, Slide 8 Kirchhoff’s Laws (1) Label all currents Choose any direction (2) Label +/- for all elements R1 A + I1 + Current goes + ⇒ - (for resistors) (3) Choose loop and direction Must start on wire, not element. 3 + B 1 - First sign you hit is sign to use. I2 I3 - + 2 (4) Write down voltage drops + R2 R3 R5 - + I4 - R4 + + I5 (5) Write down node equation Iin = Iout We’ll do calculation first today It’s actually the easiest thing to do !! Physics 212 Lecture 10, Slide 9 17 Preflight 3 DROP GAIN BB N AI G A B C D E 50 40 30 20 With the current Against the current VOLTAGE DROP 10 0 VOLTAGE GAIN Physics 212 Lecture 10, Slide 10 10 1Ω 2Ω 1Ω Calculation 2V 1V I2 In this circuit, assume Vi and Ri are known. What is I2 ?? 1V • Conceptual Analysis: – Circuit behavior described by Kirchhoff’s Rules: • KVR: ΣVdrops = 0 • KCR: ΣIin = ΣIout •Strategic Analysis – – – Write down Loop Equations (KVR) Write down Node Equations (KCR) Solve Physics 212 Lecture 10, Slide 11 11 V1 R1 ++ - -+ + R3 - I2 In this circuit, assume Vi and Ri are known. - V3 ++ I1 - V2 R2 Calculation What is I2 ?? I3 - Label and pick directions for each current Label the + and – side of each element Label This is easy for batteries For resistors, the “upstream” side is + For Now write down loop and node equations Physics 212 Lecture 10, Slide 12 12 V1 R1 ++ - -+ + R3 - I2 - V3 ++ I1 - V2 R2 Calculation BB In this circuit, assume Vi and Ri are known. What is I2 ?? I3 - • How many equations do we need to write down in order to solve for I2? (A) 1 (B) 2 (B) (C) 3 (C) (D) 4 (E) 5 • Why?? – – We have 3 unknowns: I1, I2, and I3 We need 3 independent equations to solve for these unknowns Physics 212 Lecture 10, Slide 13 13 V1 R1 ++ - -+ + R3 - I2 - V3 ++ I1 - V2 R2 Calculation I3 - BB In this circuit, assume Vi and Ri are known. What is I2 ?? • Which of the following equations is NOT correct? (A) (B) (C) (D) I2 = I1 + I3 - V1 + I1R1 - I3R3 + V3 = 0 - V3 + I3R3 + I2R2 + V2 = 0 - V2 – I2R2 + I1R1 + V1 = 0 • Why?? – – (D) is an attempt to write down KVR for the top loop Start at negative terminal of V2 and go clockwise • Vgain (-V2) then Vgain (-I2R2) then Vgain (-I1R1) then Vdrop (+V1) Physics 212 Lecture 10, Slide 14 14 R1 V1 R2 V2 R3 V3 Calculation I1 I2 BB In this circuit, assume Vi and Ri are known. I3 • We have the following 4 equations: 1. 2. 2. 3. 3. 4. I2 = I1 + I3 - V1 + I1R1 - I3R3 + V3 = 0 - V3 + I3R3 + I2R2 + V2 = 0 - V2 – I2R2 - I1R1 + V1 = 0 What is I2 ?? We need 3 equations: Which 3 should we use? A) Any 3 will do B) 1, 2, and 4 C) 2, 3, and 4 • Why?? – – – We need 3 INDEPENDENT equations Equations 2, 3, and 4 are NOT INDEPENDENT • Eqn 2 + Eqn 3 = - Eqn 4 We must choose Equation 1 and any two of the remaining ( 2, 3, and 4) Physics 212 Lecture 10, Slide 15 15 R1 R2 R3 V1 V2 V3 Calculation I1 I2 I3 In this circuit, assume Vi and Ri are known. What is I2 ?? • We have 3 equations and 3 unknowns. I2 = I1 + I3 V1 + I1R1 - I3R3 + V3 = 0 V2 – I2R2 - I1R1 + V1 = 0 R 2R R 2V V V I1 •The solution will get very messy! I2 I3 Simplify: assume V2 = V3 = V V1 = 2V R1 = R3 = R R2 = 2R Physics 212 Lecture 10, Slide 16 16 Calculation: Simplify In this circuit, assume V and R are known. R 2R R 2V V V I1 I2 What is I2 ?? • We have 3 equations and 3 unknowns. I2 = I1 + I3 -2V + I1R - I3R + V = 0 (outside) -V – I2(2R) - I1R + 2V= 0 (top) I3 • With this simplification, you can verify: I2 = ( 1/5) V/R I1 = ( 3/5) V/R I3 = (-2/5) V/R Physics 212 Lecture 10, Slide 17 17 Follow-Up 2V R I1 • We know: V 2R a • I2 = ( 1/5) V/R I1 = ( 3/5) V/R I3 = (-2/5) V/R I2 b R V I3 Suppose we short R3: (A) Vab remains the same remains (B) Vab changes sign (B) changes (C) Vab increases (C) (D) Vab goes to zero BB What happens to Vab (voltage across R2?) R 2R Why? Redraw: a 2V V I1 I2 b V d I3 c Vcd = +V Vbd = +V Vad = Vcd = +V Vab = Vad – Vbd = V – V = 0 Physics 212 Lecture 10, Slide 18 18 a b BB V R R Is there a current flowing between a and b ? Is and A) Yes B) No A & B have the same potential Current flows from battery and splits at A No current flows between A & B Some current flows down Some current flows right Physics 212 Lecture 10, Slide 19 19 Preflight 5 I1 I2 BB I I3 I4 60 I1R – I2 (2R) = 0 I1 = 2 I2 50 40 I4R – I3 (2R) = 0 I4 = 2 I3 I2 = I3 I1 = 2 I3 I = I1 - I3 = 2I3 –I3 = +I3 30 20 10 0 Physics 212 Lecture 10, Slide 20 20 Prelecture What iis the same? s Preflight Current flowing in and out of the battery Current 2R 3 2R 3 What is different? Current flowing from a to b Physics 212 Lecture 10, Slide 21 21 I 2/ R 3I 1/ 3I a V 2/ 2/ I 3 3I b R 1/ 3I V/2 2R 1/ 1/ 0I 3 2 1/ I 3 2R 1/ 3I 2/ 3I 3I Physics 212 Lecture 10, Slide 22 22 Preflight 7 IA IB c c BB 60 Current will flow from left to right in both cases 50 40 In both cases, Vac = V/2 In V/2 30 20 10 I(2R) = 2I(4R) IA = I(R) – I(2R) I(R) = I(R) – 2I(4R) I(R) 2I(4R) 0 IB = I(R) – I(4R) I(R) Physics 212 Lecture 10, Slide 23 23 Model for Real Battery: Internal Resistance + r V0 r R VL V0 R VL Usually can’t supply too much current to the supply load without voltage “sagging” Physics 212 Lecture 10, Slide 24 24 ...
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