Physics 212 Lecture 10 Today s Concept Kirchhoff s Rules

Physics 212 Lecture 10 Today's Concept: Kirchhoff’s Rules Circuits with resistors & batteries Physics 212 Lecture 10, Slide 1

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Your Comments: “This stuff is loopy” “The conventions for signs of voltage drops and gains make absolutely no sense. Why is a voltage drop considered positive and a gain negative? ” negative? “determining which way current flows and application of Kirchoff’s rules will need a ton of practice since this is all new to me. ” me. We’ll start simply, with the checkpoints, then to a calculation “Discuss multiple currents more. It is pretty confusing to understand. ” “I don't really understand how Kirchhoff's Rules apply to the blue wire problem. Could we please go over these problems in lecture? That would be a huge help!” This is a great example – we’ll do it at the end to make sure of the concepts "Fluke" is not a good brand name for a multimeter. Advertisement: "If the Physics 212 Lecture 10, Slide data is good, it must have been a Fluke. " 3

Today’s Plan: • Summary of Kirchoff’s rules – these are the key concepts • Example problem • Review Checkpoints Physics 212 Lecture 10, Slide 4

Last Time Resistors in series: Current through is same. Voltage drop across is IRi Resistors in parallel: Voltage drop across is same. Current through is V/Ri Solved Circuits R 1 V R 2 = R 3 V I 1234 R 4 Physics 212 Lecture 10, Slide 5 5

Last Time Physics 212 Lecture 10, Slide 6 5

New Circuit R 1 R 3 V 1 V 2 R 2 How Can We Solve This One? R 1 V 2 R 3 R 2 = V I 1234 R 12 THE ANSWER: Kirchhoff’s Rules Physics 212 Lecture 10, Slide 7 5

Kirchoff’s Voltage Rule 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. WHY? The potential difference between a point and itself is zero ! Physics 212 Lecture 10, Slide 8

Kirchoff’s Current Rule Kirchoff's Current Rule states that the sum of all currents entering any given point in a circuit must equal the sum of all currents leaving the same point. WHY? Electric Charge is Conserved Physics 212 Lecture 10, Slide 9

Checkpoint 1 How many potentially different currents are there in the circuit shown? A. 3 B. 4 C. 5 D. 6 E. 7 “two for the parallel branch and then one for the first and last resistors. ” A “There are four different series connections” B C “there is D potentially one between every pair of resistors. ” Physics 212 Lecture 10, Slide 10

Checkpoint 1 How many potentially different currents are there in the circuit shown? I 1 I 3 I 2 I 1 A. 3 B. 4 C. 5 I 3 D. 6 E. 7 Look at the nodes! Top node: I 1 flows in, I 2 and I 3 flow out A B Bottom C D node: I 2 and I 3 flow in, I 1 flows out That’s all of them! Physics 212 Lecture 10, Slide 11

Checkpoint 2 In the following circuit, consider the loop abc. The direction of the current through each resistor is indicated by black arrows. If we are to write Kirchoff's voltage equation for this loop in the clockwise direction starting from point a, what is the correct order of voltage gains/drops that we will encounter for resistors R 1, R 2 and R 3? A A. drop, drop B. gain, gain C. drop, gain B gain, drop D. E. drop, gain C D “going with current is drop, against current is gain” E “The voltage gains when the current is flowing with the voltage” “drops 2 times at the split then gains when merging” Physics 212 Lecture 10, Slide 12

Checkpoint 2 GAIN In the following circuit, consider the loop abc. The direction of the current through each resistor is indicated by black arrows. DROP N AI G If we are to write Kirchoff's voltage equation for this loop in the clockwise direction starting from point a, what is the correct order of voltage gains/drops that we will encounter for resistors R 1, R 2 and R 3? A A. drop, drop B. gain, gain C. drop, gain B gain, drop D. E. drop, gain C D E With the current Against the current VOLTAGE DROP VOLTAGE GAIN Physics 212 Lecture 10, Slide 13

Calculation 2 V 1 V I 2 In this circuit, assume Vi and Ri are known. What is I 2 ? ? 1 V • Conceptual Analysis: – Circuit behavior described by Kirchhoff’s Rules: • KVR: SVdrops = 0 • KCR: SIin = SIout • Strategic Analysis – – – Write down Loop Equations (KVR) Write down Node Equations (KCR) Solve Physics 212 Lecture 10, Slide 14

V 1 R 1 + + - V 2 R 2 - + + R 3 - - I 2 In this circuit, assume Vi and Ri are known. What is I 2 ? ? - V 3 + + I 1 Calculation I 3 - (1) Label and pick directions for each current (2) Label the + and – side of each element This is easy for batteries For resistors, the “upstream” side is + Now write down loop and node equations Physics 212 Lecture 10, Slide 15

V 1 R 1 + + - V 2 R 2 - + + R 3 - - I 2 - V 3 + + I 1 Calculation In this circuit, assume Vi and Ri are known. What is I 2 ? ? I 3 - • How many equations do we need to write down in order to solve for I 2? (A) 1 • Why? ? – – (B) 2 (C) 3 (D) 4 (E) 5 We have 3 unknowns: I 1, I 2, and I 3 We need 3 independent equations to solve for these unknowns (3) Choose loops and directions Physics 212 Lecture 10, Slide 16

V 1 R 1 + + - V 2 R 2 - + + R 3 - - I 2 - V 3 + + I 1 Calculation I 3 - In this circuit, assume Vi and Ri are known. What is I 2 ? ? • Which of the following equations is NOT correct? (A) (B) (C) (D) I 2 = I 1 + I 3 - V 1 + I 1 R 1 - I 3 R 3 + V 3 = 0 - V 3 + I 3 R 3 + I 2 R 2 + V 2 = 0 - V 2 – I 2 R 2 + I 1 R 1 + V 1 = 0 • Why? ? – – (4) Write down voltage drops (5) Write down node equation (D) is an attempt to write down KVR for the top loop Start at negative terminal of V 2 and go clockwise • Vgain (-V 2) then Vgain (-I 2 R 2) then Vgain (-I 1 R 1) then Vdrop (+V 1) Physics 212 Lecture 10, Slide 17

R 1 V 1 R 2 V 2 R 3 V 3 I 1 Calculation I 2 In this circuit, assume Vi and Ri are known. I 3 • We have the following 4 equations: 1. 2. 3. 4. I 2 = I 1 + I 3 - V 1 + I 1 R 1 - I 3 R 3 + V 3 = 0 - V 3 + I 3 R 3 + I 2 R 2 + V 2 = 0 - V 2 – I 2 R 2 - I 1 R 1 + V 1 = 0 • Why? ? – – – What is I 2 ? ? We need 3 equations: Which 3 should we use? A) Any 3 will do B) 1, 2, and 4 C) 2, 3, and 4 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 18

R 1 V 1 R 2 V 2 R 3 R 2 R R V 3 2 V V V I 1 I 2 I 3 Calculation In this circuit, assume Vi and Ri are known. What is I 2 ? ? • We have 3 equations and 3 unknowns. I 2 = I 1 + I 3 V 1 + I 1 R 1 - I 3 R 3 + V 3 = 0 V 2 – I 2 R 2 - I 1 R 1 + V 1 = 0 (6) Solve the equations • The solution will get very messy! Simplify: assume V 2 = V 3 = V V 1 = 2 V R 1 = R 3 = R R 2 = 2 R Physics 212 Lecture 10, Slide 19

Calculation: Simplify In this circuit, assume V and R are known. R 2 R R 2 V V V I 1 I 2 What is I 2 ? ? • We have 3 equations and 3 unknowns. I 2 = I 1 + I 3 -2 V + I 1 R - I 3 R + V = 0 (outside) -V – I 2(2 R) - I 1 R + 2 V= 0 (top) I 3 current direction • With this simplification, you can verify: I 2 = ( 1/5) V/R I 1 = ( 3/5) V/R I 3 = (-2/5) V/R Physics 212 Lecture 10, Slide 20

2 V R • We know: I 2 = ( 1/5) V/R I 1 = ( 3/5) V/R I 3 = (-2/5) V/R I 2 b R • I 1 V 2 R a Follow-Up V I 3 Suppose we short R 3: (A) Vab remains the same (B) Vab changes sign (C) Vab increases (D) Vab goes to zero What happens to Vab (voltage across R 2? ) R Why? Redraw: Bottom Loop Equation: Vab + V – V = 0 Vab = 0 2 R a 2 V V b V c I 1 I 2 d I 3 Physics 212 Lecture 10, Slide 21

A V B R R Is there a current flowing between A and B ? 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 22

Checkpoint 3 a Consider the circuit shown below. Note that this question is not identical to the similar looking one I 1 I 2 you answered in the prelecture. Which of the following best describes the current flowing in the blue wire connecting points a and b? A. Positive current flows from a to b B. Positive current flows from b to a C. No current flows between a and b “Energy flows toward least resistance, which is 1 R. ” “because b has higher voltage than a” “The top half is the same as the bottom half. ” Physics 212 Lecture 10, Slide 23

Checkpoint 3 a Consider the circuit shown below. Note that this question is not identical to the similar looking one I 1 I 2 you answered in the prelecture. I 1 I I 2 I 4 I 3 Which of the following best describes the current flowing in the blue wire connecting points a and b? Which of the following best describes the. B. Positive current flows from b connecting points a and b? current flowing in the blue wire to a A. Positive current flows from a to b C. No current flows between a and b I 1 R – I 2 (2 R) = 0 I 2 = ½ I 1 I 4 R – I 3 (2 R) = 0 I 4 = 2 I 3 a: I 1 = I + I 3 b: I + I 2 = I 4 I 1 - I 3 + ½ I 1 = 2 I 3 I = +I 3 Physics 212 Lecture 10, Slide 24

Prelecture What is the same? Checkpoint Current flowing in and out of the battery 2 R 3 What is different? Current flowing from a to b Physics 212 Lecture 10, Slide 25

I 2/ 3 I V R 1/ 3 I a 2/ 2/ I 3 2 1/ I 3 3 I 2 R b R 1/ 3 I V/2 2 R 1/ 1/ 0 I 3 1/ 3 I 2/ 3 I 3 I Physics 212 Lecture 10, Slide 26

Consider the circuit shown below. Checkpoint 3 b In which case is the current flowing in the blue wire connecting points a and b the largest? A. Case A B. Case B C. They are both the same “Case A has a lower Req” “The resistors have a value of 4 R in Case B, so the current will be more apt to bypass this section of the circuit” “The total resistance in the path taken is the same for both” Physics 212 Lecture 10, Slide 27

Consider the circuit shown below. Checkpoint 3 b IA c IB c In which case is the current flowing in the blue wire connecting points a and b the largest? A. Case A B. Case B C. They are both the same Current will flow from left to right in both cases In both cases, Vac = V/2 I 2 R = 2 I 4 R IA = IR – I 2 R = IR – 2 I 4 R IB = IR – I 4 R Physics 212 Lecture 10, Slide 28

Model for Real Battery: Internal Resistance + r V 0 r R VL V 0 R VL Usually can’t supply too much current to the load without voltage “sagging” Physics 212 Lecture 10, Slide 29

Kirchhoff’s Laws (1) Label all currents Choose any direction (2) Label +/- for all elements Current goes + - (for resistors) Battery signs fixed! (3) Choose loops and directions Must start on wire, not element. (4) Write down voltage drops First sign you hit is sign to use. (5) Write down node equations R 1 A + I 1 - + R 2 B I 2 + V 1 - + V 2 R 5 - I 6 + V 3 I 3 R 3 + I 4 R 4 + + I 5 Iin = Iout (6) Solve set of equations Physics 212 Lecture 10, Slide 30 17