Electricity & Magnetism Lecture 10: Kirchhoff s Rules Today s Concept: Kirchhoff s Rules Electricity & Magne<sm Lecture 10, Slide 1
Comments "Please explain Kirchhoff in human language. water, pipes, pumps, tanks... "Not very keen on circuits... "I like circuits :P "The Blue Wire "The whole concept of the joined 2 parallel curcuits " Direction of current flow through complicated resistor set-up - 'gains and drops'. will talk about these
Last Time Resistors in series: Current through is same. Voltage drop across is IR i Resistors in parallel: Voltage drop across is same. Current through is V/R i Solved Circuits R 1 R 2 V R 3 = V I 1234 R 1234 R 4 Electricity & Magne<sm Lecture 10, Slide 2
New Circuit I 1234 THE ANSWER: Kirchhoff s Rules Electricity & Magne<sm Lecture 10, Slide 3
Kirchhoff s Voltage Rule Kirchhoff's Voltage Rule states that the sum of the voltage changes caused by any elements (like wires, bateries, and resistors) around a circuit must be zero. WHY? The poten<al difference between a point and itself is zero! Electricity & Magne<sm Lecture 10, Slide 4
Kirchhoff s Current Rule Kirchhoff'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 Electricity & Magne<sm Lecture 10, Slide 5
Kirchhoff s Laws 1) Label all currents Choose any direc<on 2) Label +/ for all elements Current goes + (for resistors) 3) Choose loop and direc<on Must start on wire, not element. 4) Write down voltage drops First sign you hit is sign to use. A + + E 1 R 1 I 1 E 2 B + R 2 I 2 + R 5 I 3 E 3 I4 R 3 R 4 + + + I 5 + 5) Write down node equa<on I in = I out We ll do calcula<on first today It s actually the easiest thing to do! Electricity & Magne<sm Lecture 10, Slide 6
CheckPoint: Gains and Drops In the following circuit, consider the loop abc. The direc<on of the current through each resistor is indicated by black arrows. DROP GAIN GAIN If we are to write Kirchoff's voltage equa<on for this loop in the clockwise direc<on star<ng from point a, what is the correct order of voltage gains/drops that we will encounter for resistors R1, R2 and R3? A. drop, drop, drop B. gain, gain, gain C. drop, gain, gain D. gain, drop, drop E. drop, drop, gain With the current Against the current VOLTAGE DROP VOLTAGE GAIN Electricity & Magne<sm Lecture 10, Slide 7
Calculation 2V 1V I 2 In this circuit, assume V i and R i are known. What is I 2? 1V Conceptual Analysis: Circuit behavior described by Kirchhoff s Rules: KVR: Σ V drops = 0 KCR: Σ I in = Σ I out Strategic Analysis Write down Loop Equa<ons (KVR) Write down Node Equa<ons (KCR) Solve Electricity & Magne<sm Lecture 10, Slide 8
Calculation R 1 + V 1 + I 1 In this circuit, assume V i and R i are known. R 2 + V 2 + I 2 What is I 2? R 3 V 3 I 3 + + Label and pick direc<ons for each current Label the + and side of each element This is easy for bateries For resistors, the upstream side is + Now write down loop and node equa<ons Electricity & Magne<sm Lecture 10, Slide 9
Calculation R 1 + V 1 + I 1 In this circuit, assume V i and R i are known. R 2 + V 2 + I 2 What is I 2? R 3 V 3 I 3 + + How many equa<ons do we need to write down in order to solve for I 2? A) 1 B) 2 C) 3 D) 4 E) 5 Why? We have 3 unknowns: I 1, I 2, and I 3 We need 3 independent equa<ons to solve for these unknowns Electricity & Magne<sm Lecture 10, Slide 10
Calculation R 1 + V 1 + I 1 In this circuit, assume V i and R i are known. R 2 + V 2 + I 2 What is I 2? R 3 V 3 I 3 + + Which of the following equa<ons is NOT correct? A) I 2 = I 1 + I 3 B) V 1 + I 1 R 1 I 3 R 3 + V 3 = 0 C) V 3 + I 3 R 3 + I 2 R 2 + V 2 = 0 D) V 2 I 2 R 2 + I 1 R 1 + V 1 = 0 Why? (D) is an atempt to write down KVR for the top loop Start at nega<ve terminal of V 2 and go clockwise V gain ( V 2 ) then V gain ( I 2 R 2 ) then V gain ( I 1 R 1 ) then V drop (+V 1 ) Electricity & Magne<sm Lecture 10, Slide 11
Calculation R 1 V 1 I 1 In this circuit, assume V i and R i are known. R 2 V 2 I 2 What is I 2? R 3 V 3 I 3 We have the following 4 equa<ons: We need 3 equa<ons: Which 3 should we use? 1. I 2 = I 1 + I 3 2. V 1 + I 1 R 1 I 3 R 3 + V 3 = 0 3. V 3 + I 3 R 3 + I 2 R 2 + V 2 = 0 4. V 2 I 2 R 2 I 1 R 1 + V 1 = 0 Why? We need 3 INDEPENDENT equa<ons Equa<ons 2, 3, and 4 are NOT INDEPENDENT Eqn 2 + Eqn 3 = Eqn 4 A) Any 3 will do B) 1, 2, and 4 C) 2, 3, and 4 We must choose Equa<on 1 and any two of the remaining ( 2, 3, and 4) Electricity & Magne<sm Lecture 10, Slide 12
Calculation R 1 V 1 I 1 In this circuit, assume V i and R i are known. R 2 V 2 I 2 What is I 2? R 3 V 3 I 3 We have 3 equa<ons 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 R 2R 2V V I 2 I 1 The solu<on will get very messy! Simplify: assume V 2 = V 3 = V V 1 = 2V R 1 = R 3 = R R V I 3 R 2 = 2R Electricity & Magne<sm Lecture 10, Slide 13
Calculation: Simplify In this circuit, assume V and R are known. What is I 2? R 2R 2V V I 2 I 1 We have 3 equa<ons and 3 unknowns. I 2 = I 1 + I 3 2V + I 1 R I 3 R + V = 0 (outside) V I 2 (2R) I 1 R + 2V = 0 (top) R V I 3 current direc<on With this simplifica<on, you can verify: I 2 = ( 1/5) V/R I 1 = ( 3/5) V/R I 3 = ( 2/5) V/R Electricity & Magne<sm Lecture 10, Slide 14
R 2V I 1 Follow Up a 2R R b V V I 2 I 3 We know: I 2 = ( 1/5) V/R I 1 = ( 3/5) V/R I 3 = ( 2/5) V/R Suppose we short R 3 : What happens to V ab (voltage across R 2?) A) V ab remains the same B) V ab changes sign C) V ab increases Why? Redraw: a R 2R b 2V V I 2 I 1 d D) V ab goes to zerobotom Loop Equa<on: V ab + V V = 0 c V I 3 V ab = 0 Electricity & Magne<sm Lecture 10, Slide 15
Clicker Question a b V R R Is there a current flowing between a and b? A) Yes B) No a & b have the same poten<al Current flows from batery and splits at a No current flows between a & b Some current flows down Some current flows right Electricity & Magne<sm Lecture 10, Slide 16
CheckPoint: Circuits w/ Resistors and a Battery 1 Consider the circuit shown below. Which of the following statements best describes the current flowing in the blue wire connec<ng points a and b? A. Posi<ve current flows from a to b B. Posi<ve current flows from b to a C. No current flows between a and b I 1 I 2 I I 3 I 4 I 1 R I 2 (2R) = 0 I 2 = ½ I 1 I 4 R I 3 (2R) = 0 I 4 = 2 I 3 I = I 1 I 3 I + I 2 = I 4 I 1 I 3 + ½ I 1 = 2I 3 I 1 = 2I 3 I = +I 3 Electricity & Magne<sm Lecture 10, Slide 17
Prelecture CheckPoint What is the same? Current flowing in and out of the batery. 2R 3 2R 3 What is different? Current flowing from a to b. Electricity & Magne<sm Lecture 10, Slide 18
I 2 / 3 I R 1 / 3 I 2R V a b V/2 1 2R R / 3 I 2 / 3 I 2 / 3 I 12 / 3 I 1 / 3 I 0 1 / 3 I 1 / 3 I 2 / 3 I Electricity & Magne<sm Lecture 10, Slide 19
CheckPoint: Circuits w/ Resistors and a Battery 2 Consider the circuit shown below. In which case is the current flowing in the blue wire connec<ng points a and b bigger? I A I B c c A Case A B Case B C They are the same Current will flow from lem to right in both cases. In both cases, V ac = V/2 I 2R = 2I 4R I A = I R I 2R = I R 2I 4R I B = I R I 4R Electricity & Magne<sm Lecture 10, Slide 20
Model for Real Battery: Internal Resistance + r r V 0 R V L V 0 R V L Usually can t supply too much current to the load without voltage sagging Electricity & Magne<sm Lecture 10, Slide 21
Using Breadboards (protoboards)
Original Breadboards
Circuit Technique Figure 6.1: Bad and Good breadboarding technique.
Good and Bad component layout Connections among pins in the breadboard. Use horizontal rows for voltage busses: +5V, ±12V, gnd. Use vertical rows for connecting components together.
connection to scope to +5V of power supply +5V bus to gnd of power supply gnd bus