Physics 212 Lecture 10 Kirchhoffs Rules Physics 212

Physics 212 Lecture 10 Kirchhoff’s Rules Physics 212 Lecture 10, Slide 1

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 2 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 3 5

Kirchoff’s Voltage Rule Kirchoff's Voltage Rule states that the sum of the voltage changes around a circuit must be zero. WHY? The potential difference between a point and itself is zero ! 2 1 3 Physics 212 Lecture 10, Slide 4

2 1 3 True for any loop Add the voltage “drops” ( V 1 -V 2) + (V 2 -V 3) + (V 3 -V 1) = 0 or The voltage “gains” , ( V 2 -V 1) + (V 3 -V 2) + (V 1 -V 3) = 0

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 and cannot accumulate at nodes. Node I 2 I 1 I 4 I 3 Physics 212 Lecture 10, Slide 6

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 I A 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 drop, drop A. B. gain, gain C. drop, gain B D. E. gain, drop C gain, drop D E With the current VOLTAGE DROP Against the current VOLTAGE GAIN Physics 212 Lecture 10, Slide 7

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 8

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 9

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. I 3 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 Node Outer loop Bottom loop Top loop • Why is (D) wrong ? Start at negative terminal of V 2 and go clockwise around top loop: (+V 2) + (+I 2 R 2) + (+I 1 R 1) + (-V 1) = 0 This is not the same as answer (D). Physics 212 Lecture 10, Slide 10

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 What is I 2 ? • 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? – – – Node Outer Bottom Top 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 11

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 12

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 13

Follow-Up 2 V R V 2 R a • 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 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? ) Why? Redraw: Bottom Loop Equation: (Va - Vb ) + V - V = 0 Vab = 0 a R 2 V 2 R V b V c I 1 I 2 d I 3 Physics 212 Lecture 10, Slide 14

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 15

Checkpoint 3 a Consider the circuit shown below. Note that this question is not identical to the similar looking one I 2 you answered in the prelecture. I 2 I I I 2 I Which of the following best describes the current flowing in the blue wire connecting points a and b? 1. Same voltage across R, 2 R in parallel so IR = 2 I 2 R. 2. R, 2 R combinations are identical so voltage drop in each case is V/2. Therefore total current in top and bottom must be 3 I. 3. Satisfy Kirchhoff current law at node a so Iab = I Physics 212 Lecture 10, Slide 16

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 17

I 2/ 3 I V R 1/ 3 I a 2/ 2/ 3 I 2 R b R 1/ 3 I V/2 2 R 3 I 0 2/ 3 I Physics 212 Lecture 10, Slide 18

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 19

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 20

Voltage divider I = V 0 / ( r + R ) r VL = I R V 0 VL = V 0 R / ( r + R ) R VL VL R >> r V 0 Physics 212 Lecture 10, Slide 21