Physics 212 Lecture 10

1. Physics 212 Lecture 10 Kirchhoff’s Rules

2. Last Time R2 R1 V • Resistors in series: I1234 R1234 V R3 Current through is same. Voltage drop across is IRi R4 Resistors in parallel: Voltage drop across is same. Current through is V/Ri • Solved Circuits = 5

3. New Circuit V R1 R1 I1234 R12 R3 V1 V1 V2 V2 R2 R2 How Can We Solve This One? R3 = THE ANSWER: Kirchhoff’s Rules 5

4. Kirchoff’s Voltage Rule 2 1 3 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 !

5. 2 1 3 True for any loop Add the voltage “drops” ( V1-V2) + (V2-V3) + (V3-V1) = 0 or The voltage “gains” , ( V2-V1) + (V3-V2) + (V1-V3) = 0

6. Kirchoff’s Current Rule I2 I1 Node I3 I4 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.

7. Checkpoint 2 GAIN GAIN With the current VOLTAGE DROP Against the current VOLTAGE GAIN In the following circuit, consider the loop abc. The direction of the current through each resistor is indicated by black arrows. DROP 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 R1, R2 and R3? A. drop, drop, drop B. gain, gain, gain C. drop, gain, gain D. gain, drop, drop E. gain, gain, drop A B C D E

8. Calculation I1 + - - + I2 + - + - I3 - + + - (1) Label and pick directions for each current This is easy for batteries. For resistors, the “upstream” side is +. V1 R1 In this circuit, assume Vi and Ri are known. What is I2? V2 R2 R3 V3 (2) Label the + and – side of each element Now write down loop and node equations.

9. Calculation • How many equations do we need to write down in order to solve for I2? (A) 1(B) 2(C) 3 (D) 4 (E) 5 V1 R1 I1 + - - + V2 R2 I2 In this circuit, assume Vi and Ri are known. What is I2? + - + - I3 R3 V3 - + + - • Why? • We have 3 unknowns: I1, I2, and I3 • We need 3 independent equations to solve for these unknowns • (3) Choose loops and directions

10. Calculation • Which of the following equations is NOT correct? • I2 = I1 + I3 • + V1 - I1R1 + I3R3 - V3 = 0 • + V3 - I3R3 - I2R2 - V2 = 0 • + V2 + I2R2 - I1R1 + V1 = 0 V1 R1 I1 + - - + V2 R2 I2 In this circuit, assume Vi and Ri are known. What is I2? + - + - R3 V3 I3 - + + - Node Outer loop Bottom loop Top loop • Why is (D) wrong ? • Start at negative terminal of V2 and go clockwise around top loop: • (+V2) + (+I2R2) +(+I1R1)+ (-V1) = 0 • This is not the same as answer (D).

11. Calculation V1 R1 I1 V2 R2 I2 In this circuit, assume Vi and Ri are known. What is I2? R3 V3 I3 • We have the following 4 equations: We need 3 equations: Which 3 should we use? Node Outer Bottom Top 1. I2 = I1 + I3 2.+ V1 - I1R1 + I3R3 - V3 = 0 3. + V3 - I3R3 - I2R2 - V2 = 0 4. + V2 + I2R2 + I1R1 - V1 = 0 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)

12. Calculation 2V R I1 V 2R I2 R V I3 V1 R1 I1 In this circuit, assume Vi and Ri are known. What is I2? V2 R2 I2 R3 V3 I3 • We have 3 equations and 3 unknowns. • I2 = I1 + I3 • V1 - I1R1 + I3R3 - V3 = 0 • V2 + I2R2 + I1R1 - V1 = 0 (6) Solve the equations • The solution will get very messy! • Simplify: assume V2 = V3 = V • V1 = 2V • R1 = R3 = R • R2 = 2R

13. Calculation: Simplify 2V R I1 current direction V 2R I2 R V I3 In this circuit, assume V and R are known. 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) • With this simplification, you can verify: • I2 = ( 1/5) V/R • I1 = ( 3/5) V/R • I3 = (-2/5) V/R

14. Follow-Up • Suppose we short R3: What happens to Vab (voltage across R2?) • Vab remains the same • Vab changes sign • Vab increases • Vab goes to zero 2V 2V R R I1 I1 V 2R V I2 2R I2 d a b V I3 R V I3 Bottom Loop Equation: c (Va- Vb ) + V -V = 0 Vab = 0 • We know: • I2 = ( 1/5) V/R • I1 = ( 3/5) V/R • I3 = (-2/5) V/R a b Why? Redraw:

15. No current flows between A & B Some current flows down Current flows from battery and splits at A Some current flows right a b V R R • Is there a current flowing between a and b ? • Yes • No A & B have the same potential

16. Checkpoint 3a I Consider the circuit shown below. Note that this question is not identical to the similar looking one you answered in the prelecture. I2 I 2I 2I I Which of the following best describes the current flowing in the blue wire connecting points a and b? 1. Same voltage across R, 2R in parallel so IR = 2 I2R . 2. R, 2R 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

17. 2R 2R 3 3 Prelecture Checkpoint What is the same? Current flowing in and out of the battery What is different? Current flowing from a to b

18. R 2R 1/3 I 2/3I 2/3I 0 2/3I I I 1/3 2/3I R 2R V a b V/2

19. Checkpoint 3b IA IB c c Current will flow from left to right in both cases In both cases, Vac = V/2 I2R = 2I4R Consider the circuit shown below. 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 IA = IR – I2R = IR – 2I4R IB = IR – I4R

20. r V0 VL R Model for Real Battery: Internal Resistance + r VL R V0 Usually can’t supply too much current to the load without voltage “sagging”

21. r V0 VL R Voltage divider I = V0 / (r + R) VL = IR VL = V0 R/ (r + R) VLV0 R>> r