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More on Kirchhoff’s Laws

I. I. R. r. I. a. I. V. R. e. b. C. e. R. More on Kirchhoff’s Laws. Today…. Finish Kirchhoff’s Laws KCL: Junction Rule (Charge is conserved) Review KVL ( V is independent of path) Non-ideal Batteries & Power What limits the current in a real battery How to calculate power

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More on Kirchhoff’s Laws

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  1. I I R r I a I V R e b C e R More on Kirchhoff’s Laws

  2. Today… • Finish Kirchhoff’s Laws • KCL: Junction Rule (Charge is conserved) • Review KVL (V is independent of path) • Non-ideal Batteries & Power • What limits the current in a real battery • How to calculate power • Why we use high-voltage power lines • Resistor-Capacitor Circuits, qualitative

  3. What to do? I a I1 I2 • Very generally, devices in parallel have the same voltage drop V R1 R2 • But current through R1 is not I ! Call itI1. Similarly, R2«I2. I d KVL Þ I • How is I related to I1 & I2 ? • Current is conserved! a V R I d Þ Þ Resistors in Parallel

  4. Consider two cylindrical resistors with cross-sectional areas A1 and A2 A2 A1 V R1 R2 Put them together, side by side … to make one “fatter”one, Þ Another (intuitive) way…

  5. 50W a b I2 I1 20W 80W 12V c d (a)(Va-Vd) < (Va-Vc) (b)(Va-Vd) = (Va-Vc) (c)(Va-Vd) > (Va-Vc) 1B 1B • What is the relation between I1 and I2? (c)I1 > I2 (b)I1 = I2 (a)I1 < I2 Lecture 10, ACT 1 • Consider the circuit shown: • What is the relation between Va-Vd and Va -Vc ? 1A

  6. 50W a b I2 I1 20W 80W 12V c d (a)(Va-Vd) < (Va-Vc) (b)(Va-Vd) = (Va-Vc) (c)(Va-Vd) > (Va-Vc) Lecture 10, ACT 1 • Consider the circuit shown: • What is the relation between Va -Vd and Va -Vc ? 1A • Do you remember that thing about potential being independent of path? • Well, that’s what’s going on here !!! • (Va-Vd) = (Va-Vc) • Point d and c are the same, electrically

  7. 50W a b I2 I1 20W 80W 12V c d (a)(Va-Vd) < (Va-Vc) (b)(Va-Vd) = (Va-Vc) (c)(Va-Vd) > (Va-Vc) 1B • What is the relation between I1 and I2? (c)I1 > I2 (b)I1 = I2 (a)I1 < I2 • Note that: Vb-Vd = Vb-Vc • Therefore, Lecture 10, ACT 1 • Consider the circuit shown: • What is the relation between Va -Vd and Va -Vc ? 1A 1B

  8. Kirchhoff’s Second Rule“Junction Rule” or“Kirchhoff’s Current Law (KCL)” • In deriving the formula for the equivalent resistance of 2 resistors in parallel, we applied Kirchhoff's Second Rule (the junction rule). • "At any junction point in a circuit where the current can divide (also called a node), the sum of the currents into the node must equal the sum of the currents out of the node." • This is just a statement of the conservation of charge at any given node. • The currents entering and leaving circuit nodes are known as “branch currents”. • Each distinct branch must have a current, Iiassigned to it

  9. Old Preflight Two identical light bulbs are represented by the resistors R2 and R3 (R2 = R3 ). The switch S is initially open. 2) If switch S is closed, what happens to the brightness of the bulb R2? a) It increases b) It decreases c) It doesn’t change 3) What happens to the current I, after the switch is closed ? a) Iafter = 1/2 Ibefore b) Iafter = Ibefore c) Iafter = 2 Ibefore

  10. R1 I3 e 2 I2 I1 R2 R3 e 1 How to use Kirchhoff’s Laws A two loop example: • Analyze the circuit and identify all circuit nodes and use KCL. • (1)I1 = I2 + I3 • Identify all independent loops and use KVL. (2)e1-I1R1-I2R2 = 0 (3) e1-I1R1-e2-I3R3 = 0 (4) I2R2-e2-I3R3 = 0

  11. R1 I3 I2 I1 e 2 R2 R3 e 1 From eqn. (2) From eqn. (3) Þ How to use Kirchoff’s Laws • Solve the equations for I1, I2, and I3: • Firstfind I2 and I3 in terms of I1: Now solve for I1using eqn. (1):

  12. R1 I3 I2 I1 e 2 R2 R3 e 1 The sign means that the direction of I3 is opposite to what’s shown in the circuit Let’s plug in some numbers e1 = 24 V e 2 = 12 V R1= 5W R2=3W R3=4W Then I1=2.809 A andI2= 3.319 A and I3= -0.511 A See Appendix for a more complicated example, with three loops.

  13. Kirchhoff’s laws: (for further discussion see online “tutorial essay”) Summary of Simple Circuits • Resistors in series: Current thru is same; Voltage drop across is IRi • Resistors in parallel: Voltage drop across is same; Current thru is V/Ri

  14. I I R • Parameterized with "internal resistance" r V e Þ Batteries(“Nonideal” = cannot output arbitrary current)

  15. Internal Resistance DemoAs # bulbs increases, what happens to “R”?? I r V R e How big is “r”?

  16. Batteries & Resistors Energy expended chemical to electrical to heat Rate is: What’s happening? Assert: Charges per time or you can write it as Potential difference per charge For Resistors: Units okay? Power

  17. Keep R small Make Vin big Power Transmission • Transmission of power is typically at very high voltages (e.g., ~500 kV) • But why? • Calculate ohmic losses in the transmission lines • Define efficiency of transmission: • Why do we use “high tension” lines to transport power? • Note: for fixed input power and line resistance, theinefficiency µ1/V2 Example: Quebec to Montreal 1000 kmÞR=220W supposePin =500 MW WithVin=735kV,e=80%. The efficiency goes to zero quickly if Vin were lowered! Why do we use AC (60 Hz)? Easy to generate high voltage (water/steam → turbine in magnetic field → induced EMF) [Lecture 16] We can use transformers [Lecture 18] to raise the voltage for transmission and lower the voltage for use

  18. I I R Let’s try to add a Capacitor to our simple circuit Recall voltage “drop” on C? C e Consider that and substitute. Now eqn. has only “Q”: KVL gives Differential Equation ! Resistor-capacitor circuits Write KVL: What’s wrong here? We will solve this next time. For now, look at qualitative behavior…

  19. Capacitors Circuits, Qualitative Basic principle: Capacitor resists rapid change in Q  resists rapid changes in V • Charging (it takes time to put the final charge on) • Initially, the capacitor behaves like a wire (DV = 0, since Q = 0). • As current starts to flow, charge builds up on the capacitor • it then becomes more difficult to add more charge • the current slows down • After a long time, the capacitor behaves like an open switch. • Discharging • Initially, the capacitor behaves like a battery. • After a long time, the capacitor behaves like a wire.

  20. (c)I0+ = 2e /R (b)I0+ = e /2R (a)I0+ = 0 2B • What is the value of the current I¥ after a very long time? (c)I¥ > 2e /R (b)I¥ = e /2R (a)I¥ = 0 Lecture 10, ACT 2 • At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. • What is the value of the current I0+ just after the switch is thrown? 2A a I I R b C e R

  21. I a I R b C e (c)I0+ = 2e /R (b)I0+= e /2R (a)I0+ = 0 R Lecture 10, ACT 2 • At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. • What is the value of the current I0+ just after the switch is thrown? 2A • Just after the switch is thrown, the capacitor still has no charge, therefore the voltage drop across the capacitor = 0! • Applying KVL to the loop at t=0+, e-IR - 0 -IR = 0 ÞI = e /2R

  22. I a I R b C e (c)I0+ = 2e /R (b)I0+ = e /2R (a)I0+ = 0 R 2B • What is the value of the current I¥ after a very long time? (c)I¥ > 2e /R (b)I¥ = e /2R (a)I¥ = 0 Lecture 10, ACT 2 • At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. • What is the value of the current I0+ just after the switch is thrown? 2A • The key here is to realize that as the current continues to flow, the charge on the capacitor continues to grow. • As the charge on the capacitor continues to grow, the voltage across the capacitor will increase. • The voltage across the capacitor is limited to e; the current goes to 0.

  23. Summary • Kirchhoff’s Laws • KCL: Junction Rule (Charge is conserved) • Review KVL (V is independent of path) • Non-ideal Batteries & Power • Effective “internal resistance” limits current • Power generated ( ) = Power dissipated ( ) • Power transmission most efficient at low current  high voltage • Resistor-Capacitor Circuits • Capacitors resist rapid changes in Q  resist changes in V

  24. Appendix: A three-loop KVL example I3 I1 I2 I6 I1=I4 I1=I2+I3 I4+I5+I6 I4 I5 • Identify all circuit nodes - these are where KCL eqn’s are found • determine which KCL equations are algebraically independent (not all are in this circuit!) • I1=I2+I3 • I4=I2+I3 • I4=I5+I6 • I1=I5+I6 • Analyze circuit and identify all independent loops where S DVi = 0 <- KVL

  25. A three-loop KVL example I3 R1 I1 R5 R6 R4 R2a R2b R3 I2 I6 -I6R6+I5R5=0 I2R2b+I2R2a- I3R3 =0 VB-I1R1-I2R2a- I2R2b-I4R4-I5R5 = 0 Six equations, six unknowns…. I4 I5 • Here are the node equations from applying Kirchoff’s current law: I1=I4 I1=I2+I3 I4+I5+I6 • Now, for Kirchoff’s voltage law: (first, name the resistors) • There are simpler ways of analyzing this circuit, but this illustrates Kirchoff’s laws

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