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I a I I R R b C e r V e R I 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 Resistors in Parallel • What to do? V IR • Very generally, devices in parallel have the same voltage drop • But current through R1 is not I ! Call it I1. Similarly, R2 I2. KVL V I1R1 0 I a I1 V I2 R1 I d V I 2 R2 0 • How is I related to I 1 & I 2 ? R2 I a Current is conserved! V I I1 I 2 V V V R R1 R2 1 1 1 R R1 R2 R d I Another (intuitive) way… Consider two cylindrical resistors with cross-sectional areas A1 and A2 L R1 A1 A1 V A2 R1 L R2 A2 R2 Put them together, side by side … to make one “fatter”one, Reffective L A1 A2 1 Reffective 1 1 1 R R1 R2 A1 A 1 1 2 L L R1 R2 1 Lecture 10, ACT 1 • Consider the circuit shown: 1A 50W a – What is the relation between Va -Vd and Va -Vc ? b 12V (a) (Va -Vd) < (Va -Vc) (b) (Va -Vd) = (Va -Vc) (c) (Va -Vd) > (Va -Vc) 1B 1B (b) I1 = I2 20W 80W d – What is the relation between I1 and I2? (a) I1 < I2 I2 I1 (c) I1 > I2 c Lecture 10, ACT 1 • Consider the circuit shown: 1A 50W a – What is the relation between Va -Vd and Va -Vc ? (a) (Va -Vd) < (Va -Vc) (b) (Va -Vd) = (Va -Vc) (c) (Va -Vd) > (Va -Vc) b I2 I1 12V 20W 80W d c • 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 Lecture 10, ACT 1 • Consider the circuit shown: 1A 50W a – What is the relation between Va -Vd and Va -Vc ? b 12V (a) (Va -Vd) < (Va -Vc) (b) (Va -Vd) = (Va -Vc) (c) (Va -Vd) > (Va -Vc) 1B 1B 20W • Note that: Vb -Vd • Therefore, (b) I1 = I2 (c) I1 > I2 = Vb -Vc I1 (20W) I 2 (80W) 80W d – What is the relation between I1 and I2? (a) I1 < I2 I2 I1 I1 4I 2 c 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." I in I out • 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, Ii assigned to it 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 How to use Kirchhoff’s Laws A two loop example: R1 I3 e1 e2 I2 I1 R2 R3 • 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 How to use Kirchoff’s Laws R1 I3 e2 I2 I1 e1 R2 R3 • Solve the equations for I1, I2, and I3: First find I2 and I3 in terms of I1 : I 2 (e1 I1R1 ) / R2 From eqn. (2) I3 (e1 e 2 I1R1 ) / R3 Now solve for I1 using eqn. (1): I1 e1 R2 e1 e 2 R3 R1 R1 I1 ( ) R2 R3 e1 R2 From eqn. (3) e1 e 2 R3 I1 R R 1 1 1 R2 R3 Let’s plug in some numbers R1 I1 e1 e1 = 24 V Then I3 e2 I2 R2 e 2 = 12 V R3 R1= 5W I1=2.809 A The sign means that the direction of I3 is opposite to what’s shown in the circuit R2=3W R3=4W and I2= 3.319 A and I3= 0.511 A See Appendix for a more complicated example, with three loops. Summary of Simple Circuits • Resistors in series: Reffective R1 R2 R3 ... Current thru is same; • Resistors 1 in parallel: R effective Voltage drop across is IRi 1 1 1 ... R1 R2 R3 Voltage drop across is same; Current thru is V/Ri Kirchhoff’s laws: (for further discussion see online “tutorial essay”) V n 0 loop I in I out Batteries (“Nonideal” = cannot output arbitrary current) I I • Parameterized with R "internal resistance" V e Ir r V e e Ir IR 0 e I Rr R V e Rr Internal Resistance Demo As # bulbs increases, what happens to “R”?? I r V R e How big is “r”? Power Batteries & Resistors Energy expended chemical to electrical to heat Rate is: energy power sJ time What’s happening? Assert: Charges per time P VI Potential difference per charge or you can write it as For Resistors: Units okay? P IR I I R 2 P V V R V 2 R Joule Coulomb J Watt Coulomb second s Power Transmission • Why do we use “high tension” lines to transport power? –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: Pout IVin I 2 R Pin R IR Vin 1 1 2 Pin IVin Vin Vin V in Keep R small Make Vin big – Note: for fixed input power and line resistance, the inefficiency 1/V2 Example: Quebec to Montreal 1000 km R= 220W suppose Pin = 500 MW With Vin=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 Resistor-capacitor circuits I Let’s try to add a Capacitor to our simple circuit I R Recall voltage “drop” on C? V e Q C C Q Write KVL: e IR 0 C What’s wrong here? Consider that I dQ dt and substitute. Now eqn. has only “Q”: KVL gives Differential Equation ! dQ Q eR 0 dt C We will solve this next time. For now, look at qualitative behavior… 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. 2 Lecture 10, ACT 2 2A • At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. a 2B (b) I0+ = e /2R b (c) I0+ = 2e /R long time? (b) I = e /2R C e – What is the value of the current I after a very (a) I = 0 I R – What is the value of the current I0+ just after the switch is thrown? (a) I0+ = 0 I (c) I > 2e /R R Lecture 10, ACT 2 2A • At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. a (b) I0+ = e /2R I R b C e – What is the value of the current I0+ just after the switch is thrown? (a) I0+ = 0 I R (c) I0+ = 2e /R • 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 I = e /2R Lecture 10, ACT 2 a 2A • 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? I I R b C e R (a) I0+ = 0 2B (b) I0+ = e /2R (c) I0+ = 2e /R – What is the value of the current I after a very long time? (a) I = 0 (b) I = e /2R (c) I > 2e /R • 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. 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 ( I V ) = Power dissipated ( I R V 2 2 R) – Power transmission most efficient at low current high voltage • Resistor-Capacitor Circuits – Capacitors resist rapid changes in Q resist changes in V Appendix: A three-loop KVL example • Identify all circuit nodes these are where KCL eqn’s are found I3 I1 I2 – 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 I1=I4 I1=I2+I3 I4+I5+I6 • Analyze circuit and identify all independent loops where S DVi = 0 < KVL I6 I4 I5 A three-loop KVL example • 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) I6R6+I5R5=0 I2R2b+I2R2a- I3R3 =0 VB-I1R1-I2R2a- I2R2b-I4R4-I5R5 = 0 Six equations, six unknowns…. I3 R1 I1 I2 R3 R2b I6 R6 R5 • There are simpler ways of analyzing this circuit, but this illustrates Kirchoff’s laws R2a I5 R4 I4