2Resistors in Parallel I R1 R2 I1 I2 V I R V a d KVL Þ a d Þ What to do?Very generally, devices in parallel have the same voltage dropadIR1R2I1I2VBut current through R1 is not I ! Call it I1. Similarly, R2 «I2.KVL ÞIadRVHow is I related to I 1 & I 2 ?Current is conserved!Þ
3Another (intuitive) way… Consider two cylindrical resistors withcross-sectional areas A1 and A2VR1R2A1A2Put them together, side by side … to make one “fatter”one,Þ
4Consider the circuit shown: What is the relation between Va -Vd and Va -Vc ?12VI1I2abdc50W20W80W1A(a) (Va -Vd) < (Va -Vc)(b) (Va -Vd) = (Va -Vc)(c) (Va -Vd) > (Va -Vc)(a) I1 < I2(b) I1 = I2(c) I1 > I21BWhat is the relation between I1 and I2?
5Consider the circuit shown: What is the relation between Va -Vd and Va -Vc ?12VI1I2abdc50W20W80W1A(a) (Va -Vd) < (Va -Vc)(b) (Va -Vd) = (Va -Vc)(c) (Va -Vd) > (Va -Vc)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
6Consider the circuit shown: What is the relation between Va -Vd and Va -Vc ?12VI1I2abdc50W20W80W1A(a) (Va -Vd) < (Va -Vc)(b) (Va -Vd) = (Va -Vc)(c) (Va -Vd) > (Va -Vc)1B1BWhat is the relation between I1 and I2?(a) I1 < I2(b) I1 = I2(c) I1 > I2Note that: Vb -Vd = Vb -VcTherefore,
7Kirchhoff’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, Ii assigned to it
8Old PreflightTwo 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 change3) What happens to the current I, after the switch is closed ?a) Iafter = 1/2 Ibeforeb) Iafter = Ibeforec) Iafter = 2 Ibefore
9How to use Kirchhoff’s Laws A two loop example:e 1e 2R1R3R2I3I1I2Analyze the circuit and identify all circuit nodes and use KCL.(1) I1 = I2 + I3Identify all independent loops and use KVL.(2) e1 - I1R1 - I2R2 = 0(3) e1 - I1R1 - e2 - I3R3 = 0(4) I2R2 - e2 - I3R3 = 0
10How to use Kirchoff’s Laws Solve the equations for I1, I2, and I3:First find I2 and I3 in terms of I1 :From eqn. (2)From eqn. (3)Now solve for I1 using eqn. (1):Þ
11Let’s plug in some numbers e1 = 24 Ve 2 = 12 VR1= 5W R2=3W R3=4WThen I1=2.809 A and I2= Aand I3= AThe sign means that the direction of I3 is opposite to what’s shown in the circuitSee Appendix for a more complicated example, with three loops.
12Summary of Simple Circuits Resistors in series:Current thru is same; Voltage drop across is IRiResistors in parallel:Voltage drop across is same; Current thru is V/RiKirchhoff’s laws: (for further discussion see online “tutorial essay”)
14Internal Resistance Demo As # bulbs increases, what happens to “R”?? VReHow big is “r”?
15Power Batteries & Resistors Energy expended What’s happening? Assert: chemicalto electricalto heatWhat’s happening?Assert:Rate is:Charges per timePotential difference per chargeUnits okay?For Resistors:or you can write it as
16Power TransmissionWhy 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 linesDefine efficiency of transmission:Keep R smallMake Vin bigNote: for fixed input power and line resistance, the inefficiency µ 1/V2Example: Quebec to Montreal1000 km Þ R= 220Wsuppose Pin = 500 MWWith 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
17Resistor- capacitor circuits Let’s try to add a Capacitor to our simple circuitRecall voltage “drop” on C?Write KVL:What’s wrong here?Consider thatand substitute. Now eqn. has only “Q”:KVL gives Differential Equation !We will solve this next time. For now, look at qualitative behavior…
18Capacitors Circuits, Qualitative Basic principle: Capacitor resists rapid change in Q resists rapid changes in VCharging (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 downAfter a long time, the capacitor behaves like an open switch.DischargingInitially, the capacitor behaves like a battery.After a long time, the capacitor behaves like a wire.
19At 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?2AaIIRbCeR(a) I0+ = 0(b) I0+ = e /2R(c) I0+ = 2e /R(a) I¥ = 0(b) I¥ = e /2R(c) I¥ > 2e /RWhat is the value of the current I¥ after a very long time?2B
20At 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?2AabeRCI(a) I0+ = 0(b) I0+ = e /2R(c) I0+ = 2e /RJust 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 IR = 0 Þ I = e /2R
212A (a) I0+ = 0 (b) I0+ = e /2R (c) I0+ = 2e /R (a) I¥ = 0 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) I0+ = 0(b) I0+ = e /2R(c) I0+ = 2e /R(a) I¥ = 0(b) I¥ = e /2R(c) I¥ > 2e /RWhat is the value of the current I¥ after a very long time?2BThe 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.
22Summary Kirchhoff’s Laws Non-ideal Batteries & Power KCL: Junction Rule (Charge is conserved)Review KVL (V is independent of path)Non-ideal Batteries & PowerEffective “internal resistance” limits currentPower generated ( ) = Power dissipated ( )Power transmission most efficient at low current high voltageResistor-Capacitor CircuitsCapacitors resist rapid changes in Q resist changes in V
23Appendix: A three-loop KVL example Identify all circuit nodes - these are where KCL eqn’s are founddetermine which KCL equations are algebraically independent (not all are in this circuit!)I1=I2+I3I4=I2+I3I4=I5+I6I1=I5+I6I1=I4I1=I2+I3I4+I5+I6Analyze circuit and identify all independent loops where S DVi = 0 <- KVL
24A three-loop KVL example I1I2I3I4I5I6Here are the node equations from applying Kirchoff’s current law:R1R2aR2bR3R4R6R5I1=I4I1=I2+I3I4+I5+I6Now, for Kirchoff’s voltage law: (first, name the resistors)-I6R6+I5R5=0I2R2b+I2R2a- I3R3 =0VB-I1R1-I2R2a- I2R2b-I4R4-I5R5 = 0Six equations, six unknowns….There are simpler ways of analyzing this circuit, but this illustrates Kirchoff’s laws