Direct-Current Circuits Physics 231 Lecture 6-1 Fall 2008 Resistors in Series and Parallel As with capacitors, resistors are often in series and parallel configurations in circuits Series Parallel The question then is what is the equivalent resistance Physics 231 Lecture 6-2 Fall 2008 Resistors in Series Since these resistors are in series, we have the same current in all three resistors I1 = I2 = I3 = I We also have that the sum of the potential differences across the three resistors must be the same as the potential difference between points a and b Vab =Vax +Vxy +Vyb Physics 231 Lecture 6-3 Fall 2008 Resistors in Series Then using Vax = I R1; Vxy = I R2; Vyb = I R3 We have that Vab = I (R1 + R2 + R3 ) Now the equivalent resistor, R, will also have the same potential difference across it as Vab, and it will also have the same current I Vab = I R Equating these last two results, we then have that R = R1 + R2 + R3 = ∑ Ri i The equivalent resistance for a sequence of resistors in series is just the sum of the individual resistances Physics 231 Lecture 6-4 Fall 2008 Resistors in Parallel Here we have that the voltage across each resistor has to be the same (work done in going from a to b is independent of the path, independent of which resistor you go through) V1 =V2 =V3 =Vab Physics 231 Lecture 6-5 Fall 2008 Resistors in Parallel We now deal with currents through the resistors At point a the current splits up into three distinct currents We have that the sum of theses three currents must add to the value coming into this point I = I1 + I2 + I3 Vab Vab Vab We also have that I1 = ; I2 = ; I3 = R1 R2 R3 The equivalent resistor, R, will have also have the current I going through it Physics 231 Lecture 6-6 Fall 2008 Resistors in Parallel V Using I = ab R and combining with the previous equations, we then have V V V V ab = ab + ab + ab R R1 R2 R3 or 1 1 1 1 1 = + + = ∑ R R1 R2 R3 i Ri The inverse of the effective resistance is given by the sum of the inverses of the individual resistances Physics 231 Lecture 6-7 Fall 2008 Solving Resistor Networks Make a drawing of the resistor network Determine whether the resistors are in series or parallel or some combination Determine what is being asked Equivalent resistance Potential difference across a particular resistance Current through a particular resistor Physics 231 Lecture 6-8 Fall 2008 Solving Resistor Networks Solve simplest parts of the network first Then redraw network using the just calculated effective resistance Repeat calculating effective resistances until only one effective resistance is left Physics 231 Lecture 6-9 Fall 2008 Solving Resistor Networks Given the following circuit What is the equivalent resistance and what is the current through each resistor We see that we have two resistors in parallel with each other and the effective resistance of these two is in series with the remaining resistor Physics 231 Lecture 6-10 Fall 2008 Solving Resistor Networks Step 1: Combine the two resistors that are in parallel 1 1 1 1 = + = ; Reff = 2 Ω Reff 6 Ω 3 Ω 2 Ω yielding Step 2: Combine the two resistors that are in series Reff = 4 Ω + 2 Ω = 6 Ω yielding Physics 231 Lecture 6-11 Fall 2008 Solving Resistor Networks Current through this effective resistor is given by V 18 I = = = 3 Amps Reff 6 The current through the resistors in the intermediate circuit of Step 1 is also 3 Amps with the voltage drop across the individual resistors being given by V4Ω = 3⋅ 4 =12Volts; V2Ω = 3⋅ 2 = 6Volts Physics 231 Lecture 6-12 Fall 2008 Solving Resistor Networks To find the current through the resistors of the parallel section of the initial circuit, we use the fact that both resistors have the same voltage drop – 6 Volts 6Volts I = =1Amp; 6Ω 6Ω 6Volts I = = 2 Amps 3Ω 3Ω Physics 231 Lecture 6-13 Fall 2008 Consistency Check There is a check that can be made to see if the answers for the currents make sense: The power supplied by the battery should equal the total power being dissipated by the resistors The power being supplied by the battery is given by P = IV where I is the total current P = IV = 3⋅18 = 54Watts The power being dissipated by each of the resistors is given by P = I 2 R 2 2 P4Ω = 3 ⋅ 4 = 36Watts; P3Ω = 2 ⋅3 =12Watts; 2 P6Ω =1 ⋅ 6 = 6Watts; PTotal = 54Watts Physics 231 Lecture 6-14 Fall 2008 Example 1 Two identical light bulbs are represented by the resistors R2 and R3 (R2 = R3 ). The switch S is initially open. 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 2 The power dissipated in R2 is given by P =V R When the switch is closed neither V nor R changes So the brightness does not change Physics 231 Lecture 6-15 Fall 2008 Example 2 Two identical light bulbs are represented by the resistors R2 and R3 (R2 = R3 ). The switch S is initially open. What happens to the current I, after the switch is closed ? a) Iafter = 1/2 Ibefore b) Iafter = Ibefore c) Iafter = 2 Ibefore Initially the current is given by Ibefore = ε R2 After the switch is closed the net resistance is given by 1 1 1 2 R = + = since R = R R = 2 2 3 net 2 Rnet R2 R3 R2 ε ⎛ ε ⎞ The new current is then Iafter = = 2⎜ ⎟ = 2Ibefore Rnet ⎝ R2 ⎠ Physics 231 Lecture 6-16 Fall 2008 Kirchoff’s Rules Not all circuits are reducible There is no way to reduce the four resistors to one effective resistance or to combine the three voltage sources to one voltage source Physics 231 Lecture 6-17 Fall 2008 Kirchoff’s Rules First some terminology A junction, also called a node or branch point, is is a point where three or more conductors meet A loop is any closed conducting path Physics 231 Lecture 6-18 Fall 2008 Kirchoff’s Rules Kirchoff’s Rules are basically two statements 1. The algebraic sum of the I = 0 currents into any junction is zero ∑ A sign convention: A current heading towards a junction, is considered to be positive, A current heading away from a junction, is considered to be negative I1 + I2 − I3 = 0 Be aware that all the junction equations for a circuit may not be independent of each other Physics 231 Lecture 6-19 Fall 2008 Kirchoff’s Rules 2. The algebraic sum of the potential differences in any loop including those associated with emfs and those of resistive elements must equal zero ∑V = 0 Procedures to apply this rule: Pick a direction for the current in each branch If you picked the wrong direction, the current will come out negative Physics 231 Lecture 6-20 Fall 2008 Kirchoff’s Rules Pick a direction for traversing a loop – this direction must be the same for all loops Note that there is a third loop along the outside branches As with the junction equations not all the loop equations will be independent of each other. Physics 231 Lecture 6-21 Fall 2008 Kirchoff’s Rules Starting at any point on the loop add the emfs and IR terms An IR term is negative if we traverse it in the same sense as the current that is going through it, otherwise it is positive An emf is considered to be positive if we go in the direction - to +, otherwise it is negative Need to have as many independent equations as there are unknowns Physics 231 Lecture 6-22 Fall 2008 Kirchoff’s Rules For loop I we have − I1R1 − I1R2 − I3R4 + ε1 − ε3 = 0 For loop II we have − I2R3 + I3R4 − ε 2 + ε3 = 0 Junction equation at a gives us I1 − I2 − I3 = 0 We now have three equations for the three unknown currents Physics 231 Lecture 6-23 Fall 2008 Kirchoff’s Rules Assume that the batteries are: ε1 = 19 V; ε2 = 6 V; ε3 = 2 V and the resistors are: R1 = 6Ω; R2 = 4Ω; R3 = 4Ω; R4 = 1Ω you should end up with: I1 = 1.5 A; I2 = -0.5 A; I3 = 2.0 A The minus sign on I2 indicates that the current is in fact in the opposite direction to that shown on the diagram Complete details can be found here Physics 231 Lecture 6-24 Fall 2008 RC Circuits Up until now we have assumed that the emfs and resistances are constant in time so that all potentials, currents, and powers are constant in time However, whenever we have a capacitor that is being charged or discharged this is not the case Now consider a circuit that consists of a source of emf, a resistor and a capacitor but with an open switch With the switch open the current in the circuit is zero and zero charge accumulates on the capacitor Physics 231 Lecture 6-25 Fall 2008 RC Circuits Now close the switch Initially the full potential will be across the resistor as the potential across the capacitor is zero since q is zero Initially the full potential is across the resistor The initial current in the circuit is then given by I0 = ε / R As the current flows a charge will accumulate on the capacitor At some time t, the current in the circuit will be I and the charge on the capacitor will be q Physics 231 Lecture 6-26 Fall 2008 RC Circuits According to Kirchoff’s 2nd rule we have Using a counterclockwise loop ε −Vresistor −Vcapacitor = 0 q ε − IR − = 0 C ε q Solving for the current I = − R RC As time increases, the charge on the capacitor increases, therefore the current in the circuit decreases Current will flow until the capacitor has a charge on it given by Q = C ε Physics 231 Lecture 6-27 Fall 2008 RC Circuits d q We remember that I = d t dq ε q 1 So we then have = − = − ()q − Cε dt R RC RC d q d t Rearranging we have = − q − Cε RC q d q t d t Setting up the integration we have = − ∫ q − Cε ∫ RC 0 0 ⎛ q − Cε ⎞ 1 The resultant integration yields ln⎜ ⎟ = − ⎝ − Cε ⎠ RC Physics 231 Lecture 6-28 Fall 2008 RC Circuits We exponentiate both sides of this last equation and rearrange to obtain −t / RC −t / RC q = Cε (1− e )= Q f (1− e ) where Qf is the final charge on the capacitor given by Cε The constant RC is known as the time constant of the circuit We see that the charge on the capacitor increases exponentially Physics 231 Lecture 6-29 Fall 2008 Example 3 I1 At t = 0 the switch is closed in I2 I3 the circuit shown.