# Direct-Current Circuits. Physics 231 Lecture 6-1

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1 Direct-Current Circuits Physics 231 Lecture 6-1

2 esistors 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

3 esistors in Series Since these resistors are in series, we have the same current in all three resistors I I = I = 1 = 2 3 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 V = V + V + V ab ax xy I yb Physics 231 Lecture 6-3

4 Then using We have that esistors in Series V ax = I 1 Vxy = I 2; Vyb = ; I ( + ) V ab = I Now the equivalent resistor,, will also have the same potential difference across it as V ab, and it will also have the same current I V ab = I Equating these last two results, we then have that = = The equivalent resistance for a sequence of resistors in series is just the sum of the individual resistances i i 3 Physics 231 Lecture 6-4

5 esistors 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) V V = V = 1 = 2 3 V ab Physics 231 Lecture 6-5

6 esistors 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 We also have that I = I + I 1 + I2 I3 V V 1 = ab ; I2 = ab ; I3 = 1 2 The equivalent resistor,, will have also have the current I going through it V ab 3 Physics 231 Lecture 6-6

7 Using esistors in Parallel I V = ab and combining with the previous equations, we then have or 1 V V V V ab = ab + ab ab = = The inverse of the effective resistance is given by the sum of the inverses of the individual resistances 3 i 1 i Physics 231 Lecture 6-7

8 Solving esistor 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

9 Solving esistor Networks Solve simplest parts of the network first Then redraw network using the just calculated effective resistance epeat calculating effective resistances until only one effective resistance is left Physics 231 Lecture 6-9

10 Solving esistor 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

11 Solving esistor Networks Step 1: Combine the two resistors that are in parallel 1 eff = + = ; eff 6 Ω 3 Ω 2 Ω = 2 Ω yielding Step 2: Combine the two resistors that are in series eff = 4 Ω + 2 Ω = 6 Ω yielding Physics 231 Lecture 6-11

12 Solving esistor Networks Current through this effective resistor is given by I = V eff 18 = = 6 3 Amps 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 V 4Ω V 2Ω = 3 4 = 12Volts; = 3 2 = 6Volts Physics 231 Lecture 6-12

13 Solving esistor 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 I 6Ω = 6Volts 6Ω = 1Amp; I 3Ω = 6Volts 3Ω = 2 Amps Physics 231 Lecture 6-13

14 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 where I is the total current P = IV = 3 18 = 54Watts The power being dissipated by each of the resistors is 2 given by P = I P P 4Ω 6Ω = 3 = = 36Watts; 6 = 6Watts; P P 3Ω = 2 2 Total = 54 3 = 12Watts; Watts P = IV Physics 231 Lecture 6-14

15 Two identical light bulbs are represented by the resistors 2 and 3 ( 2 = 3 ). The switch S is initially open. Example 1 If switch S is closed, what happens to the brightness of the bulb 2? a) It increases b) It decreases c) It doesn t change The power dissipated in 2 is given by P = V When the switch is closed neither V nor changes So the brightness does not change 2 Physics 231 Lecture 6-15

16 Two identical light bulbs are represented by the resistors 2 and 3 ( 2 = 3 ). The switch S is initially open. Example 2 What happens to the current I, after the switch is closed? a) I after = 1/2 I before b) I after = I before c) I after = 2 I before Initially the current is given by I before = ε 2 After the switch is closed the net resistance is given by = + = since 2 = net The new current is then 2 3 net = 2 I after = ε = 2 = I ε net before Physics 231 Lecture 6-16

17 Kirchoff s ules 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

18 First some terminology A junction, also called a node or branch point, is is a point where three or more conductors meet Kirchoff s ules A loop is any closed conducting path Physics 231 Lecture 6-18

19 Kirchoff s ules Kirchoff s ules 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 I 1 + I2 I3 = Be aware that all the junction equations for a circuit may not be independent of each other 0 Physics 231 Lecture 6-19

20 Kirchoff s ules 2. The algebraic sum of the potential differences in any loop including those associated with emfs and those of resistive elements must equal zero Procedures to apply this rule: V = 0 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

21 Kirchoff s ules 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

22 Kirchoff s ules Starting at any point on the loop add the emfs and I terms An I 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

23 Physics 231 Lecture 6-23 Kirchoff s ules For loop I we have = + ε ε I I I For loop II we have = + + ε ε I I Junction equation at a gives us = I I I We now have three equations for the three unknown currents

24 Kirchoff s ules Assume that the batteries are: ε 1 = 19 V; ε 2 = 6 V; ε 3 = 2 V and the resistors are: 1 = 6Ω; 2 = 4Ω; 3 = 4Ω; 4 = 1Ω you should end up with: I 1 = 1.5 A; I 2 = -0.5 A; I 3 = 2.0 A The minus sign on I 2 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

25 C 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

26 C 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 I 0 = ε / 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

27 C Circuits According to Kirchoff s 2 nd rule we have Using a counterclockwise loop ε V resistor V capacitor ε I Solving for the current q C = = 0 0 I = ε As time increases, the charge on the capacitor increases, therefore the current in the circuit decreases q C Current will flow until the capacitor has a charge on it given by Q = C ε Physics 231 Lecture 6-27

28 We remember that So we then have C Circuits dq dt earranging we have d q I = d t = ε q C d q = q Cε Setting up the integration we have 1 = C d t C q d q = q Cε ( q Cε ) 0 t 0 d t C The resultant integration yields ln q Cε 1 = Cε C Physics 231 Lecture 6-28

29 C Circuits We exponentiate both sides of this last equation and rearrange to obtain q = Cε ( t / C ) ( t / C 1 e = Q 1 e ) where Q f is the final charge on the capacitor given by Cε The constant C is known as the time constant of the circuit f We see that the charge on the capacitor increases exponentially Physics 231 Lecture 6-29

30 Example 3 At t = 0 the switch is closed in the circuit shown. The initially uncharged capacitor then begins to charge. I 1 I 3 I 2 ε C 2 1 What will be the voltage across the capacitor a long time after the switch is closed? (a) V C = 0 (b) V C = ε 2 /( ) (c) V C = ε After a long time the capacitor is completely charged, so no current flows through it The circuit is then equivalent to a battery with two resistors in series The voltage across the capacitor equals the voltage across 2 (since C and 2 are in parallel) Physics 231 Lecture 6-30

31 C Circuits The current in the circuit is given by and looks like I = d q d t = ε e t / C t = I0 e / C Note that is also how the voltage across the resistor behaves V resistor = I Physics 231 Lecture 6-31

32 C Circuits Charging Summary For the simple C circuit we have the following for the voltage drops across the capacitor and the resistor Physics 231 Lecture 6-32

33 C Circuits We now start from a situation where we have a charged capacitor in series with a resistor and an open switch The capacitor will now act as a source of emf, but one whose value is not constant with time Physics 231 Lecture 6-33

34 C Circuits We now close the switch and a current will flow Kirchoff s 2 nd rule gives us I q C = 0 earranging we have d q d t = q C To find q as a function of time we integrate the above equation Physics 231 Lecture 6-34

35 C Circuits q Q 0 d q' q' 1 = d t' C t 0 ln q Q 0 = t C Exponentiation of both sides of the equation on the right yields q = Q t / C 0e We see that the charge on the capacitor decreases exponentially Physics 231 Lecture 6-35

36 C Circuits The current in the circuit is obtained by taking the derivative of the charge equation I dq dt Q = e C = 0 The quantity Q 0 / C is just the initial voltage, V o, across the capacitor t C But then V 0 / is the initial current I 0 So we then have that I = I 0 e t C The voltage across the resistor is given by V = V 0 e t C Physics 231 Lecture 6-36

37 Example 4 The two circuits shown below contain identical fully charged capacitors at t = 0. Circuit 2 has twice as much resistance as circuit 1. Compare the charge on the two capacitors a short time after t = 0 a) Q 1 > Q 2 b) Q 1 = Q 2 c) Q 1 < Q 2 Initially, the charges on the two capacitors are the same. But the two circuits have different time constants: τ 1 = C and τ 2 = 2C Since τ 2 > τ 1 it takes circuit 2 longer to discharge its capacitor Therefore, at any given time, the charge on capacitor 2 is larger than that on capacitor 1 Physics 231 Lecture 6-37

38 Example 5 The capacitor in the circuit shown is initially charged to Q = Q 0. At t = 0 the switch is connected to position a. At t = t 0 the switch is immediately flipped from position a to position b. a C b 3 a) Which of the following graphs best represents the time dependence of the charge on C? Q Q0 a) b) c) Q Q0 Q Q0 t 0 time t 0 time t 0 time b) Which of the following correctly relates the value of t 0 to the time constant τ a while the switch is at a? (a) t 0 < τ a (b) t 0 = τ a (c) t 0 > τ a Physics 231 Lecture 6-38

39 Example 5 The capacitor in the circuit shown is initially charged to Q = Q 0. At t = 0 the switch is connected to position a. At t = t 0 the switch is immediately flipped from position a to position b. a C b 3 a) Which of the following graphs best represents the time dependence of the charge on C? Q Q0 a) b) c) Q Q0 Q Q0 t 0 time t 0 time t 0 time For 0 < t < t 0, the capacitor is discharging with time constant t = C For t > t 0, the capacitor is discharging with time constant τ = 3C, i.e., much more slowly Therefore, the answer is a) Physics 231 Lecture 6-39

40 Example 5 The capacitor in the circuit shown is initially charged to Q = Q 0. At t = 0 the switch is connected to position a. At t = t 0 the switch is immediately flipped from position a to position b. a C b 3 Q Q0 a) b) c) Q Q0 Q Q0 t 0 time t 0 time t 0 time b) Which of the following correctly relates the value of t 0 to the time constant τ a while the switch is at a? (a) t 0 < τ a (b) t 0 = τ a (c) t 0 > τ a We know that for t = τ a, the value of the charge is e -1 = 0.37 of the value at t = 0 Since the curve shows Q(t 0 ) ~ 0.6 Q 0, t 0 must be less than τ a Physics 231 Lecture 6-40

41 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 ( V = 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 decreases 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. Physics 231 Lecture 6-41

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