Physics 115. General Physics II. Session 24 Circuits Series and parallel R Meters Kirchoff s Rules

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1 Physics 115 General Physics II Session 24 Circuits Series and parallel R Meters Kirchoff s Rules R. J. Wilkes phy115a@u.washington.edu Home page: 5/15/14 Phys 115 1

2 Lecture Schedule Today 5/15/14 2 Phys 115

3 Voltmeters vs. Ammeters A voltmeter is connected across circuit elements to measure the potential difference between two points in the circuit. An ideal voltmeter has infinite internal resistance, so it draws no current from the circuit. An ammeter is inserted by breaking a circuit connection, to measure the current flowing through that connection in the circuit. An ideal ammeter has zero internal resistance, so it does not affect the current passing through it. 5/15/14 Phys I V A X I

4 What s inside ammeter, voltmeter Galvanometer deflects in proportion to current through it Assume its coil has negligible resistance, R c ~ 0 1. Ammeter: Voltage drop across shunt resistor R sh V sh =I R sh determines current through coil Want R sh as small as possible 2. Voltmeter: Voltage drop V a V b determines current through R S and coil Want series resistance R S as large as possible 5/15/14 Phys 115 4

5 More about resistor circuits: ground points Earth = infinite charge reservoir Ground wire = connected to the earth Ground = zero potential for circuits This circuit is grounded at the junction between the two resistors. This becomes the zero for the V scale, rather than the negative terminal of the battery. (With no ground, we say the circuit is floating ) Find the potential difference ΔV across each resistor: I = E R = 10 V 8 Ω+12 Ω = 0.5 A E+ ΔV 1 + ΔV 2 = 0 Ohm s Law Cons. of energy Example: A Grounded Circuit Δ V 1 = (8 Ω )(0.5 A) = 4 V Δ V = (12 Ω )(0.5 A) = 6 V Notice: no current flows through the ground wire: no ΔV across it (Point A is +6V relative to battery s terminal, but 0V relative to ground) Choice of zero doesn t matter: only ΔV s have physical meaning 5/15/14 Phys I ΔV 1 A ΔV 2 Symbol for ground

6 Example: Analyzing a more complex R circuit (1) R 1 Four resistors are connected to a 2 V battery as shown. Find the current through the battery. R 4 R 2 R 3 Step 1: combine parallel resistors into R eq1 : Step 2: combine series resistances R eq1 and R 3 into an equivalent R: 1 R eq1 = 1 R R 2 R eq1 = 1 1/ R 1 +1/ R 2 = R 1 R 2 R 1 + R 2 = 600Ω(400Ω) ( )Ω = 240Ω R eq2 = R eq1 + R 3 = 240Ω+560Ω = 800Ω 5/15/14 Phys 115 6

7 Analyzing a more complex circuit (2) Now we have this equivalent circuit: Step 3: combine these parallel resistors into one R eq : 800 R R eq2 R eq3 = R 4 R eq2 R 4 + R eq2 = 800Ω(800Ω) ( )Ω = 400Ω I battery = E / R eq3 =12V / 400Ω = 0.03A 5/15/14 Phys 115 7

8 How to analyze even more complex circuits To deal with greater complexity, we can use physical laws to organize our work: As usual: first, some definitions Junction = place where wires connect Closed loop = part of a circuit that forms a single closed path for current Then 1. Conservation of matter (charge): for any junction in a circuit, current in must equal current out: net current = 0 2. Conservation of energy: for any closed loop in a circuit, the sum of potential differences must be 0: net ΔV around the loop = 0 = Kirchoff s rules for electric circuits Gustav Kirchoff, /15/14 Phys 115 8

9 About Circuit Elements & Diagrams These are some of the symbols commonly used to represent components in circuit diagrams. Other components (coming soon) and their symbols: inductance, transformer, diode, transistor 5/15/14 Phys 115 9

10 In discussing circuits, we will make the following assumptions: Actual Circuit Circuit Diagrams 1. Wires have very small resistance, so that we can take R wire =0 and ΔV wire =0 in circuits. All wire connections are ideal. 2. Resistors have constant resistance values, regardless of current Circuit Diagram 3. Insulators are ideal non-conductors, with R= and I=0 through the insulator. 5/15/14 Phys

11 Kirchhoff s Rules for Circuits -> I in = I out Kirchoff: 1. for any junction in a circuit, current in must equal current out: net current = 0 2. for any closed loop in a circuit, the sum of potential differences must be 0: net ΔV around the loop = 0 If there is a battery (source of EMF) in the circuit, ΔV loop = (ΔV ) i = 0 = +ΔV bat ΔV R i 5/15/14 Phys

12 Kirchhoff s Laws for Multi-loop Circuits 1. Redraw to make a minimum set of current loops. Label all elements. 2. Write a loop equation for each loop. Take into account orientation of sources : add V s algebraically. R 1 = 3. Take into account direction of currents where 2 loops share a side: add currents algebraically V bat = 4. Solve equations for currents* Loop equations: Loop 1: V = Ri + R ( i i ) + R ( i i ) bat i 1 R 3 = i 2 R 2 = R 4 = R 5 = i 3 Loop 2: 0 = R ( i i ) + R ( i i ) Loop 3: 0 = R ( i i) + Ri + R ( i i ) *(If any current comes out negative, that means it actually goes opposite to the direction you assumed) 5/15/14 Phys

13 Solving the example circuit: 1. Solve loop equations for currents* 24V = 4Ω i 1 + 6Ω (i 1 i 3 ) +8Ω (i 1 i 2 ) 0 = 8(i 2 i 1 ) + 24(i 2 i 3 ) 0 = 6(i 3 i 1 ) + 24i (i 3 i 2 ) R 1 = R 2 = i 3 8i i 3 = 32i 2 i 2 = 1 4 i i 3 12i 1 = 36i 3 i 3 = 1 3 i 1, i 2 = 1 2 i 1 V bat = i 1 R 3 = R 4 = R 5 = i 2 24 = 4i 1 + 6( 2 3 i 1 ) +8(1 2 i 1 ) =12i 1 i 1 = 2A, i 2 =1A, i 3 = 2 3 A Exercise: check that these satisfy the loop equations *(If any current comes out negative, that means it actually goes opposite to the direction you assumed) HERE: all currents turned out as shown 5/15/14 Phys

14 Applying Kirchhoff s Junction Law to Multi-Junction circuit Alternatively, solve a set of junction equations: same number as loop eqns 1. Define a minimum set of junction potentials. You can choose one R 1 = J 1 R 5 = ground point, defining it as 0V. V Label all elements. bat 2. Write a junction equation for each unknown junction: I net = 0 3. Solve these equations for the unknown junction potentials. Junction equations (in this example, we already know V bat ): V 1 R 2 = R 3 = R 4 = J 2 Vbat V 0 V V V V V J1: 0 = R R R R V=0 V 2 J2: 0 0 V V V V V R R R = + + For each junction: V i,away V J = 0 i R i 5/15/14 Phys

15 Circuits with Capacitors: C s in Parallel The connected plates are at the same potential, so they form one effective capacitor, with a plate area that is the sum of the two plate areas. Since C~A, C eff = C 1 +C 2, For capacitors in parallel: add capacitances (like resistors in series). 5/15/14 15 Phys 115

Example: Capacitors in Parallel A 6.0 µf and a 12.0 µf capacitor are in parallel (terminals joined), and in series with (terminals end-to-end) a 12 V battery and a switch.

16 Example: Capacitors in Parallel A 6.0 µf and a 12.0 µf capacitor are in parallel (terminals joined), and in series with (terminals end-to-end) a 12 V battery and a switch. Initially, the switch is open and the capacitors are uncharged. The switch is then closed. When the capacitors are fully charged (a) What is voltage across each capacitor in the circuit? (b) What is the amount of charge on each capacitor plate? (c) What total charge has passed through the battery? V a =12 V; V b = 0 V. Both C's have same ΔV. Q CV µ ( = a = F)(12 V) = C = 72 C 6 6 = a = = = Q2 CV 2 ( F)(12 V) C 144 µ C Qtotal = Q1 + Q 2 = (72 µ C) + (144 µ C) = 216 µ C d) What is effective net capacitance seen by the battery? C = Q V eff total / = (216 µ C) / (12 V) = 18 µ F a 5/15/14 16 Phys 115

17 Capacitors in Series The connected string of capacitors all have the same charge Q on their plates, and the voltage drops across the invididual capacitors add. Q= CV = CV = C ( V + V ) eff 1 2 V = V( C / C ) C eff CV 1 1 CV 1 1 = = V + V V + C C [ 1 ( / )] CC 1 C + C 1/ C + 1/ C 1 2 = = Capacitors in series: add inversely (like parallel resistors). 5/15/14 17 Phys 115

18 Example: Capacitors in Series Now the 6.0 µf and 12.0 µf capacitor, 12 V battery and switch are all in series. Initially, the switch is open and the capacitors are uncharged. The switch is then closed. When the capacitors are fully charged (a) What is voltage across each device in the circuit? (b) What is the amount of charge on each capacitor plate? (c) What total charge has passed through the battery? (d) What is the effective capacitance across the battery? Va = 12 V; Vb = 0 V; Vm =? V1 = Va Vm; V2 = Vm Vb. = = ( ) Q = CV = C ( V V ) Q CV C V V a m m b ( ) ( ) Q1 = Q2 ( why?) C1 Va Vm = C2 Vm Vb C1 C2 1 2 Vm = Va + Vb = ( 12 V) + ( 0 V) = 4 V C + C C + C Q = Q = Q = C V V = (6.0 µ F) (12 V) (4 V) = 48 µ C 1 2 total 1( a m ) [ ] C /( ) eff = Qtotal Va Vb = (48 µ C) /(12 V) = 4 µ F 5/15/14 18 Phys 115

19 Summary: Combining Capacitors Parallel: Same ΔV, but different Qs. C parallel = Q ΔV C = Q 1 + Q 2 + Q 3 + ΔV C = C 1 +C 2 +C 3 + Series: Same Q, but different ΔVs. C series = Q ΔV C = = = Q ΔV 1 + ΔV 2 + ΔV ( ΔV 1 / Q) + ΔV 2 / Q ( ) + ( ΔV 3 / Q) + 1 1/ C 1 +1/ C 2 +1/ C 3 + 5/15/14 19 Phys 115

20 RC circuit : battery, R, and C in series Valve Constriction P 1 P 2 Pump Rubber Bladder Plumber s analogy of an RC circuit: a pump (=battery) pushing water through a closed loop of pipe that includes a valve (=switch), a constriction (=resistor), and a rubber bladder. When the valve starts the flow, the rubber stretches until the pressure difference across the pump (P 1 -P 3 ) equals that across the constriction (P 1 -P 2 ) + bladder (P 2 -P 3 ). P 3 Pump = Battery Valve = Switch Constriction = Resistor Capacitor= Rubber bladder Pressure = Potential Water Flow = Current 5/15/14 Phys

21 Capacitors: Time-varying current When switch closes there is a potential difference of 0 across an uncharged capacitor. After a long time, the capacitor reaches its maximum charge and there is no current flow through the capacitor. Therefore, at t=0 the capacitor behaves like a short circuit (R=0), and at t= the capacitor behaves like an open circuit (R= ). Example: before t=0 (C is uncharged) At t=0 (C starts charging) at t=, when C is fully charged: I = 0 again What s happening? Voltage drop across C at any time must be V C =Q/C. Current I at any time must be such that E = V C + IR 5/15/14 Phys

22 Discharging a fully charged capacitor Through the magic of calculus... After switch closes, current flows: ΔV C + ΔV R = Q C IR = Q C + R ΔQ Δt = 0 dq 1 = dt Q RC Q f dq = 1 Q RC Q 0 t 0 dt We find : Q(t) Q 0 " = exp t % $ ' # RC & Q f = Q 0 e t/rc Exponential behavior! 5/15/14 Physics

23 RC Exponential Decay Qt () = Qe = Qe τ 0 t/ RC t/ 0 0 Define RC time constant: τ RC 1/ e = 1/ = dq() t Q It () = = e dt RC = Ie = Ie τ t/ RC t/ 0 0 I 0 = Q 0 /RC t RC 5/15/14 Physics

Chapter 28. Direct Current Circuits

Chapter 28. Direct Current Circuits Chapter 28 Direct Current Circuits Circuit Analysis Simple electric circuits may contain batteries, resistors, and capacitors in various combinations. For some circuits, analysis may consist of combining