Physics 115 General Physics II Session 24 Circuits Series and parallel R Meters Kirchoff’s Rules • R. J. Wilkes • Email: phy115a@u.washington.edu • Home page: http://courses.washington.edu/phy115a/ 5/15/14 1 Phys 115
Lecture Schedule Today 5/15/14 2 Phys 115
Voltmeters vs. Ammeters A voltmeter is connected across circuit elements to measure the potential difference between two V 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 A connection in the circuit. An ideal ammeter has zero internal resistance, so it does not I I X affect the current passing through it. 5/15/14 3 Phys 115
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 4 Phys 115
More about resistor circuits: ground points Example: A Grounded Circuit Earth = infinite charge reservoir Ground wire = connected to the earth Ground = zero potential for circuits This circuit is grounded at the I junction between the two resistors. Δ V 1 This becomes the “zero” for the V scale, rather than the negative A terminal of the battery. (With no ground, we say the circuit is “floating”) Find the potential difference Δ V Symbol Δ V 2 across each resistor: for ground I = E 10 V Ohm’s 8 Ω + 12 Ω = 0.5 A R = Law V (8 )(0.5 A) 4 V Δ = Ω = 1 Cons. of energy E + Δ V 1 + Δ V 2 = 0 V (12 )(0.5 A) 6 V Δ = Ω = 2 Notice: no current flows through the ground wire: no Δ V across it (Point A is +6V relative to battery’s – terminal, but 0V relative to ground) Choice of zero doesn’t matter: only Δ V’s have physical meaning 5/15/14 5 Phys 115
Example: Analyzing a more complex R circuit (1) R 1 Four resistors are connected to a 2 V battery as shown. R 2 Find the current through the battery. R 4 R 3 1 = 1 + 1 Step 1: combine parallel resistors into R eq 1 R R 2 1 R eq1 : = R 1 R 2 1 = 600 Ω (400 Ω ) R eq 1 = (600 + 400) Ω = 240 Ω 1/ R 1 + 1/ R 2 R 1 + R 2 Step 2: combine series resistances R eq1 and R 3 R eq 2 = R eq 1 + R 3 = 240 Ω + 560 Ω = 800 Ω into an equivalent R: 5/15/14 6 Phys 115
Analyzing a more complex circuit (2) Now we have this equivalent circuit: 800 800 Step 3: combine these parallel R 4 R eq2 resistors into one R eq : R eq 3 = R 4 R eq 2 = 800 Ω (800 Ω ) (800 + 800) Ω = 400 Ω R 4 + R eq 2 I battery = E / R eq 3 = 12 V / 400 Ω = 0.03 A 5/15/14 7 Phys 115
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 Δ V around the loop = 0 = Kirchoff’s rules for electric circuits Gustav Kirchoff, 1824-1887 5/15/14 8 Phys 115
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 9 Phys 115
Circuit Diagrams In discussing circuits, we will make the following Actual Circuit assumptions: 1. Wires have very small resistance, so that we can take R wire =0 and Δ V wire =0 in circuits. All wire connections are ideal. Circuit Diagram 2. Resistors have constant resistance values, regardless of current 3. Insulators are ideal non-conductors, with R= ∞ and I=0 through the insulator. 5/15/14 10 Phys 115
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 Δ V around the loop = 0 If there is a battery (source of EMF) in the circuit, Δ V loop = ( Δ V ) i = 0 = + Δ V bat − Δ V R ∑ ∑ i 5/15/14 11 Phys 115
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. R 1 = Take into account orientation of sources : add V ’ s algebraically. 3. Take into account direction of i 3 R 2 = currents where 2 loops share a side: i 1 add currents algebraically V bat = 4. Solve equations for currents* R 4 = Loop equations: R 3 = R 5 = i 2 Loop 1: V Ri R i ( i ) R i ( i ) = + − + − bat 1 1 2 1 3 3 1 2 Loop 2: 0 R i ( i ) R i ( i ) = − + − 3 2 1 5 2 3 *(If any current comes out negative , Loop 3: 0 R i ( i ) R i R i ( i ) = − + + − that means it actually goes opposite to 2 3 1 4 3 5 3 2 the direction you assumed) 5/15/14 12 Phys 115
Solving the example circuit: 1. Solve loop equations for currents* 24 V = 4 Ω⋅ i 1 + 6 Ω⋅ ( i 1 − i 3 ) + 8 Ω⋅ ( i 1 − i 2 ) R 1 = 0 = 8( i 2 − i 1 ) + 24( i 2 − i 3 ) 0 = 6( i 3 − i 1 ) + 24 i 3 + 24( i 3 − i 2 ) i 3 R 2 = 8 i 1 + 24 i 3 = 32 i 2 → i 2 = 1 4 i 1 + 3 i 1 4 i 3 V bat = R 4 = 12 i 1 = 36 i 3 → i 3 = 1 i 2 = 1 3 i 1 , 2 i 1 R 3 = R 5 = i 2 24 = 4 i 1 + 6(2 3 i 1 ) + 8(1 2 i 1 ) = 12 i 1 i 3 = 2 *(If any current comes out negative , i 1 = 2 A , i 2 = 1 A , 3 A that means it actually goes opposite to Exercise: check that these the direction you assumed) satisfy the loop equations HERE: all currents turned out as shown 5/15/14 13 Phys 115
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 J 1 R 1 = R 5 = potentials. You can choose one ground point, defining it as 0V. V bat Label all elements. V 1 2. Write a junction equation for each unknown junction: I net = 0 R 2 = R 3 = 3. Solve these equations for the unknown junction potentials. R 4 = Junction equations (in this example, we already know V bat ): J 2 V 2 V V 0 V V V V V V=0 − − − − J1: 0 bat 1 1 2 1 2 1 = + + + R R R R 1 2 3 5 V i , away − V J 0 V V V V V − − − J2: 0 2 1 2 1 2 For each junction: ∑ = 0 = + + R R R R i 4 3 5 i 5/15/14 14 Phys 115
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 µ F and a 12.0 µ F 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 d) What is effective net the battery? capacitance ”seen” by the battery? V a = 12 V; V b = 0 V. Both C's have same Δ V . C Q total / V 6 6 = Q CV (6.0 10 F)(12 V) 72 10 C 72 C − − = = × = × = µ eff a 1 1 a 6 6 Q C V (12.0 10 F)(12 V) − 144 10 C − 144 C (216 C) / (12 V) = = × = × = µ = µ 2 2 a Q Q Q (72 C) (144 C) 216 C 18 F = + = µ + µ = µ = µ total 1 2 5/15/14 16 Phys 115
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 C V C ( V V ) = = = + 1 1 2 2 eff 1 2 V V C ( / C ) = 2 1 1 2 CV CV C 1 1 1 1 = = eff V V V [ 1 ( C / C ) ] + + 1 2 1 1 2 C C 1 1 2 = = C C 1/ C 1/ C + + 1 2 1 2 Capacitors in series: add inversely (like parallel resistors) . 5/15/14 17 Phys 115
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