Tody s gend: Potentil Chnges Around Circuit. You must e le to clculte potentil chnges round closed loop. Electromotive force (EMF), Terminl Voltge, nd Internl Resistnce. You must e le to incorporte ll of the ove quntities in your circuit clcultions. Electric Power. You must e le to clculte the electric power dissipted in circuit components. Exmples.
Kirchhoff s loop rule Recll (lecture 7): voltge drops cross circuit components connected in series dd V C 1 C 2 C 3 V = V 1 + V 2 + V 3 V 1 V 2 V 3 Generliztion: Loop rule round ny closed loop: V i = 0 i strictly: reflects energy conservtion (round ny closed pth, U f =U i ) i ΔV = 0 i
Recipe for Kirchhoff s loop rule 1. Assign (guess) direction of current I positive I: current flows in this direction negtive I: current flows in opposite direction 5 Ω 10 Ω 2. Pick trvel direction I (clockwise or counterclockwise) 9 V 3. Add up voltge differences V i, set sum to zero sign rules: ttery voltge counts positive if you trvel from to + resistor voltge counts negtive of you trvel with current strt t point : +9V I 5Ω - I 10Ω =0 I=9V / 15Ω = 0.6A Ohm s lw V=I R
Exmple: Loop rule in circuit of 3 resistors nd 2 tteries R 1 R 2 R 3 V 1 V 2 V 3 I - + V B V A 1. Assign current direction 2. Pick trvel direction: clockwise 3. Add voltges - V 1 - V 2 - V 3 + V A - V B = 0 - IR 1 - IR 2 - IR 3 + V A - V B = 0 Recommendtion: choose your pth round the circuit in the sme direction s your guessed current.
Exmple: clculte I, V, nd V for the circuit shown. To e worked t the lckord in lecture. I 5 Ω 9 V 10 Ω +9 5 I 10 I = 0 15 I = +9 I = +9/15 = 0.6 A
Exmple: clculte I, V, nd V for the circuit shown. I = 0.6 A 5 Ω 9 V 10 Ω V + 9 5 (0.6) = V V = V V = 9 + 5 (0.6) = -6 V
Exmple: clculte I, V, nd V for the circuit shown. I = 0.6 A 5 Ω 9 V 10 Ω V 10 (0.6) = V V = V V = + 10 (0.6) = +6 V
Exmple: clculte I, V, nd V for the circuit shown. I = 0.6 A 5 Ω 9 V 10 Ω Note: V = +6 V = - V, s expected
Exmple: clculte I, V, nd V for the circuit shown. Grph the potentil rises nd drops in this circuit. I = 0.6 A 5 Ω 9 V 10 Ω ε = 9V 5 I = 3V 10 I = 6V
Exmple: clculte I, V, nd V for the circuit shown. To e worked t the lckord in lecture. 5 Ω I - + 6 V 9 V 10 Ω + 9 6 5 I 10 I =0 0 15 I = 3 I = 0.2 A
Exmple: clculte I, V, nd V for the circuit shown. 5 Ω I = 0.2 A - + 6 V 9 V 10 Ω V + 9 6 5 (0.2) = V V = V V = 9 + 6 + 5 (0.2) = 2 V
Exmple: clculte I, V, nd V for the circuit shown. The smrt wy: V = - V = +2 V 5 Ω I = 0.2 A - + 6 V 9 V 10 Ω V 10 (0.2) = V V = V V = + 10 (0.2) = + 2 V
Exmple: clculte I, V, nd V for the circuit shown. Wht if you guess the wrong current direction? 5 Ω I - + 6 V 9 V 10 Ω 10 I 5 I +6 9 = 0 15 I = 3 I = 0.2 A oops, guessed wrong direction, no ig del!
DC Currents in Physics 2135, (lmost) ll currents re direct current. d.c. currents flow in one direction, from + to -. we will not encounter c (lternting currents) much nlysis of c current is more complex household current is c, ssuming dc will e close enough to give you feel for the physics.
Tody s gend: Potentil Chnges Around Circuit. You must e le to clculte potentil chnges round closed loop. Electromotive force (Emf), Terminl Voltge, nd Internl Resistnce. You must e le to incorporte ll of the ove quntities in your circuit clcultions. Electric Power. You must e le to clculte the electric power dissipted in circuit components, nd incorporte electric power in work-energy prolems. Exmples.
Idel versus rel voltge sources idel ttery (or other voltge source): voltge does not depend on the current flowing rel ttery: voltge does depend on current, typiclly voltge decreses with incresing current (lod) How to model rel ttery? rel ttery consists of idel ttery + internl resistnce Voltge of idel ttery is clled electromotive force (emf) ε r ε internl resistnce r
EMF nd terminl voltge The electromotive force (emf) of voltge source is the potentil difference it produces when no current is flowing. Cn the emf e mesured? emf is not force! hook up voltmeter: ε (emf) s soon s you connect the voltmeter, current flows you cn only mesure terminl voltge V, ut not emf ε I An idel voltmeter would e le to mesure ε. The ttery is everything inside the green ox.
To model ttery, simply include n extr resistor to represent the internl resistnce, nd lel the voltge source* s n emf insted of V (units re still volts): r ε If the internl resistnce is negligile, simply don t include it! If you re sked to clculte the terminl voltge, it is just V = V V, clculted using the techniques I m showing you tody. (Terminl voltge is usully expressed s positive numer, so it is etter to tke the solute vlue of V.) *Rememer, ll sources of emf not just tteries hve n internl resistnce.
Exmple: ttery is known to hve n emf of 9 volts. If 1 ohm resistor is connected to the ttery, the terminl voltge is mesured to e 3 volts. Wht is the internl resistnce of the ttery? Loop rule: R=1 Ω ε - Ir - IR = 0 I emf terminl voltge V = V -V = IR r ε terminl voltge V
ε - Ir - IR = 0 Ir = ε - IR V = IR V I = R r = r = ε - IR I ε - R I I R=1 Ω emf r = εr V - R ε r = R - 1 V 9 r = 1-1 = ( 3-1 ) = 2Ω 3 A rther unrelisticlly lrge vlue for the internl resistnce of 9V ttery.
Tody s gend: Potentil Chnges Around Circuit. You must e le to clculte potentil chnges round closed loop. Emf, Terminl Voltge, nd Internl Resistnce. You must e le to incorporte ll of the ove quntities in your circuit clcultions. Electric Power. You must e le to clculte the electric power dissipted in circuit components, nd incorporte electric power in work-energy prolems. Exmples.
Electric Power power in terms of work done y force: P F = dw dt work done y electric force moving chrge q through potentil difference V for infinitesiml chrge dq instntneous power: W = i f U i U = f q(v i V f). dw = i f dq (V i V f). dw dq (V V ) dq V dt dt dt i f i f P = = =. F
dw dq P = = V = IV. dt dt P = IV P<0 mens loss of electric energy P>0 mens gin of electric energy For resistors: using Ohm s lw V=IR, we cn write P = I 2 R = V 2 /R.
Exmple: n electric heter drws 15.0 A on 120 V line. How much power does it use nd how much does it cost per 30 dy month if it opertes 3.0 h per dy nd the electric compny chrges 10.5 cents per kwh. For simplicity ssume dc current. Household current is c rther thn dc. Our clcultion will e resonle pproximtion to relity.
An electric heter drws 15.0 A on 120 V line. How much power does it use. P = IV P = ( 15 A)( 120 V) = 1800 W = 1.8 kw How much does it cost per 30 dy month if it opertes 3.0 h per dy nd the electric compny chrges 10.5 cents per kwh. 3 h $0.105 cos t = ( 1.8 kw)( 30 dys ) dy kwh cos t = $17.00
How much energy is kilowtt hour (kwh)? ( 1 kw)( 1 h) = ( 1000 W)( 3600 s) J = 1000 3600 s s ( ) 6 = 3.6 10 J So kwh is funny unit of energy. K (kilo) nd h (hours) re lowercse, nd W (Jmes Wtt) is uppercse.
How much energy did the electric heter use? P verge W done y force = = time Energy Trnsformed time Energy Trnsformed = ( Pverge )( time) J 3 h used 3600 s Energy Trnsformed = 1800 ( 30 dys) s dy h Energy Trnsformed = 583, 200, 000 Joules used
Exmple: A 12 V ttery with 2 Ω internl resistnce is connected to 4 Ω resistor. Clculte () the rte t which chemicl energy is converted to electricl energy in the ttery, () the power dissipted internlly in the ttery, nd (c) the power output of the ttery. R=4Ω I r=2ω ε = 12 V
Clculte () the rte t which chemicl energy is converted to electricl energy in the ttery. I R=4Ω V = r=2ω ε = 12 V 0 round ny closed circuit loop Strt t negtive terminl of ttery ε - I R 2Ω - I R 4Ω = 0 I = ε / (R 2Ω + R 4Ω ) = 12 V / 6 Ω = 2 A Energy is converted t the rte P converted =Iε=(2 A)(12 V)=24W.
Clculte () the power dissipted internlly in the ttery. R=4Ω I=2A r=2ω ε = 12 V P dissipted = I 2 r = (2 A) 2 (2 Ω) = 8 W. Clculte (c) the power output of the ttery. P output = P converted - P dissipted = 24 W - 8 W = 16W.
Clculte (c) the power output of the ttery (doule-check). R=4Ω I=2A r=2ω ε = 12 V The output power is delivered to (nd dissipted y) the resistor: P output = P resistor = I 2 R = (2 A) 2 (4 Ω) = 16W.
Exmple: 3 volt nd 6 volt ttery re connected in series, long with 6 ohm resistor. The tteries* re connected the wrong wy (+ to + or - to -). Wht is the power dissipted in the resistor? In the 3 volt ttery? 6 Ω I - + 3 V 6 V Strting t point... + 6 3 6 I = 0 I = (6 3) / 6 = 0.5 A *Assume zero internl resistnce unless the prolem suggests otherwise.
Wht is the power dissipted in the resistor? 6 Ω I = 0.5 A - + 3 V 6 V P R = I 2 R = (0.5) 2 (6) = 1.5 W Wht is the power dissipted in the 3 volt ttery? P 3V = IV = (0.5) (3) = 1.5 W Note: P 6V = IV = (0.5) (6) = 3 W = P R + P 3V