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Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Phy2049 •Circuits have capacitors and resistors.•Capacitors in parallel in series•Resistors in parallelin Series •Kirchhof’s two rules: voltage around the loop and currents at a junction.•P = iV = i2R = v2/R21CCCeq21111CCCeq21RRReq21111RRReqIn the diagram R1 > R2 > R3. Rank the three resistors according to the current in them, least to greatest.A. 1, 2, 3B. 3, 2, 1C. 1, 3, 2D. 3, 1, 3E. All are the sameI. In the diagram, the current in the 3-resistor is 4A. The potential difference between points 1 and 2 is:A. 0.75VB. 0.8VC. 1.25VD. 12VE. 20VA. 5 Ω B. 7 Ω C. 2 Ω D. 11 Ω E. NOTConsider the circuit shown in the figure. We assumethat the emf device is ideal and that the connectingwires have negligible resistance. A current flowsthrough the ciCurrent in a single loop circuitircuit in the clockwise direction.In a time interval a charge passes through the circuit. The battery is doing work . Using energy conservation we can set this amount of work equal to the rate at which heat isdt dq idtdW dq idt== =E Ei i2 generated on R. Kirchhoff put the equation above in the form of a rule known as Kirchhoff's l0oop rule(KLR for short)idt Ri dtRi iR= �= � - =EiE Ei i0iR- =EiThe algerbraic sum of the changes in potential encountered in a completetraversal of any loop in a circuit is equal to zero. KLR : The rules that give us algebraic sign of the charges in potential through a resistorand a battery are given on the next page.(27 – 5)Consider the circuit shown in the figure. There are three brances in it: Branch bad, bcd, and bd.We assign currents for each branch and define the current directions arbitrarily.Multiloop circuits : The method is selfcorrecting. If we have made a mistake in the directionof a particular current the calculation will yield a negative value and thus provide us with a warning. 1 2 31 3 21We asign current for branch bad, current for branch bcd, and current for branch bd. Consider junction d. Currents and arrive, while leaves.Charge is conserved thus we have : i i ii i ii +3 2. This equation can be fomulated as a more general principle knwon as Kirchhoff's junction rule (KJR)i i= The sum of the currents entering any junction is equal to the sum of the currentsleaving the junctionKJR : (27 – 11)Rimotion-V iRD =RimotionV iRD =+motion-VD = E+-motionVD =+E+-For a move through a resistancein the direction of the current, the change in the potential For a move through a resistance in the directionopposite to that of the current, th V iRD =-Resistance Rule : e change in the potential V iRD =+For a move through an ideal emf device in the direction of the emf arrow, the change in the potentialFor a move through an ideal emf device in a directionopposi te to that of the VD =+EMF Rule : Eemf arrow, the change in the potenti al VD =- E(27 – 6)R1 = 1 and R2 = 2 ὨE1 = 2, E2 = E3 = 4 VSize and direction of current in E1and E2Va-Vb. (a) We note that the R1 resistors occur in series pairs, contributing net resistance 2R1 in each branch where they appear. Since e2 = e3 and R2 = 2R1, from symmetry we know that the currents through e2 and e3 are the same: i2 = i3 = i. Therefore, the current through e1 is i1 = 2i. Then from Vb – Va = e2 – iR2 = e1 + (2R1)(2i) we get 2 11 24.0 V 2.0 V0.33A.4 4 1.0 2.0iR R       Therefore, the current through e1 is i1 = 2i = 0.67 A.(b) The direction of i1 is downward. (c) The current through e2 is i2 = 0.33 A.(d) The direction of i2 is upward.(e) From part (a), we have i3 = i2 = 0.33 A.(f) The direction of i3 is also upward.(g) Va – Vb = –iR2 + e2 = –(0.333 A)(2.0 W) + 4.0 V = 3.3 V.Kirchhof’s rules:Choose current (magnitude and direction) in all branches. At junction a (or b), I1 = I2+I3Consider the right box loop, go clockwise.I3(2R1)-E3+E2-I2R2 = 0  I3 = I 2 for the numbers above. And I1 = 2I2Now the left loop, clockwise:2 + 2I1 +2I2 - 4 = 0  I2 = 0.33 A = I3 I1 = 0.67 A The current directions are correct as chosen at the outset.R = 4 Ω, V = 4 V, what is the current through R?R = 4 Ω, V = 4 V, what is the current through R?4+4+4-4-4I=0 I = 2AObtain current in all branches of this circuit by repeated application of Kirchhof’s loop and junction rules•What is i? All V = 10 Volts and all R = 4 Ohms•What is i? All V = 10 Volts and all R = 4 Ohms?•What is j? What is k?jk•What is i? All V = 10 Volts and all R = 4 Ohmsjk-10 +10+10-10+10-10-10-10-10-10 +10 I = 0  I = 4A•Three circuits are connected to a battery. Rank them by the –final charge–Time


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UF PHY 2049 - Circuits

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