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a Jonathan Dowling Physics 2102 Lecture jdowling/PHYS21024SP07/lectures/lecture11.pdf · PDF file lecture11.ppt Author: Jonathan Dowling Created Date: 2/11/2007 9:25:58 PM

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  • Physics 2102Physics 2102 Lecture 11Lecture 11

    DC CircuitsDC Circuits

    Physics 2102

    Jonathan Dowling

    b

    a

  • Incandescent light bulbsIncandescent light bulbs (a) Which light bulb has a smaller resistance: a 60W, or a 100W one? (b) Is the resistance of a light bulb different when it is on and off? (c) Which light bulb has a larger current through its filament: a 60W one, or a

    100 W one? (d) Would a light bulb be any brighter if used in Europe, using 240 V outlets? (e) Would a US light bulb used in Europe last more or less time? (f) Why do light bulbs mostly burn out when switched on?

  • b

    a

    The battery operates as a “pump” that moves positive charges from lower to higher electric potential. A battery is an example of an “electromotive force” (EMF) device.

    These come in various kinds, and all transform one source of energy into electrical energy. A battery uses chemical energy, a generator mechanical energy, a solar cell energy from light, etc.

    The difference in potential energy that the device establishes is called the EMF and denoted by Ε.

    EMF devices and single loop circuitsEMF devices and single loop circuits

    Ε = iR

    a b c d=a

    Va

    Ε iR

    ba dc − + i

    i

  • Given the emf devices and resistors in a circuit, we want to calculate the circulating currents. Circuit solving consists in “taking a walk” along the wires. As one “walks” through the circuit (in any direction) one needs to follow two rules:

    When walking through an EMF, add +E if you flow with the current or −E otherwise. How to remember: current “gains” potential in a battery.

    When walking through a resistor, add -iR, if flowing with the current or +iR otherwise. How to remember: resistors are passive, current flows “potential down”.

    Example: Walking clockwise from a: + E-iR=0. Walking counter-clockwise from a: − E+iR=0.

    Circuit problemsCircuit problems

  • If one connects resistors of lower and lower value of R to get higher and higher currents, eventually a real battery fails to establish the potential difference Ε, and settles for a lower value. One can represent a “real EMF device” as an ideal one attached to a resistor, called “internal resistance” of the EMF device:

    The true EMF is a function of current: the more current we want, the smaller Etrue we get.

    Ideal batteries vs. real batteriesIdeal batteries vs. real batteries

    Etrue = E –i r

    E –i r − i R=0 → i=E/(r+R)

  • Resistors in series and parallelResistors in series and parallel An electrical cable consists of 100 strands of fine wire, each having 2 Ω resistance. The same potential

    difference is applied between the ends of all the strands and results in a total current of 5 A.

    (a) What is the current in each strand? Ans: 0.05 A

    (b) What is the applied potential difference? Ans: 0.1 V

    (c) What is the resistance of the cable? Ans: 0.02 Ω

    Assume now that the same 2 Ω strands in the cable are tied in series, one after the other, and the 100 times longer cable connected to the same 0.1 Volts potential difference as before.

    (d) What is the potential difference through each strand? Ans: 0.001 V

    (e) What is the current in each strand? Ans: 0.0005 A

    (f) What is the resistance of the cable? Ans: 200 Ω

    (g) Which cable gets hotter, the one with strands in parallel or the one with strands in series? Ans: each strand in parallel dissipates 5mW (and the cable dissipates 500 mW); each strand in series dissipates 50 µW (and the cable dissipates 5mW)

  • DC circuits: resistances in seriesDC circuits: resistances in series Two resistors are “in series” if they are connected such that the same current flows in both. The “equivalent resistance” is a single imaginary resistor that can replace the resistances in series.

    “Walking the loop” results in : E –iR1-iR2-iR3=0 → i=E/(R1+R2+R3)

    In the circuit with the equivalent resistance, E –iReq=0 → i=E/Req

    Thus,

    ! =

    = n

    j

    jeq RR 1

  • Multiloop Multiloop circuits: resistors in parallelcircuits: resistors in parallel Two resistors are “in parallel” if they are connected such that there is the same potential drop through both. The “equivalent resistance” is a single imaginary resistor that can replace the resistances in parallel.

    “Walking the loops” results in : E –i1R1=0, E –i2R2=0, E –i3R3=0 The total current delivered by the battery is i = i1+i2+i3 = E/R1+ E/R2+ E/R3. In the circuit with the equivalent resistor, i=E/Req. Thus,

    ! =

    = n

    j jeq RR 1

    11

  • Resistors and CapacitorsResistors and Capacitors

    Resistors Capacitors

    Key formula: V=iR Q=CV

    In series: same current same charge Req=∑Rj 1/Ceq= ∑1/Cj

    In parallel: same voltage same voltage 1/Req= ∑1/Rj Ceq=∑Cj

  • A V

    R

    V i 5.1

    8

    12 =

    ! ==

    ExampleExample

    Bottom loop: (all else is irrelevant)

    8Ω12V

    Which resistor gets hotter?

  • ExampleExample

    a) Which circuit has the largest equivalent resistance?

    b) Assuming that all resistors are the same, which one dissipates more power?

    c) Which resistor has the smallest potential difference across it?

  • ExampleExample Find the equivalent resistance between points (a) F and H and (b) F and G. (Hint: For each pair of points, imagine that a battery is connected across the pair.)

  • Monster MazesMonster Mazes If all resistors have a resistance of 4Ω, and all batteries are ideal and have an emf of 4V, what is the current through R?

    If all capacitors have a capacitance of 6µF, and all batteries are ideal and have an emf of 10V, what is the charge on capacitor C?

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