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    Teacher Notes Ohm'S Law

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    Alib Budiyanto

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    Teacher Notes Ohm'S Law

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    Alib Budiyanto
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    tchr_ohm

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    31 views8 pages

    Teacher Notes Ohm'S Law

    Uploaded by

    Alib Budiyanto

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    © All Rights Reserved
    Download as DOC, PDF, TXT or read online from Scribd
    Download as doc, pdf, or txt
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    TEACHER NOTES Ohm’s Law

    Georg(sic) Ohm 1787-1854(see Encarta)


    Professor Laithwaite of Imperial College was the inspiration behind this
    program. He encouraged the use of plumbing analogies for teaching
    electricity, and this program is dedicated to his memory.
    The plumbing analogy is designed to try and encourage class discussion
    and to consider the many similarities between the way electricity and
    water flow. By toggling between the circuit and water screen the
    similarities can be emphasised. There is a similarity to the plumbing
    circuit and a central heating system, i.e. there is a pump and the
    radiators act as resistors, and also there is a flow and return. A lot of
    physics can be taught by considering analogies and models.
    Students also find it interesting to consider where the analogy
    breaks down such as if there were a hole in the water pipe. If
    you look at the water circuit, water would flow out from the
    top of the pressure pipes, so they should be longer………
    No. Answer Description
    1 3(C) Definition and hint available
    2 3(C) If the voltage is doubled the current doubles
    3 4(D) If 3V applied to 3 W, I = V/R = 3/3 = 1 A. Hint
    4 2(B) Temperature kept constant for Ohm’s Law. Hint
    5 1(A) Gradient gives value of resistance R = V/I
    6 2(B) If 2mA through 3 kW, V = IR = 2x10-3x3x103= 6 V needed. Hint
    7 2BFalse If the temperature rises, the resistance rises
    8 4(D) R = V/I Rearranging equations and magic triangle. Hint
    9 3(C) If 6V applied to 2 W, I = V/R = 6/2 = 3 A. Hint
    10 1(A)Yes Graph linear i.e. a straight line
    11 2(B) Ammeter should have zero resistance and be in series. Hint
    12 4(D) Voltmeter should have infinite resistance and be in parallel.Hint
    13 1(A) Same current flows through both resistors in series. Hint
    14 2(B) Same voltage applied across both resistors in parallel. Hint
    15 2(B) Fuse melts and breaks the circuit
    16 2(B) Bulbs A and B will light, the switch for C is open
    17 4(D) Total resistance is 6+4=10 W and I = V/R = 10/10 = 1 A
    18 4(D) Two batteries and a single bulb will be the brightest
    19 1(A) Diode acts in a similar way to a heart valve. Hint
    20 1(A)Yes Diode rectifies (allows current flow in one direction)
    Some web pages about electricity are available at:

    Copyright-free from www.physconcepts.co.uk


    http://www.bbc.co.uk/schools/gcsebitesize/physics/electricity/index.shtml

    Copyright-free from www.physconcepts.co.uk


    Ohm’s Law Worksheet
    1. Define Ohm’s Law
    The current flowing through a resistor is directly proportional to the potential
    difference across its ends, provided the temperature remains constant.
    2. Why is it better for the batteries to use the kilo ohm rather than
    an ohm range of resistors?
    If the resistance is much higher, the current will be much lower and the
    batteries will not go flat very quickly e.g. with 6V and 1the current will be
    6A. This could lead on to a discussion of fuses in the home, battery life for
    walkmen, minidisks, mobiles etc. and the advantages and disadvantages of
    rechargeable batteries.
    3. Calculate data for the following chart and use the program to
    check the values, use I=V/R
    Resistor/ 1 k 2 k 3 k
    Voltage/V Current/A Current/A Current/A
    6.0 6.0 3.0 2.0
    4.5 4.5 2.25 1.5
    3.0 3.0 1.5 1.0
    1.5 1.5 0.75 0.5
    This can be done as a class exercise or individually.
    4. Now try plotting a graph of Voltage against Current for each
    resistor on the same graph.
    In this graph plotting exercise the weaker students could have a simple grid
    photocopied for them with the axes already drawn.

    For the steepest line


    R=6.0/2 = 3 W = 3 W
    For the middle line
    R=6.0/3 = 2 W = 2 W
    For the least steep line
    R=6.0/6 = 1 W = 1 W

    Note X axis in Amps


    5. Measure the gradient of each line, do you notice anything?
    R=V/I, so the gradient gives the value of the resistor provided Ohm’s Law is
    obeyed. This is the main reason why the graph is drawn this way round. Some
    textbooks draw I against V so the straight-line gradient is 1/R, which can be
    confusing.

    Copyright-free from www.physconcepts.co.uk


    Ohm’s Law Practical
    1. Connect up the circuit and take readings of current and voltage.
    Stress laying out the components first, and connecting the ammeter in series
    and the voltmeter in parallel. A train-set/Scalectric analogy might be useful for
    the series part of the circuit.
    N.B. To keep the voltage as constant as possible use low internal resistance
    batteries such as Duracell.

    2. Now change the values of the resistor and the number of batteries
    and complete the following table

    Approx. Resistor
    Values 1 k 2 k 3 k
    Number of Voltage Current Current Current
    Batteries /V /mA /mA /mA
    4 6.0 6.0 3.0 2.0
    3 4.5 4.5 2.25 1.5
    2 3.0 3.0 1.5 1.0
    1 1.5 1.5 0.75 0.5
    0 0.0 0.0 0.0 0.0

    3. Are the values exactly the same as the ones in the worksheet or
    program, if not can you explain this?
    No. In this table the resistors are a thousand times bigger, and the current is
    not in amps but in milliamps which are a thousand times smaller. The resistors
    are normally accurate to about ±10%, also the multimeter will probably have a
    little inaccuracy, so some variation in the readings will be normal. Some variation
    in battery voltage is also inevitable, especially when they start running down.

    Copyright-free from www.physconcepts.co.uk


    4. Now try plotting a graph of Voltage against Current for each
    resistor on the same graph. Reasonably able students can use graph
    paper.

    N.B. X axis in milliamps

    Measure the gradient of each line, do you notice anything?

    R=V/I, so the gradient gives the value of the resistor


    For the steepest line R=6.0/2x10 -3 = 3000 W = 3 kW
    For the middle line R=6.0/3x10 -3 = 2000 W = 2 kW
    For the least steep line R=6.0/6x10 -3 = 1000 W = 1 kW

    The circuit below can also be used for a full investigation of Ohm’s
    Law.

    Other components can be used such as a bulb (probably with a 25 W variable


    resistor) or a resistance wire. This can be done as coursework.

    Copyright-free from www.physconcepts.co.uk


    Diode Practical

    Take a reading of the milliammeter, reverse the direction of the


    diode and then take another reading. What did you notice?
    In the forward direction the current should be about 1mA, and when reverse
    biased it should be 0mA. The more able students could investigate the
    characteristics of a diode.
    Diode Characteristics

    Connect up the circuit as shown, and vary the value of the


    variable resistor. Draw up a table of current and voltage.
    Reverse the battery and record these readings. Plot the
    current against forward and reverse voltage. Check with
    your teacher to see what the graph should look like.
    The diode should trigger at about 0.6 V, and the current should shoot up
    to about 10 mA or more. When reverse biasing the voltage can be –6.0V
    and the current should be 0 mA.

    Very able students could consider the


    ‘actual’ resistance of the diode. This
    will not be the gradient, but the ratio
    of V/I at any given point.
    This concept is normally covered in AS
    physics.

    Copyright-free from www.physconcepts.co.uk


    ALTERNATING CURRENT A.C.
    No. Answer Description
    1 2(B) A.C. flows in BOTH directions
    2 4(D) Diode rectifies A.C.
    3 4(D) D.C. comes from batteries
    4 1(A) Yes. Hint shows that flow can be stationary
    5 3(B) Sine curve represents current and/or voltage against time
    This latest simulation uses the concept of a piston executing S.H.M. (simple
    harmonic motion). The flow (shown by the bending reed) and the voltage (shown by
    the pressure pipes) varies according to the sine curve which can be hidden or
    shown.
    Discussion
    (1) Sketch what you think A.C. would look like after it has been
    passed through a diode.
    Half-wave rectified.

    (2) How long does it take to charge up a typical mobile ‘phone?

    Normally about 2 hours.

    (3) What is the big heavy object inside a mobile ‘phone charger?

    Transformer. This might be a good place to talk about power supplies, step up
    and step down transformers and efficient power transmission and electrical
    safety.

    (4) Why does the charger get hot at the beginning of the charging
    cycle?

    Initially a large current flows onto the flat battery causing heating of the
    transformer coils as they have some resistance (I 2R factor).

    (5) Where in the home would you find D.C. power supplies?

    Radio, stereo, answer-‘phone, television, video, DVD player, computer, lots of


    devices have their own external power supplies.

    Copyright-free from www.physconcepts.co.uk


    Electrical Problems/Homework/Test
    1. A bulb of 10 W carries a current of 1.2 A, what voltage is
    across the bulb?
    V=IR
    V = 1.2 x 10 = 12 V
    2. A car's headlamps pass a current of 15 A. If the car's
    battery is 12 V, what is the overall resistance of the
    headlamp circuit?
    R = V/I
    R = 12/15 = 0.8 W
    3. A 1 MW resistor is attached across a 12 kV power supply,
    what current will flow?
    I = V/R
    I = 12000/1000000 = 0.012 A = 12 mA
    Chance to discuss milli and Mega together with use of calculators,
    and mention that both forms of answer acceptable.
    4. A hand-held neon screwdriver passes a current of
    0.01 mA when a 240 V terminal is touched, what is the
    total resistance of the circuit?
    R = V/I
    R = 240/0.01x10-3 = 24000000 W = 24.0 MW

    5. A 12 V battery is connected across a pair of 4 W and 2 W


    resistors in series. What current flows in each resistor,
    and what is the voltage across each one?
    Total resistance = 4+2 = 6 W Total current I = V/R = 12/6 = 2 A
    For 4W resistor V = IR = 2 x 4 = 8 V
    For 2W resistor V = IR = 2 x 2 = 4 V

    Copyright-free from www.physconcepts.co.uk

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