black box guessing

Discussion in 'Homework Help' started by prescott2006, Oct 6, 2009.

  1. prescott2006

    Thread Starter Active Member

    Nov 8, 2008
    72
    1
    if a black box was given,with two wire pull out from the box.no other additional info was given,then how will you determine the impedance in the circuit?what is the resistance?if given that is a passive circuit,what will the circuit look like?
    seem the task quite challenging.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    Depending on the device inside, what do you think a varying frequency signal might do? Especially if you have a fixed resistor in series with the black box?
     
  3. ELECTRONERD

    Senior Member

    May 26, 2009
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    You couldn't, you would need information such as the inductance, capacitance, and resistance of the box. Using the following equations, you can calculate the reactance and add that with the resistance to form the overall impedance of the black box.

    XL = 2∏fL - Where ∏ = Pi or 3.14, f = frequency, and L = Inductance value.

    XC = 1 / 2∏fC - Where C = Capacitance value.

    You can use either equation, the black box might have a negligible amount of inductance, nothing worth being concerned about, or vice verse.
     
  4. prescott2006

    Thread Starter Active Member

    Nov 8, 2008
    72
    1
    i think you might be misunderstanding my question.we are given a plastic black box(just prevent us from see the circuit) ,then two wire pull out,inside the box have a circuit.then resistance and impedance has to be found.if given the circuit is passive circuit,what will the circuit may look like?
     
  5. RiJoRI

    Well-Known Member

    Aug 15, 2007
    536
    26
    No, you'll need to (a) find the resistance of what is in the box (Good Ol' Multimeter), then (b) watch the impedance as an alternating current at different frequencies passes through the box.

    From this you can work out the box's impedance, and capacitance and/ or inductance.

    Or, if your GOM has a capacitance meter and/or an inductance meter, that can make it a little easier.

    --Rich
     
  6. elirentz

    New Member

    Oct 6, 2009
    1
    0
    It may help if you worded your question more clearly.

    I'll give it a shot though. To find out what the circuit inside could possibly look like you must apply a signal or input and measure the current coming out of or voltage accross the element (black box) to determine the resistance or impedance. Then based on how the box effects the input you can get an idea of what could be going on in there. There are many possibilities of what the circuit can look like for each output characteristic.

    This is basically the same answer others have given but without the math.
     
  7. ELECTRONERD

    Senior Member

    May 26, 2009
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    Yep, that's what I tried to explain. Was it clear enough for the op?
     
  8. GetDeviceInfo

    Senior Member

    Jun 7, 2009
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    I would pulse the box and look for a time constant that you could calculate back to an RC or RL load equation. If you know that it's simply resistance, measure it.
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Suppose it's an incandescent light globe with a non-linear resistance .....

    Would one anticipate non-linearity as a possible (unknown) condition?

    So a multimeter observation followed by an single point AC or pulse analysis may leave one none the wiser. Multiple observations at different conditions may then be required to eliminate or confirm the presence of non-linearity.
     
  10. prescott2006

    Thread Starter Active Member

    Nov 8, 2008
    72
    1
    could we probably know the circuit configuration in the box by guessing?given the circuit inside is passive circuit.
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Suppose you are given a clue which states that the circuit comprises 10 linear resistors of equal resistance. You measure 10 Ω across the two wires. Is there a unique circuit configuration which satisfies this result - given the aforementioned "clue"?
     
  12. GetDeviceInfo

    Senior Member

    Jun 7, 2009
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    That is most true. In pulsing however, one would not terminate until a steady state was achieved. Not much sense in trying to evaluate a time constant if you haven't let it time out.
     
  13. syed_husain

    Active Member

    Aug 24, 2009
    61
    5

    what i understand, this is a thevenin ( voltage source and resistor in series) or norton equivalent ( a current source & a resistor in parrallel) circuit. just measure the open cct voltage (Voc) and then short the two wire and measre the short circuit current (Isc). So, the resistance will be Voc/Isc. but, for impedance u need a meter like Fluke that is capable of measuring power factor. with that u can able to measure the impedance.
     
  14. beenthere

    Retired Moderator

    Apr 20, 2004
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    Simply apply an AC signal through the fixed external resistor and the box. Vary the frequency.
     
  15. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    My turn,

    How about,
    If it is purely passive,
    He only wants to know what impedance is being reflected not the actual reactances themselves.

    1. put an ohm meter across the wires. record the value. To check if it's a pure resistance or not.

    2. Put a POT.in series with the wires to ground and place a voltmeter across the POT. and supply the network with a freq. varying the POT. until a voltage drops across the POT. is around 1/2 the supplied voltage.
    May also have to vary the frequency to get a reading that's obtainable, to use.

    But if he needs to know what components are being used, then:

    when the voltage drops by 1/2, then
    1. put a variable inductance and a ammeter in series with this network to check for series resonance.

    2.Then put a variable capacitor in series and do the same to check for series resonance.

    3. then put a inductor in parralell to check for parrallel resonance, and do the same with a capacitor in parrallel.
     
    Last edited: Oct 11, 2009
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