Study help only, not test help.

Discussion in 'Homework Help' started by pfelectronicstech, Apr 13, 2012.

  1. pfelectronicstech

    pfelectronicstech Thread Starter Member

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    I'm confident in that i ruled out the other choices. So i am right?
  2. Ron H

    Ron H E-book Developer

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    You are, but can you name each device?
  3. mlog

    mlog Member

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    I figure the answer was B. You definitely want to turn off the power in the circuit, because it could damage the meter or cause an error in the resistance reading by producing an unwanted current in the meter.

    As for whether you start with a high or low scale, I would start with the low scale and work my way up. Why? If there are semiconductors in the circuit, then the lower scale produces a lower voltage, which is less likely to force a semiconductor junction into conduction. Forward biased semiconductors can result in an error in resistance measurement, and that was the basis of the question.
  4. Ron H

    Ron H E-book Developer

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    Lower resistance scales force higher current, not lower.
  5. mlog

    mlog Member

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    I always thought that the higher resistance ranges often use a higher voltage battery than the one used for the low resistance ranges. At least that's how many analog meters work.

    I found this reference on a website.

    http://www.electronics-radio.com/articles/test-methods/meters/multimeter-diode-transistor-test.php

    "The Avo has three resistance ranges and no diode marking. However it is important that you only use the middle of the three ranges (usually marked x100 on modern meters). The reason for this is that the x1 range does not produce enough current for testing junctions, and more importantly the x10K range uses a 15V battery to supply test voltages to the probes. Putting 15V across a semiconductor junction whose forward voltage is only 0.6V will more than probably destroy it!"

    Anyway, my point was not so much about the current, but about the voltage that could forward bias a semiconductor junction if it's in the circuit under measurement.

    I looked up the specs for my Fluke 87 DMM, and it says the short circuit current (SCC) for the 400 ohm range is 200 uA. The SCC for the 4k range is 80 uA, 40k range is 12 uA, 400k range is 1.4 uA, and so on. The DMM current does go down in the higher ohm ranges, but my point was about an analog meter that uses a higher voltage in the higher ohms ranges. The test material seemed to be oriented around older technology, so it wouldn't be a stretch to think they used analog meters when the test was written.
    Last edited: Apr 17, 2012
  6. pfelectronicstech

    pfelectronicstech Thread Starter Member

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    Ok this is the last one I think I need help with. I've had some trouble in the past with knowing exactly what happens if a certain switch is pressed. So please any help will be of course appreciated.

    what will happen if switch S1 is momentarily pressed?
    A. Bulb B1 will light up for as long as the switch is pressed.
    B. Bulb B1 will go out and remain out after S1 is released.
    C. Bulb B1 will light up and remain lit after S1 is released.
    D. Bulb B1 will go out, and then light up after S1 is released.

    I think its A, but as I said I have some trouble with knowing what happens when certain switches are pressed.

    All of the components in Figure A-6 are operating correctly, and Bulb B1 is off. Which of the following
    actions must you take to make Bulb B1 light up?
    A. Momentarily press switch S2 to de-energize RL1.
    B. Momentarily press switch S1.
    C. Press and hold switch S1.
    D. Momentarily press switch S2 to energize RL

    Again I think its C, but what do you think?


    Thanks again for any help, any guidance.

    Attached Files:

  7. mlog

    mlog Member

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    No, the first question is not A. Look at the initial state before you push anything; lamp B1 is already powered, but the relay is in its "relaxed" state.

    Battery current flows through B1 through the relay contacts and back to the battery. Since the lamp is already lighted, that eliminates two of the answers.

    Pressing switch S1 simply provides current through the relay coil RL1. When the relay coil is powered, the relay contacts change state, which means the switch moves to the down position in the picture.

    As for the second question, it's not C. If the bulb B1 is already off -- and I'm assuming the user is taking no action, which means it's found a stable operating condition -- then you know the relay contact must be in the DOWN position in order to interrupt current to B1. What could make the relay contact be down? Current through the relay coil would activate the relay, so the likelyhood is that coil of RL1 is active, which mean it's latched with its contacts in the lower position.

    Let's talk about that "latch" condition. Notice there are two ways to power the coil of RL1: (a) Momentarily pressing S1 or (b) The relay contact is in the down position.

    If S1 is pressed, it causes current to flow through the coil, thus closing the contact downward. You now have a source of coil current even after you release S1. In other words, you have a latched relay. The contact will stay in the DOWN position until the battery depletes or until the coil current is somehow interrupted (think about switch S2).

    Finally, while the relay is latched, only the relay coil is powered. There is not current path through B1. What's your answer now?
    Last edited: Apr 17, 2012
  8. mlog

    mlog Member

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    Going back to the question about using a low resistance scale vs. a high resistance scale in a circuit, I found a manual on the venerable Simpson 260. Those of you who are old enough should remember that meter well. I have used it many times, either in school or at the work place.

    The Simpson 260 Series 8 VOM has 3 resistance ranges, Rx1, Rx100, and Rx10,000. The usable ranges in ohms are 1-2k, 0-200k, and 0-20M, respectively. The nominal short circuit current is 125ma, 1.25mA, and 75uA, respevitly. Here's the interesting part. The nominal open circuit voltage is 1.5V, 1.5V, and 9V, respectively.

    So while the current is higher in the lower resistance ranges, the open circuit voltage is lower in the lower ranges. If the VOM was used with the Rx10,000 range, then the 9V battery is in circuit, and it could forward bias any semiconductors that might happen to be in the circuit.

    What many of us were taught in technical school was to use a low to mid resistance range if possible while measuring resistances in circuit, especially if the circuit might include diodes or transistors. That way the chance of turning on a junction and producing a measurement error is reduced.
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