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Simple logic diagram-with switch and resistor

  1. Jun 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Why is it that the output is equal to V+ (when the switch is open) in diagram A, if there is no current flowing?

    Why is it that A (1/0) is used by convention? I mean the lamp I got at home has a "0" for off and "1" for on, on the switch. This is confusing me.

    2. Relevant equations


    High= 1
    Low = 0

    3. The attempt at a solution

    Not sure at all.
     

    Attached Files:

  2. jcsd
  3. Jun 19, 2010 #2
    Suggestion: Apply the voltage divider rule assuming the switch to be a resistor which can have one of two values (∞ or 0) depending on whether it's open or closed.
     
  4. Jun 20, 2010 #3
    leave voltage diving rules and others.

    very simply,
    if upper resistor (called Pull up resistor)-----is present while lower (called Pull down)
    resister is not present,
    that means that total amount of current coming from power supply V+,
    will pass through the o/p node.
    thus,
    o/p node gets full amount of current coming from power supply V+.


    ** the reason of your confusion is,
    there is no connection made to ground, then how does the current flows through network.
    but, there is always a load (or any other network, that is connected to GND obviously.)
    so, current flows through that path.

    Hope your confusion will be solved.
    Best of Luck.
     
  5. Jun 20, 2010 #4
    Oh thanks. I also want to know if there is only 1 resistor in a circuit, why is it that the total V+ will be dropped across the resistor, even if V=100V, I mean what if that resistor is small and only consumes a bit of V?
     
  6. Jun 20, 2010 #5
    Resistors don't consume V. They flow a current proportional to whatever V is applied to their terminals. They consume power which is V*I.

    The voltages in a circuit loop must total to 0.
     
  7. Jun 20, 2010 #6

    yes.
    that's right.
    whenever that type of ckt you will see
    you think like a o/p network is attached with it.
    this type of ckt is made for this purpose
    you think, that without any network or resister at o/p ,
    how can you measure the voltage.
    even if you measyre with a voltmeter or multi-meter, there will also be a resitor to measure it.
    and low resistance case is discussed in just above post.
     
  8. Jun 20, 2010 #7
    While resistors don't "consume" voltage they do "drop" voltage. That is why I suggested using the voltage divider rule (substituting a resistance for the switch) and see what happens as you vary the bottom resistance from some finite value down to zero and then up to infinity.
     
  9. Jun 24, 2010 #8
    Why also is it that current is defined as I = V/R all over a circuit? As in why is it that the current where the circuit has no resistance different? e.g. voltage line between 2 resistors and line between earth and a resistor.
     
  10. Jun 24, 2010 #9

    berkeman

    User Avatar

    Staff: Mentor

    A real wire has a real (but very small) resistance. So the current flowing through it does generate a small voltage drop. You can figure out how much the drop will be, based on wire resistance tables and the current that is flowing.

    Wire resistance table: http://en.wikipedia.org/wiki/American_wire_gauge

    .
     
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