Voltage After a Single Resistor or Between Two?

Discussion in 'Homework Help' started by speedster239, Jun 9, 2010.

  1. speedster239

    Thread Starter New Member

    Nov 24, 2009
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    I understand how voltage drop is calculated when two or more resistors are in series and you're attempting to calculate the voltage drop between a resistor, but what about in these two situations when you're trying to calculate the voltage drop after or between two resistors?

    This might seem like a silly question to those of you with tons of experience with electricity, but I'm at a loss to figure this out. I really appreciate your help.

    [​IMG]

    or

    [​IMG]
     
  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    An open circuit has no current, which results in having the source voltage at the open circuit

    which basically means that if you open a circuit and measure voltage at that point, u will get the voltage as same as the source voltage.
    Does not matter where you open the circuit.
     
  3. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    On the other hand, you cannot measure a voltage on a conductor.
    Unless you have a highly sensitive milivoltmeter
     
  4. Ghar

    Active Member

    Mar 8, 2010
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    Voltage drops add as you go around a loop.

    In your second picture (assuming it's a closed circuit, not as drawn) you have two 6 ohm resistors.
    The current in your circuit is 12 V / (6 + 6) = 1 A.

    The voltage drop across each resistor is 1A * 6 ohms = 6 V.

    The voltage in the middle is the battery voltage 12, minus one drop, 6, giving you 6 V in the middle node.
     
  5. speedster239

    Thread Starter New Member

    Nov 24, 2009
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    Yes, both of the schematics are meant to show a closed circuit, sorry.

    Why is it only minus one drop? I don't understand that fully. Why aren't both resistors accounted for? Or are they?
     
  6. Ghar

    Active Member

    Mar 8, 2010
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    There are 3 voltage 'drops'.

    6 across the first resistor, 6 across the second resistor, and -12 across the battery (so it's actually a voltage rise)

    The sum of the 3 is 0:
    6 + 6 - 12 = 0

    This is true around any closed loop.

    So if you put the negative of the battery as 0, the positive of the battery is +12.
    Go through one resistor, it's 12 - 6 = 6.
    Go through both resistors and you're back at the negative of the battery, 12 - 6 - 6 = 0.

    This is Kirchoff's voltage law and is one of the basic tools in circuit analysis.

    This is the AAC chapter you want:
    http://www.allaboutcircuits.com/vol_1/chpt_6/index.html
     
  7. speedster239

    Thread Starter New Member

    Nov 24, 2009
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    Thank you guys for your help, I figured that last question out. Now I have another one...

    I know that you can use a transistor to regulate current by allowing current to flow between the base and emitter terminals, which will in turn allow a larger current to flow between the collector and emitter terminal.

    Is there a way that I can wire up such a "switch" in the exact opposite sense. Ie, if I allow electricity to flow between the base and emitter terminals, current can NOT flow between the collector and emitter terminal, but while current isn't flowing between base and emitter it CAN flow between the collector and emitter.

    Is there a way to wire this up?

    Thanks again,
    Vaughan
     
  8. Ghar

    Active Member

    Mar 8, 2010
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    Not like how you explained because that's just how a transistor works. However you can have the same functional result by using the transistor in a different way or by using a combination of transistors.
    I'm assuming you just want a switch that's off when the input signal is high?
     
  9. speedster239

    Thread Starter New Member

    Nov 24, 2009
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    Yes indeed. And when there is no input the current does flow. How would I accomplish such a task?
     
    Last edited: Jun 11, 2010
  10. Ghar

    Active Member

    Mar 8, 2010
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    The best circuit depends on what exactly you're trying to do...

    A simple approach is just using two of these back to back (collector output of first goes to base input of second):
    http://www.ecelab.com/switch-bjt.htm

    Overall you get the behaviour you're asking for.
     
  11. Georacer

    Moderator

    Nov 25, 2009
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    If you have small current demands, you could use a PNP transistor. Unlike NPN, PNP transistors, allow current to flow from the Emitter to the Collector when the Base is kept at the Low voltage. If you want to stop current from flowing, raise the Base to High voltage. Be careful of the difference in pin layout between PNP and NPN transistors. And remember that the maximum current you can draw through the transistor is a manufacurer parameter you cannot overcome.
     
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