Simple Diode Circuit

Discussion in 'Homework Help' started by marineoceania, Dec 1, 2008.

  1. marineoceania

    Thread Starter New Member

    Dec 1, 2008
    2
    0
    hey everybody, I have a simple question for you guys. On my homework I need to find the current flowing in the circuit.

    [​IMG]


    The battery = 5 volts

    R1 = 250 Ω
    R2 = 180 Ω

    I know the first (top left) corner is 5 volts and the bottom two are 0 volts (?). How do I find the current and the corner before the diode?

    Thanks all!
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    If you assume the diode has a voltage drop of 0.7V then the current through th circuit is:

    I=(5-0.7)/(R1+R2)
     
  3. marineoceania

    Thread Starter New Member

    Dec 1, 2008
    2
    0
    hmm makes sense, but why would you add both resistors? Just curious haha
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    It's a series circuit.
     
  5. leftyretro

    Active Member

    Nov 25, 2008
    394
    2
    Because there is only one current path and it has to pass through both resistors. The current doesn't know how many resistors there are, 1 or many, it just feels the total series resistance. One of the Kirchhoff's circuit laws covers the topic.
     
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