Question about voltage drop

Discussion in 'General Electronics Chat' started by fr0sty, Apr 6, 2008.

  1. fr0sty

    Thread Starter New Member

    Apr 1, 2008

    What exactly is 'voltage drop' in terms of a resistor? Is it the amount of voltage the resistor is using up from the voltage source? I'm playing around with a simulator and with a simple circuit that has just a battery (V1=9V) and a resistor (R1), the voltage across R1 is 9V. But when I add a second resistor, the voltage changes (I'd expect this from my reading, but am trying to understand why). Is it because the presence of the new resistor in series (R2) causes less current to flow through R1?
  2. bloguetronica

    AAC Fanatic!

    Apr 27, 2007
    You can say that it is the voltage "used" by the resistor. In other words, the voltage drop across any device is the voltage measured between the terminals of that device (resistor, diode, etc...).
  3. hgmjr

    Retired Moderator

    Jan 28, 2005
    You may find it helpful to review the material in the AAC ebook at this link.

  4. Papabravo


    Feb 24, 2006
    Ohm's Law REQUIRES, that any time a current flows through a resistance that there will be different voltages on the two ends. If you are an imaginary Maxwell's demon travelling around a loop in a circuit keeping track of voltages then going from a higher voltage to a lower voltage looks like a ..."voltage drop", and a battery or power supply looks like a "voltage rise". Imagine what happens if you travel around the loop in the reverse direction. The physics is the same; only the signs have changed. Kirchoff's voltage law still holds because if you reverse the signs of each voltage in a loop that summed to zero it will still sum to zero. Isn't that amazing?!
  5. fr0sty

    Thread Starter New Member

    Apr 1, 2008
    So why is there then less of a voltage drop on the first resistor in the circuit when the second resistor is present in series, but when you take the second resistor away the voltage drop goes up to the full voltage of the battery? Is that because the second resistor affects the circuit as a whole, so that there's now less current being forced through the first resistor (making it work less hard than if it was just in the circuit by itself?) ?
  6. beenthere

    Retired Moderator

    Apr 20, 2004
    That's just Ohm's law - E = IR. The current through the resistor determines the voltage across it.
  7. fr0sty

    Thread Starter New Member

    Apr 1, 2008
    After reading this thread and a few of the others, I think I'm starting to 'get it' slowly. Less current flowing through the same resistance = less voltage, the resistors limit the current flowing through the whole circuit and hence the two resistors have less current flowing through them than one alone :)

    Thanks a lot for the help.