Kirchoffs Law

Discussion in 'General Electronics Chat' started by horsebox, Feb 14, 2008.

  1. horsebox

    Thread Starter Active Member

    Jun 9, 2007
    How does the sum of all the voltage drops in a circuit equal zero?

    Lets say I have a 5V power source. 5V on the negative terminal 0V on the positive terminal. I have 2 resistors both of which cause the voltage to drop 2V. At the end I have 1V flowing between the second resistor and the positive terminal.

    If I add those 2 voltage drops together I get 4V not 0V. Am I getting this mixed up?
  2. thingmaker3

    Retired Moderator

    May 16, 2005
    Actually, you don't.:( The sum of the voltage drops across the two resistors will equal the applied voltage.:)

    Also, voltage doesn't flow, current flows.
    For more information, peruse
  3. mozikluv

    AAC Fanatic!

    Jan 22, 2004
    have you tried browsing thru the forum E books on Vol I - DC?

    scroll all the way down and you will find the link

  4. horsebox

    Thread Starter Active Member

    Jun 9, 2007
    I've read all this theory. You say the sum of the voltage drops across the 2 resistors will equal the applied voltage. What if I lower the resistances of those 2 resistors?

    I'll use another example. I have a circuit with a power source of 5V. There is 1 LED in the circuit and a total resistance of 2 ohms therefore theres 2.5A flowing through the circuit. Lets say the LED has 1 ohm of resistance and the wire accounts for the other 1 ohm.

    Are you telling me that the voltage will drop 5V to 0V after the LED?
  5. horsebox

    Thread Starter Active Member

    Jun 9, 2007
    When I see that resistor symbol I usually think of a load like an LED or something.

    In the tutorial are they just referring to the length of wire between the points? For example are they just saying theres a 5 Ohm resistance between point 2 and 3? They're showing negative values on the multimeter pictures. Are they just trying to explain that there will be a voltage drop of 10V between point 2 and 3, 20V between point 3 and 4 etc.?

    Does Kirchoffs voltage law basically just state that the voltage of a circuit will always drop to that of the positive terminal of its power supply?

    For example the negative terminal of the power supply is 25V and the positive terminal is -25V giving the circuit a potential difference of 50V. The voltage just before the first point of resistance past the negative terminal is 25V. After the very last point of resistance in the circuit the voltage will have dropped down to -25V? Is that whats meant by Kirchoffs voltage law?

    Also I don't mean to complicate matters but what if you have a superconductor with no resistance. You apply 50V to it and technically theres nothing causing the voltage to drop.

    Maybe I'm too dumb to learn electronics or maybe I'm reading this the wrong way but either way this makes absolutely no sense to me. Maybe I just need to calm down and read it in a relaxed manner so I don't jump to the wrong conclusions. I spent about 4 hours trying to figure out what they meant by voltage dividers one day and when I finally figured out what they were talking about I was ready to punch myself in the face for wasting 4 hours pondering the stupidest little concept that I grasped in 10 seconds the first day I started learning electronics. I just had the idea that they were referring to something highly complex that I had to put alot of thought into in order to learn it. I had no idea that I already knew what a voltage divider was and it was so simple a handicapped 6 year old could understand it.
  6. Dave

    Retired Moderator

    Nov 17, 2003
    You've pretty much got it. Kirchoff's Voltage Law merely states that the algebraic sum of voltages in a closed circuit will equal zero - therefore if the source is 50V, 50V must be dropped across the components in the circuit. If the source increases to 60V, then the voltage drops in the circuit will increase to reflect this.

    Think about I = V/R

  7. horsebox

    Thread Starter Active Member

    Jun 9, 2007
    Ah it was that simple all along. There I go again complicated things unnecessarily. When I heard sum I thought they meant the sum of the voltage drops alone not including the input voltage but what they mean is the algebraic sum of the input voltage and all the voltage drops. Thanks for clearing that up.