Volume I - basic concepts of electricity

Discussion in 'General Electronics Chat' started by niallofnine, Jul 19, 2013.

  1. niallofnine

    Thread Starter New Member

    Jul 19, 2013
    hello im new here and so far learning alot about electronics, would like to thank anyone who contributed to making this site..

    Anyways i think i found something in volume 1 basic concepts of electricity

    "In a normal lamp circuit, the resistance of a lamp will be much greater than the resistance of the connecting wires, so we should expect to see a substantial amount of voltage between points 2 and 3, with very little between points 1 and 2, or between 3 and 4."

    to me (a newbie) it seems to say that the lamp actually makes the circuit produce More voltage. I thought that voltage is directly correlated with current (it says current is constant in a simple circuit) so how could you have "very little" voltage between points 1-2,3-4... unless im not reading it right, maybe it was supposed to say "we should expect to see a substantial amount of voltage drop? between points 2 and 3"

    Hopefully its not a stupid thread and im not wasting someone's time haha
  2. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    Voltage is analogous to a pressure

    current is what flows and the 'pressure' is what causes it.

    The more 'pressure', voltage you apply to a 1 ohm resistor the larger the 'flow', current of electricity.

    1 volt across 1 ohm creates 1 amp of current.

    100 volts across 1 ohm creates 100 amps of current.
  3. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    study ohms law.

    any value of voltage, current or resistance can be found when two of three values is given.
  4. WBahn


    Mar 31, 2012
    A phrase like, "substantial amount of voltage between points 2 and 3," means that the magnitude of the voltage difference between points 2 and 3 is significant compared to the other voltages being discussed. When you specify "drop" or "rise", then you really need to specify a polarity, but in most informal discussions this is not necessary and all three are effectively equivalent.

    There is nothing that is implying that the lamp generates voltage. The question of whether the current results in a voltage difference or the voltage difference results in the current is semantics and philosphical. From a practical standpoint, you can think of it either way and which way makes sense depends on the circuit and what you are trying to do at the time.
    cabraham likes this.
  5. niallofnine

    Thread Starter New Member

    Jul 19, 2013
    so what your saying is that there is a voltage change in the circuit? so the power source produces more voltage near to a component like a lamp to "push" the current through the higher resistance filament? is it only "before" or "after" the lamp that the voltage difference can be measured?
  6. vk6zgo

    Active Member

    Jul 21, 2012
    (1)if you completely remove everything but the voltage source & measure its voltage with an ideal voltmeter, you will read all the voltage that can ever be read in that circuit.

    (2)If you now read across the lamp,while everything is still disconnected,you will see zero volts,as you would probably expect.

    (3)Now,remove the lamp & just reconnect the wires then read the voltage between the two ends where the lamp was.
    It will be the same as the voltage you read in (1).

    (4) Reconnect the lamp.
    The voltage across the lamp will be very close to that which you read in (1) & (3).----Note I said "very close"!

    What does this imply about the voltages from end to end of each of the connecting wires?

    Does this agree with the description you were questioning?
  7. WBahn


    Mar 31, 2012
    Voltage, by definition is a DIFFERENCE in electric potential between two points. You can only measure the voltage at one point relative to the voltage at another point. It's just like talking about the height of a building. It only makes sense to talk about the height of the top of the building relative to the height of some other point, whether it be the base of the building or sea level or some other point.

    Let's say that you have a 12V battery connected to a wire that is 1Ω, that then goes to a 9Ω lamp filament, that then goes to a 2Ω wire that then goes back to the battery.

    The voltage ACROSS any one of these elements is the current THROUGH that element multiplied by the resistant OF that element.

    Since they are in series, whatever current is flowing in one has to be the same current that is flowing in each of the others.

    Just as you could compute the height of the roof of a building relative to the ground by measuring the heights of any sequence of steps that gets you from the ground to the roof, so too can you find the voltage between any two points in a circuit by adding up the voltage differences along any path between those two points. And, just as in the case of the building, you have to keep track of the signs of the differences. For instance, going from the ground to the top of the antenna mounted on the roof might be a difference of 100m and going from the top of the antenna to the roof might be -10m, making the height of the roof above the ground

    h = 100m + -10m = 90m

    In the case of the circuit I described, we can calulate the voltage from the bottom (negative side) of the battery to the top (positive side) of the battery by following two paths. The first is through the battery itself. By definition of an (ideal) battery, the voltage along this path will be a rise of 12V. Let's call that Vb.

    Vb = 12V

    But we can also use the path that goes through the wires and the filament and add them all up. Let's call that Vp. That leads to

    Vp = I*2Ω + I*9Ω + I*1Ω = I*(2Ω + 9Ω + 1Ω) = I*12Ω

    Since we know that the voltage between two points doesn't depend on the path taken, these two voltages have to be equal, so

    Vb = Vp
    12V = I*12Ω

    Solving for the current, we get

    I = 12V/12Ω = 1A

    The voltage across the individual elements is, as already indicated, the product of the current flowing in that element and the resistance of that elements:

    1Ω wire: 1A*1Ω = 1V
    9Ω lamp: 1A*9Ω = 9V
    2Ω wire: 1A*2Ω = 2V

    Now, change the value of the lamp filament to 3Ω and repeat. Then change it to 21Ω and repeat. Then change it to 117Ω.

    Hopefully you will start to see the general concept of what is going on.

    What's important to keep in mind is that the battery has no idea what elements are hooked up to it or what their resistances are. All it sees is the TOTAL effective resistance connected to it and it delivers whatever current into that resistance is needed to result in the total voltage through the attached circuit to have a difference across the entire circuit that is equal to the battery voltage. How that current gets distributed is determined by the details of the circuit.