Trying to understand current and ohms law.

Discussion in 'General Electronics Chat' started by Padapolis, Apr 11, 2013.

  1. Padapolis

    Thread Starter New Member

    Feb 21, 2013
    I’ve been doing a lot of reading lately trying to understand the basics of electronic circuits. I can build a circuit from a schematic, but I want to understand why a capacitor goes here and a resistor goes there.
    But even with all this reading I still find myself confused about what could be the most basic principle of electronic circuits, ohms law.
    My confusion comes down to this, is current being pushed or is it pulled through a circuit?
    And here is what is causing my apparent lack of understanding, a simple AC to DC wall adaptor. Let me try to explain.
    I can buy two wall adaptors. The first one converts to 12V 3.7A, and the second one converts to 12V 1.5A. Now if I plug each of those into their own circuit with a 300 ohm resistor, ohms law says that the current flowing in each circuit is 40mA and the power dissipated by the resistor is 480mW. Now if that is true, then to me 40mA of current is being pulled through the circuit, and the rest is just sitting around waiting to be used. As long as I don’t pull more current than there is available I’m fine.
    Or, I’m totally wrong here and I’m going to start blowing up poor helpless resistors? Because the amps from the wall adaptors are 3.7A and 1.5A respectively. And if I use that current rating in the ohms law calculations, then my circuits power dissipation is going to be a great deal more than a little 1 watt resistor can handle.
    Hopefully that explains my confusion.
  2. crutschow


    Mar 14, 2008
    The current rating of a power supply is commonly misunderstood. The value should be more properly stated as 12V at 1.5A maximum. The ampere rating is the maximum current the device can safely deliver, not the current that it will deliver under any circumstances. The wall adapter just provides a constant voltage (like a battery). The current is determined by the resistance of the load and this voltage, as calculated by Ohm's law. The difference between the load current and the rating of the wall adapter is just the current that can be drawn by another load in parallel or a lower resistance load. If it's not being used then this difference in current between the load and the maximum rating is just current that can be used if needed, but does not exist anywhere.
    screen1988 and PackratKing like this.
  3. WBahn


    Mar 31, 2012
    Think of a car engine that is rated at 400hp. Is the engine actually putting out 400hp when it is idling? No. It is merely CAPABLE of delivering up to 400hp on demand. If you only demand 40hp while crusing down the highway, the engine will only put out 40hp at that moment in time. So, where is all this remaining power when the car is idling? It's tied up in the chemical energy stored in the fuel, which is burned as needed to deliver the power that is being demanded at the moment.

    It's the same with the wallwart. It is capable of delivering up to its rated current on demand, but at any time it will only deliver what is being demanded of it at that time. So where is all the remaining energy that is not being put out? It's back at the powerplant tied up in the chemical energy stored in the fuel.
    screen1988 and PackratKing like this.
  4. #12


    Nov 30, 2010
    I wrote a very beginner piece about Ohm's Law on this site. Please don't take it as an insult, but it might help you picture the very basic aspects of electron flow. I realize you need help in "circuits", and they are more advanced than Ohm's Law, but it's a start.
    screen1988 likes this.
  5. Padapolis

    Thread Starter New Member

    Feb 21, 2013
    Thanks for the help, I think I finally put it all together in my head.

    And I did read your post #12, but I still couldn't figure it out. And no I'm not insulted, I hit a mental block and needed help [​IMG]. I think my problem was the wall adaptors listed rating. I kept thinking that the current listed on the adaptor was being pushed out.
    The swimming pool analogy worked great for a battery, but I couldn't make the mental leap to a wall adaptor. It might seem silly, but there you go.
    I suppose that if I keep using the swimming pool analogy in my head; I could imagine that a wall adaptor is a pool, which is connected to the water main, and is continually being replenished. Where a battery is a pool not connected to a main source.