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Really basic circuit-building help?

  1. Jan 27, 2012 #1
    Hi everyone,

    Unlike a lot of the other threads and people I've seen here, I'm just playing around with trying to build circuits and have some really basic questions. Before I get any further, is that ok for here as well? (like, is it still considered "electrical engineering")

    Anyway, I started out like most people do, with the simple LED circuit: battery -> resistor -> LED and that worked fine. I bought some other stuff, intending to play around with it later, and I finally got around to it. One of those things is some thing that I guess has 2 LED's in the same casing, one in each direction. I'm not sure if that's the correct way to describe it, because I'm not completely sure how it works, but it has 2 leads, and if current goes through it in one direction it lights up red and if it goes through it in the other direction it lights up green. It is 30 mA, with voltages of 2.0 V for the red direction and 2.1 V for the green direction. For 6 V of battery, I used a 121 Ω resistor (obtained a number slightly less than that using a couple equations from my E&M class) and it worked fine.

    My first question is about what I should do with the resistors when I put it in a circuit that allows the current to change direction using switches. What I have are 2 on-on toggle switches, and if they are both to the left the current will flow one way, and if they are both to the right the current will flow the other way. If there is one in each direction, the batteries will not be in the circuit. When I did this, because I wanted to be safe, I put a 121 Ω resistor on each side of the LED thing. I don't know if it's important for the resistor to always come before the LED first though. I know that in a single loop circuit like this, the current is the same throughout the entire thing, and the resistor lowers the current flowing through the circuit. So if I had this LED, the 6 V of batteries, and resistors, would the circuit still function all the same if I put the resistor so that current flowed battery -> LED -> resistor? What if I used 2 60 Ω resistors and had the current flow as battery -> resistor -> LED -> resistor? I'm afraid that by doing any of the other ways, the LED will get too much voltage, and I don't know if too much voltage will break it like too much current will. If not, then in my other circuit, I can get rid of one of the resistors to make the LED light up brighter, right?

    For my second question, let's say I wanted to use a 2-loop circuit with the batteries on the first branch, the capacitor, LED thing and resistors in the middle, and nothing on the bottom, wired up with an on-on toggle switch in such a way that either the top two branches are in the circuit (so basically all components are in the circuit), or that the bottom two branches are in the circuit (excluding the battery, allowing the capacitor to discharge and have current flow in the other direction). First, would this work? Second, how would I choose a capacitor? I remember some equations from E&M class involving capacitors as well, and from what I remember, everything depended on the capacitance of the capacitor and the resistance of the circuit. So let's say I wanted it to take about 2 seconds for the capacitor to charge and discharge, so you are able to watch the LED light up and fade away each time you flip the switch. Is this possible? If so, could you please point me in the direction of some equations to use to find a capacitor to use here (and resistors to use so that the LED doesn't break and the capacitor takes some time to discharge)?

    Thank you very much :)
  2. jcsd
  3. Jan 27, 2012 #2
    As for your first question. All you need is one resistor. It's really doesn't matter when we have a diode or resistor first. It is all act the same.
    In all this circuit the current will be the same.
    And you should choose your resistor more wisely.
    LED current should not exceed 20mA If you want long life for your LED.
    R = (Vbat - Vled)/ Iled = ( 6V - 2V) / 15mA = 260 ≈ 220Ω or 330Ω

    As for your second question
    Simply build this circuit
    And the time constant is equal to T = R*C

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