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another very basic question

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can some one please explain very simplily how current works and how it changes in a circuit because i cant seem to get a grasp on it
do resisitors change current and voltage? if so how to work it out an a calculator
does a LED run on current or voltage.....if i have a 10k resistor and a 9 volt battery how much current/volts is left.

if these questions sound dumb :oops: try not to laugh at me just lend a hand. and also thanks for the replies to my other questions they ave been very helpfull.


Current, voltage, and resistance share a dependent relationship. This relationship is expressed, in its various forms, by the statement of Ohm's Law. The three basic forms of Ohm's Law are:

  • I=E/R where I is current in amperes, E is electromotive force in volts, and R is resistance in ohms;

    E=IxR where I is current in amperes, E is electromotive force in volts, and R is resistance in ohms; and

    R=E/I where I is current in amperes, E is electromotive force in volts, and R is resistance in ohms.

A handy device for remembering these equations is illustrated below. Cover the symbol of the value to be calculated and the proper formula will remain. For example, to calculate resistance, cover the "R" and "E over I" will show.

Another such tool is one which also includes the formulas for power (P):

Simply put, voltage is electrmotive force, which is the force that causes electrons to move. The movement of electrons is current, and resistance is the opposition to the movement of electrons.

All of the current that flows into a given circuit will also flow out of that circuit -- current is not "lost" or consumed by a circuit. On the other hand, voltage is "dropped" by the various elements in a circuit. The sum of all of the individual voltage drops in a given circuit is equal to the source voltage applied to that circuit.

Consider your question -- the 9V source and the 10K-ohm load. In this simple circuit, a current of 0.9mA (0.0009A) will flow, based upon Ohm's Law stating that the current is equal to the voltage divided by the resistance. The entire 9 volts will be present across (dropped by) the 10K-ohm load. If that 10K-ohm load is a single resistor, the full 9 volts will be dropped across it. If, on the other hand, the 10K-ohms is comprised of two separate 5K-ohm resistors in series, we will have a slightly different situation. The circuit current will remain the same, as the total resistance is the same. The voltage, however, will be divided by the two resistors, with 4.5V being dropped across each one. Looking at Ohm's Law again, this time for voltage, will will see that 0.0009 x 5000 = 4.5 -- the current multiplied by the resistance gives the voltage across that resistance.

I hope that some of this makes sense to you... and keep on posting questions if you have them! :)


BTW -- your other question:
does a LED run on current or voltage.....
is best answered by saying BOTH!

As should already be obvious based upon my last reply, the voltage causes the current to flow through the LED. The LED appears to the circuit as a load, and voltage will be dropped across it. The current through that load is what actually does the work, but its presence would be impossible without the associated forces required to move the necessary electrons.

Consider a device like the L53SGD super bright green LED from Kingbright. The specs on this LED show that it is a 20mA/2.2V LED. Suppose that you wanted to operate this LED from a 9V source, as shown below:

As already stated, the LED is a 2.2V device, and the source is a 9V source. We want to use a resistor in series with the LED to limit the current through the circuit so as not to destroy the LED with too high a current. The question is, what value should the resistor be?

The first thing that we need to know is what effective resistance the LED will have in this simple series circuit. So, based upon what we know -- the forward voltage across the LED (2.2V) and the design forward current of the LED (0.02A) -- we can calculate that the LED will effectively be a 110 ohm resistance (2.2V divided by 0.02A). Remember that...

We know that we want approximately 20mA (0.02A) as our circuit current, right? So all that we need to do is to determine what resistance will cause (or allow, if we are thinking in terms of current limiting) the desired current when supplied by a 9V source.

Using Ohm's Law: R=E/I -- R = 9/0.02 -- R is determined to be 450 ohms. Subtract from that the effective resistance of the LED (110 ohms), and we come up with a calcualted value of 340 ohms for the series limiting resistor. The closest standard resistor to this value in a common 5% tolerance would be 330 ohms. 330 + 110 = 440 ohms, which would allow a circuit current of 20.454mA.

Now let's see how the circuit voltages break down. Again using Ohm's Law, and based upon a current of 0.020454A, we see that the voltage across the 330 ohm resistor would be 6.74982V (0.020454 x 330), leaving 2.25018V across the LED (9.0 - 6.74982). Of course, as it is a simple series circuit, the current through the LED is the same as that through the resistor at 20.454mA.

BTW -- the resistor used here should be at least a 1/4-watt type, as the calculated continuous power to be dissipated in the resistor is 0.1380614319 watts (6.74982V x 0.020454A), which is greater than the power capacity of a 1/8-watt resistor. Also, should the LED short, causing the full source voltage to be dropped across the resistor, the 1/4-watt rating would still be (just) adequate, as the calculated power then would be 0.245 watts.
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