voltage and amps

[ACE37918]

I am trying to find out more about voltage and amps , as you see on wall warts and such.

1- How can you have 12v @ 2 amps and have 12v at .5 amps?<p>2-How do you keep the same voltage and change the amps, what in the design would have to change?<p>3-I know that resistors can be used to reduce voltage but I am lost when it comes to reducting amperage and keeping the same voltage.<p> Thanks,
Rick

[greg123]

Ok Rick, good questions. Here is a basic summary to start. I will answer as if you have no knowledge of electronics, or a limited knowledge.<p>Voltage, or EMF is the potential energy due to the "positing of charge" between two points in a circuit, the unit is VOLTS and expressed as V. The current is due to the impedance that "resists" the voltage between those two points. The current is basically the "electron flow" through a conductor, and is measured in AMPS and represented by the letter A.<p>The current/voltage relationship is as follows:<p>Voltage = Current * Resistance<p>This is known as Ohm's Law<p>Meaning that the voltage accross a resistor is a product of the current through the resistor and the resistor resistance. <p>So....<p>
1- How can you have 12v @ 2 amps and have 12v at .5 amps?<p>From the formula, you can have 12V at 1 billion amps with a resistor of 12 pico Ohms. This however is not practical (as the resistor dosn't exist) and the heat from the power generated will be too much for the circuit to handle. <p>I am assuming you read the above spec's off your wall wart. All the rating means is that the wall wart will output 12V at a max rating of 2 Amps or 12V at a rating of .5A. The circuit will only draw as much current as it neads. If you were to use the 12V .5A wall wart with a unit that required 12V 2A, the unit would NOT work as intended because there is not enough current available to the unit. <p>This is actually what happens when you home circuit breaker trips. The circuit is trying to source more current than the breaker or FUSE will allow and BANG the fuse or breaker trips and the circuit is shut down. This is for protection.<p>
2-How do you keep the same voltage and change the amps, what in the design would have to change?<p>Again, the voltage is ouputted by the supply at a max current. This current could be the max allowed by the regulator or the max allowed by the transformer and the max allowed by the capacitor. You can't really change the rating on the wall warts as they are encased in the "so hard to open" plastic coffin so changing parts is hard. A certain amount of reverse engineering is also reguired.<p>
3-I know that resistors can be used to reduce voltage but I am lost when it comes to reducting amperage and keeping the same voltage.<p>This is basically a series and parallel quiestion. <p>In a circuit, the total current is a the quotient of the Total Voltage devided by the Total resistance. <p>In a series circuit the voltage drops accross all the resistors will be different as the resistances changes (The sum of all the voltages will be the total supply voltage). The total current through each resistor will be the same (the same as the total voltage devided by the total resistance). <p>In a parallel circuit, the voltage accross each branch will be the same (same as the supply), while the current through each branch will be different (but will add up to the total current).<p>You can use resistors in a voltage devider network to devide the voltage to a certain level and the total current will be determined by the total resistance. <p>For example, lets say you wanted 5 volts at 2 amps.<p>The voltage devider looks like this:<p><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">code:</font><hr><pre><p>10V supply------/\/\/\-------/\/\/\-----ground
R1 | R2
|
0

output<p>**not code, only way to make ascii art look good<p></pre><hr></blockquote><p>where 10V is the supply and /\/\ is a resistor<p>Now the "tricky part":<p>If you want the circuit to have 2 amps, using the Ohm's Law formula:<p>Resistance = Volts / Amps
R = 10 / 2
R = 5 ohms<p>Meaning with one five ohm resistor you will have 2 Amps at 10 Volts. But we need to devide the voltage. So by using two 2.5 ohm resistors and if we take the output off the middle of the two resistors, we will have (from ohm's law):<p>Volts = Resistance * current
V = 2.5 ohms * 2 Amps
V = 5 volts @ 2 Amps<p>You can adjust the values accordingly to suit your needs.<p>Hope this helps....<p>
Greg<p>
Have fun in electronics<p>[ January 18, 2003: Message edited by: Greg ]</p>

[ACE37918]

Greg,
Thanks for your time,you have removed the fog and now I think I get it.
I didn't get the amp. / current relationship and you have cleared that up for me.<p> Thanks,again
Rick <p>[ January 18, 2003: Message edited by: ACE37918 ]</p>

[Dean Huster]

Nice job, Greg!<p>Dean

[Dimbulb]

An illustration may help visualize this.<p>Lets say you are trying to describe the nature of Beer as it flows out of a 50 gallon drum.<p>You observe a tiny stream from a 1/8 inch pipe mounted at the base of the barrel and do some tests and record the results and give them names.<p>Next you observe nature of the the massive flow coming out of a 4 inch pipe. After concidering the output difference between the two pipes you decide that you need a standard size pipe should be 1 inch.<p>Current is the amount of electrons at a given point. (cross section of pipe for water)<p>Voltage is the electrical pressure or force. (another term is potential helpful when you use resistors)<p>Later you get tired of having to describe both so you multiply voltage and amperage and to get power. This extra step is puts them together is simple and is the unit everyone can guage from.

12V * 2A = 24 watts
12V * .5 = 6 watts<p>If you have 12 Volts at 2 Amps and you would like only .5 amps then using one resistor you can inhibit the flow.<p>P = (I * I) * R
solving for R
6/.25 = 25 Ohms 6 watt
The current is now limited to 1/2 Amp.<p>using a 1.2K resistor on a 12V 2A supply
12/1200 = .010 or 10 mA ( good for one small LED )
.5 amp or 500 mA could run 50 10 mA LEDs<p>Now you want to reduce the voltage using resistors. One basic method is to use two resistors. Lets say you use a 1.2 K and a 600 ohm in series then when you tap into them from center and the ends you should have divided the voltage.
The potential is said to be less from the center.<p>If you measure the voltage across just one resistor the voltage will be divided but If you
break the circuit and measure the amperage the larger value resistor limits more current.
Which resistor dou you think will be warmer ?<p>[ January 18, 2003: Message edited by: dim bulb ]</p>

[greg123]

See Dean,<p>I could have written for your mag!<p>Greg

[greg123]

I figure i should add a bit on power.....<p>Power is the rate of doing work, or rather the amount of work that can be done in a specific time. The unit for power is the WATT which is measured in Joules per Second (J/S).<p>Power with regards to electronics can be found the following way:<p>Power = Voltage * Current<p>Power = Voltage^2/Resistance<p>Power = Current^2 * Resistance<p>So by knowing the value of voltage accross a resistor and the current through it, we can find the power dissipated by the resistor. As dim bulb mentioned, this is important when determining the size of resistor needed (either .5W or 1W....)<p>I always found the first formula helpful in high school theatrical plays when i wanted to know how many lights i could plug into one outlet.<p>BY the formula<p>Power = Voltage * Current
P = 120 (approx) * 15A (max due to 15A breaker)
P = 1800 Watts<p>That means there is 1800 Watts (ideal) available from one circuit at the theater. So if i had 500 Watt Lights, i could plug 3 into the outlet and not pop the breaker. This is because 3 lights * 500 Watts each is 1500 W (300 less than the max).<p>By the way, if you want to convert watts to horsepower, 746 watt is approx equal to 1 Horsepower.<p>Greg<p>
[ January 18, 2003: Message edited by: Greg ]<p>[ January 18, 2003: Message edited by: Greg ]</p>

[Chris Smith]

1- How can you have 12v @ 2 amps and have 12v at .5 amps? <p>Amps is the amount of working current, like horse power, that is available, but not necessarily used at all times.<p>Voltage is the pressure in a water hose, while amps is the total volume available, over time.

2-How do you keep the same voltage and change the amps, what in the design would have to change?

A Stronger motor, delivers more Horse power, but can be used one HP at a time, or all at once. <p>3-I know that resistors can be used to reduce voltage but I am lost when it comes to reducting amperage and keeping the same voltage.<p>Resistors reduce Current, and the voltage follows.

[ACE37918]

Thanks guys,<p> You have giving me some great insite on how volts and amps work, and thanks for adding power and watts that would have been my next question. Thanks for your time. Your help is what makes this hobby fun!<p>
Thanks again,
Rick