Noob questions about ampers and volts. - Page 1
Forum section: electronics / beginners

author: Monkfish
author_rank: Newbie
timestamp: January 02, 2013, 04:22:55 pm
content: Greetings all, new guy, interested in electronics and electricity, been browsing for info and managed to accumulate a bit questions about amperes and volts. If you guys would mind answering them for me, I would be one step closer to not killing myself in a freakish electrical accident.1) I see alot of electrical products giving its recommended input voltage for its powering up, which is fine n dandy, but almost never do they give you the recommended ampers ? Why is it so ? Doesnt the 2 of those give you the power (watts) for it ? What if you buy yourself an adapter and overshoot or undershoot with the power, and you get excessive heat or it not working at all ?2) My next main question would be, does the ampere value drop in a circuit when passing resistors ? I know for sure voltage does. We can just calculate the total resistor value and find the amperes with the voltage source value ? Where exactly is the calculated voltage at on the resistor ? At the resistor start, the middle, the end, or just barely past it ?3) From mains wiring, what exactly dictates how many amps were getting from it ? Voltage sources come with a fixed voltage value, do they also come with a fixed amperage value ? How do we calculate them ?4) Linear power supplies often have an amps setting in them aswell, how exactly is the value being controlled ? Was i wrong with the ampers not dropping in a circuit ? If I was wrong then looking back at question number 2, how do we calculate the ampers on each resistor in a circuit ?These are some of the more confusing ones I could thought out of my head, the wording may not be the best, but oh well...Thanks for reading

author: Chipset
author_rank: Contributor
timestamp: January 02, 2013, 04:31:34 pm
content: Most (if not all) of your questions seem to stem from a poor understanding of exactly how volts and amps work. It's okay, we've all been there and I know how frustrating it is. The best advice I can give for understanding these concepts is to start using them in practice, building up simple circuits like LED blinkers and simple adjustable power supplies and whatnot. I remember when I first started with electronics, no amount of textbooks or online tutorials managed to give me an intuitive understanding of the subject. Only when I started using them in practice (and blowing up a lot of components in the process) did I start to understand.I'm a strong proponent of learning by doing and honestly I don't think there's a better substitute in this situation. That said people do learn differently so if you prefer it, there's about a million and one tutorials on and off youtube explaining this stuff in varying detail and using a load of different analogies. I'm not gonna try it here because plenty of these tutorials do it a lot better than I could on short notice.Here are some I recall being helpful:Good luck!

author: Kremmen
author_rank: Super Contributor
timestamp: January 02, 2013, 05:36:55 pm
content: You should get familiar with Ohm's law for starters. Once there you understand basic current flow.Having said that, a shortcut is to think about the water analogy: voltage is pressure and current is - surprise - current. Now imagine you connect a lawn watering attachment to your garden hose. Open the valve and water sprays forth. Now, does it make sense to ask how much (water) current the sprayer pulls from the hose? Or does it make more sense to think about how much water the water supply manages to push through the hose and sprayer? That is how you should think about it. The attachments you fix to the hose each resist water flow according to their internal structure. If you connect nothing, the only thing limiting the water flow is the internal friction of the hose and water mains. These are analogous to electrical resistance. If you plug the hose, that is infinite resistance. If you leave the hose open, it is a short circuit. If you connect an attachment, the flow is defined by the flow friction of the attachment (electrical resistance) and pressure (voltage) in the hose. A higher pressure pushes more flow (current) through the same friction (resistance) than a lower pressure.In electrical terms (Ohm's law): I = U/R, or current equals voltage over resistance.

author: olsenn
author_rank: Frequent Contributor
timestamp: January 02, 2013, 06:09:32 pm
content: As "Chipset" mentioned, voltage is the force pushing charged particles (electrons) through a circuit, and current is the rate at which charged particles are being pushed through the circuit. You may think there is a relationship between how hard you push and how much you get out, and that's because there is! Ohm's Law (which is actually a definition and not a law) states that voltage is proportional to current... that is if you push those electrons through a circuit twice as hard, the electrons will be pushed through twice as fast.The problem with this is that not all circuits obey this false law; only linear elements. Since you're just starting out I'll leave out reactance and complex impedance, and instead focus on resistance. A resistor is by definition a linear element that obeys the above relationship (voltage accross is proportional to current through). The proportionality constant (voltage divided by current) is called "resistance". Resistance is the measure of the ability to limit the current to voltage ratio.Here's how all this can be useful: Current passes through components, and voltage occurs accross components! So to answer your question, no, there is no drop in current as it passes through a resistor -- all the current that enters a resistor must exit that resistor. However, if you have two resistors connected one after another in series, then the current passing through that second resistor must be the same as the current that passed through the first! If you know the current, you can multiply that value by the value of resistance for each resistor and that will give you the voltage drop over that resistor. Otherwise, you can calculate the current with voltage and resistance.

author: FenderBender
author_rank: Super Contributor
timestamp: January 03, 2013, 02:42:19 am
content: Personally the water analogies never worked for me too much. If they do for you, great.Here are some things that might clear some of your misunderstandings up, or at least worked for me...Voltage describes the difference in electrical potential between two points in a circuit. Voltage does not "move" through a circuit. It is a stationary thing. Current on the other hand refers to the charge moving through the circuit.An analogy I like is a simple physics analogy with dropping a ball from a height. Think of the height above the ground as the voltage. As you raise the ball higher and higher, you increase it's potential energy. You are increasing the ability for the ball to do work. If you dropped a ball from 1ft off the ground, it will not move very fast or have any noticeable impact. On the other hand, if you raised the ball 10ft off the ground, the ball will hit the ground with a much larger velocity and it might bounce really high or something.With electricity, it's pretty similar. Think of voltage as the distance the ball is from the ground. When you raise the voltage, you increase the ability for the electrons in a circuit to do work. A higher voltage means that the "ball goes faster" when you drop it, or in electrical terms, there will be a greater current (flow of electric charge). If you increase the voltage, you increase the flow of charge in the circuit, just as if you were to raise the ball higher, the ball would go faster and have a bigger impact.Then if you'd like to go further, suppose that there was a fan placed under the ball as it is dropped blowing upwards as it falls. You can think of the fan as a resistance. If the fan is on a low speed, it will not interfere too much with the ball falling. It might slightly slow down the fall, but maybe not too much. Think of the fan on a low speed as a low resistance. On the other hand, if you put the fan on full blast, it will push up against the ball as it falls, and it may take a very long time for the ball to hit the ground because of the wind resistance. Think of the fan on full speed as a high resistance.We can say that the current is the speed at which the ball hits the ground or its acceleration. If we increase the height of the drop, the ball can go very fast, BUT, that's provided there is also a low resistance (fan is on low). You can drop the ball from a 100ft up, but if the fan is on full blast, or you have a couple fans on full blast (high resistance), the ball will not hit the ground very quickly because there is such a high resistance.I think if you understand these concepts, all of your questions will pretty much be answers. Most sources of electricity are voltage sources.Phew I just put a good 10 minutes into writing that and I tried to make it as clear as possible so give it a look.

author: Kremmen
author_rank: Super Contributor
timestamp: January 03, 2013, 04:32:24 pm
content: And you think this is somehow clearer than the water modelSorry - i feel i can say this because the water analogy was by no means invented by me - it is a commonly used one and to be frank, IMO far superior to this one...

author: Unknown
author_rank: Unknown
timestamp: January 03, 2013, 09:19:11 pm
content: Q: I see alot of electrical products giving its recommended input voltage for its powering up, which is fine n dandy, but almost never do they give you the recommended ampers ? Why is it so ?A: if you know the working voltage, you know how much amps pass trough. Then it depends if the object powered up is "linear" on not linear, but assume it linear(a resistor) and powered it up in DC, you have a 10 ohm resistor and you power it up at 20 V, you get 2A.Q: Doesnt the 2 of those give you the power (watts) for it ? What if you buy yourself an adapter and overshoot or undershoot with the power, and you get excessive heat or it not working at all ?A: Depends. P=V*I only in DC circuits. In AC circuits, you have to care about the power factor(which is between 0 and 1). If you buy an adapter and overshot/undershot the equipment you get less or higher value of current, and precisely, if you overshot the circuit and you get some high value of current you will break the product: for instance, if you have a DC motor and it works whit 3.3V and you power it up to 9 V, you will see the motor working, and rotating very fast, then it will start to heat up and then you get a short circuit because of the heat melting the insulation between the wires.... The problem in cables, products, motors and all things like that is the current: current produce heat, and heat is not so good for the insulation.... If you undershot the equipment you get the object working, In some cases won't work, but if you have a resistor and its powered it up at 10 V or 40 V and nominally works at 100V the current will change of course, but it is still working like a resistor... but it's like having a Ferrari and drive to 5 mile/h...Q: My next main question would be, does the ampere value drop in a circuit when passing resistors ? I know for sure voltage does. We can just calculate the total resistor value and find the amperes with the voltage source value ? Where exactly is the calculated voltage at on the resistor ? At the resistor start, the middle, the end, or just barely past it ?A: If we measure voltage, we measure ti between two points: it's called potential difference between two points. At the same potential we have 0V (for remember it just think: if you make a step are you dead? no, so you are at 0V). so, if you have a drop it between the resistor. You should know the ohm law: V=R*I dimensionally speaking [ Voltage ] = [ohms ] *[ amps] . This formula can be always reversed, so if you know the voltage and the current, you know the resistance (between the current path)Q:From mains wiring, what exactly dictates how many amps were getting from it ? Voltage sources come with a fixed voltage value, do they also come with a fixed amperage value ? How do we calculate them ?A: No: in fact you can plug to your mains a load witch is 24K VA and for a certain time it will work, then what happen? the fuses will open the circuit: you request too much power and before you house start to burn it's better open the circuit.... But it's better you let go the mains: it's better for you starting to know basics principles of the electricity/electronics instead complicating your way..... i told this to you for experience.Q: Linear power supplies often have an amps setting in them aswell, how exactly is the value being controlled ? Was i wrong with the ampers not dropping in a circuit ? If I was wrong then looking back at question number 2, how do we calculate the ampers on each resistor in a circuit ?A: As I know, you control what's on the LCD. If you have a circuit which in the paper is a resistor of 8 ohm and you power it up whit 8V, you get 1A. If you get 10A or the max current something is wrong. Depends how much resistor you have: remember la large number of same resistor in series = an open circuit (0A), and a large number of same resistor in parallel = larger current. so, if you have a circuit whit a larger number of resistor in parallel, you get that. For calculating the drop on resistor or the current value on one of them you should see the series/ parallel formulas, + the voltage/current divider.Hope I answered.

author: FenderBender
author_rank: Super Contributor
timestamp: January 03, 2013, 10:50:25 pm
content: Quote from: Kremmen on January 03, 2013, 04:32:24 pmWell thanks for the facepalm asshole, I said this worked for me and I liked it better. Not everyone learns the same. If he likes the water analogy better, good for him. I'm giving another option.

author: Kremmen
author_rank: Super Contributor
timestamp: January 04, 2013, 06:52:18 am
content: You are free to your opinions, as i am to mine.Now at least i know what kind of person you are.

author: FenderBender
author_rank: Super Contributor
timestamp: January 04, 2013, 01:47:15 pm
content: I wasn't the one with the poor attitude.