What does it mean by a battery is of 1.5 Volt ? Does it mean that it would take 1.5 Joules of energy to transfer -1 columb of charge from the negative end to the positive end of the battery ?  What is the reason for it to take 1.5 Volt of energy to transfer the charge - is it needed to do work to move the charge against the electric field generated by the positive and negative sides of the battery ? Then would the Volt increase if the distance between the positive and the negative sides are increased (as it would have to do more work against the field)?  Then if you join a broken piece of wire to one side and a broken piece of wire to another side of the battery, why the potential difference between the two ends of the battery is still 1.5 Volt ?
121247
So this is a question with which I have been stuck for a while and it all revolves around the concept of volt.  I have posted a LOT of questions on physics forums while trying to understand the concept of volt and potential difference. This is hopefully going to be the last one.  What do you exactly mean when you say that a battery is 2.5 V. I know that it says that the potential difference between the 2 terminals of the battery is 2.5 V  but then what is the potential of the 2 ends of the battery? Is there a unit of potential and not just potential difference?  Also volt = work/charge but how can work/charge express the difference of potentials across the 2 terminals?
116781
So I know that if you increase the voltage across a wire then the current will increase. But an increase in current leads to a increase in heat production though $P=I^2R$, but as the temperature increase the vibrations of the metal ions increase and so the current is more restricted implying the resistance increases. But as the resistance has increased, the heating has increased again and this will go in a cycle leading to infinite resistance and heat production so where am I going wrong?
28762
A uncharged capacitor $C$ is connected to a battery with potential $V$. It becomes fully charged and has a charge $Q=CV$ stored on it.   Now the polarity of the battery is reversed. The capacitor will have the charge $Q$ still but with polarity reversed too.  My question is: What is the work done by the the battery?  According to me, during 1st charging it will do a work of $QV$. The energy of the capacitor is also not changing, then what is the work done to change its polarity?
61248
Imagine a battery with the positive terminal connected to ground.  The positive side has a high eletrical potential, ground obviously has a lower electric potential.  Electricity wants to seek a lower potential, and there's no significant resistance between the terminal and the ground.  Shouldn't electricity flow?  I know it doesn't but I don't understand why.  I know you have to close the loop, but why is this necessary?
63750
I would like to know why does electrons flow through a wire connected to a battery as said in .It tells that "the electrons that are stripped from the carbon electrode is collected on the zinc can" and then tells     The electrons at the negative terminal want to go to positive terminal, they just need a way to get there. In our light bulb circuit, the way to get there is through the wire.   Why couldn't the electrons go to the positive terminal through the electrolyte(potassium hydroxide) instead of going through the wire?
121650
let me say if I have a pack of lithium batteries connected in series and parallels, and i want to connect it to an Inverter, BUT...BEFORE I connect it to inverter, lets say it gives out 12 volts "hypothetically" some informations like 3.7 volt batteries, and 2500 mAh,  lets say then the pack gives 12 volts in DC and hypothetically again I calculate it ((still in DC)) is .....  W = V* Amp ... then it means wattage = 11.1 V * 5 Ah which is 55.5 W. then we got 55.5 watts, THEN....after we want to change the dc into ac it gives you 120 volts "please hypothetically" just answer me this, ARE the Wattage And Ampere are remains the same as DC? what would it be the result after AC?
84095
Conceptual question:  Suppose we have a configuration of point charges. If the potential of the energy of the system is negative, this means work is positive.   I'm kind of rusty with my mechanics, does the fact that the work is positive mean that to arrange the configuration you have to do work against the electric field? I.e. supply energy via some external force?  In terms of electrostatics, is there a general way of interpreting the sign of potential energy and work?
41239
So I have been mulling over Ohm's law for hours today. It makes sense to me that the current gets bigger as voltage rises, since the voltage is supplying more electrons and current is the number of electrons flowing. However, I'm confused by the theory that as resistance drops, current increases, even when voltage stays constant. Such as:  $$V/R = I$$ $$\frac{1V}{0.0001 \Omega} = 10,000,000 mA!!$$  That's a big current from such a small voltage!  My mental picture of a current is more electrons flowing. Electrons are supplied by the voltage, right? If we start off with a small voltage, say 1V, but infinitely decrease the resistance, we'll get a bigger and bigger current. How can this be? Isn't current the number of electrons flowing? With fewer electrons (because of less voltage), how can less resistance infinitely increase the current?
78706
I'm new to electrical engineering, and probably asking the wrong question. Not sure how else to phrase it.  For the question, I'm going to use an example layout of simply a 9v battery attached to a toy DC motor.  From my understanding of my , the voltage coming from the battery through the motor will be around 9v. The current will be from 0amps to however much the battery can supply without frying. What decides how much current goes through the motor?
87469
As I remembered, at the 2 poles of a battery, positive or negative electric charges are gathered. So there'll be electric field existing inside the battery. This filed is neutralized by the chemical power of the battery so the electric charges will stay at the poles.  Since there are electric charges at both poles, there must also be electric fields outside the battery. What happens when we connect a metal wire between the 2 poles of a battery? I vaguely remembered that the wire has the ability to constrain and reshape the electric field and keep it within the wire, maybe like an electric field tube. But is that true?
83004
I would like to know how potential difference is created between two points?Is potential difference created by adding more electrons at one side?If that's the case how to add more electrons at one side?  I'm really looking to know what really constitutes potential difference?
111141
when we use a DC battery into a circuit the flow of electrons in wire creates a clear picture as they flow from negative terminal to positive terminal. under AC(Alternating current) we know how energy flows and what does it mean by $I_{rms}$ or average current etc. But what about the pattern in which electrons move?
36568
So I started off with electrostatics and everything seemed nice and mathematical and justified and then "DC circuits" happened!  I just cannot understand the model of electron flow in electrical circuits.  Here are my specific doubts-:  1) If potential difference across a tiny cross section of conducting wire is zero, then why on earth does electron flow across that cross section at all? Never mind potential difference across the whole circuit.   2) Is there a constant electric field across a wire connected to a battery? If yes then how is potential difference across a zero resistance wire constant?   Shouldnt it be increasing?  Doesnt it violate ohms law?  If no,  then why do electrons flow at all?  Please take time to consider these doubts and relieve me of my frustration.  I havs searched through the net for this but every answer seems like beating around the bush.  All of the 4 books I have consulted do not address these facts to my satisfaction.   Frankly I think nobody understands this.
23496
I'm having trouble conceptualizing why the voltage drop between two points of an ideal wire (i.e. no resistance) is $0~V$. Using Ohm's Law, the equation is such:  $$ V = IR \\ V = I(0~\Omega) \\ V = 0$$  However, conceptually I can't see how there is no change in energy between these two points.  It is my understanding that the electrical field of this circuit produces a force running counterclockwise and parallel to the wire which acts continuously on the electrons as they move through the wire. As such, I expect there to be a change in energy equal to the work.  Voltage drop is the difference in electric potential energy per coulomb, so it should be greater than $0~V$:  $$ \Delta V = \frac{\Delta J}C \\ \Delta J &gt; 0 \\ \therefore \Delta V &gt; 0 $$  For example, suppose I have a simple circuit consisting of a $9~V$ battery in series with a $3~k\Omega$ resistor:    If the length from point 4 to point 3 is $5~m$, I would expect the following:  $$ W = F \cdot d \\ W = \Delta E \\ F &gt; 0 \\ d = 5 &gt; 0 \\ \therefore W &gt; 0 \\ \therefore \Delta E &gt; 0$$  Since work is positive for any given charge, the change in energy for any given charge is positive -- therefore the voltage drop must be positive. Yet, according to Ohm's Law it is $0~V$ since the wire has negligible resistance.  Where is the fault in my logic?
97290
I bought this new battery really cheap (it is really small and i needed a small 12v battery) but i am not sure of some things. First of all, it says it is 1.3Ah/20HR. Does that mean that it can deliever 1.3Ah for 20 hours? Like, i could power a 12v 1A light bulb (estimately...) for 20+ hours from it? Second of all, is this a deep cycle battery? ( i do not speak polish, so i do not understand the label). If i fully discharge the battery (9-10 volts) will it lose its capacity to store charge, or there is no problem? What is its optimal voltage? I used some rough google translate and it turns out that the "standby voltage" is 13.6-13.8 volts and the "usage" voltage is 14.5-14.9 volts. Isn't that too much for a 12v battery? Right now it is at 12.6V. Is it undercharged? Also, what is the voltage needed to charge it and is it a bad idea to charge it with a stepped up 5v power supply? (Phone charger, stepped up to the battery voltage required for charging)
83072
Now, I know that emf is measured using an ammeter, while potential difference is measured using a voltmeter. But why is that so? What do we mean when we say the emf of the battery is say, 3V, and the potential difference is 2V or anything? How do we define these terms? What is emf and the current supplied by the battery? Are they the same thing?
52457
Imagine I have a 10V battery with a capacity of 1Ahrs connected to a 100 Ohm resistor.  I'll get 0.1A (V/R) at a power of 1W (current x voltage) for a total of 10hrs. As a watt is defined as J/s my battery produces 3,600J (1x60x60) in total.  Now, imagine I get a second, identical battery and connect it in series to the first one. I will effectively have a battery of 20V with a capacity of 2Ahrs, all connected to the same 100 Ohm resistor.  I'll get 0.2A at a power of 4W (current x voltage) for a total of 10hrs. This will generate 14,400J (4x60x60).  So, from doubling my energy source I have quadrupled the energy output!  I've obviously made an error somewhere either in my calculations or conceptual understanding, and I would be most grateful to anyone who could set me right.  Thanks!