What is the internal resistance of a battery? What is the origin of it?
913372
It doesn't really have one. Its just a useful way of characterising the battery's behaviour.\n\nBatteries approximate to perfect voltage sources - in other words, the voltage across them does not drop no matter how much current you draw. (By contrast, solar cells approximate to perfect current sources).\n\nOf course, in reality the voltage does drop a little, and drops more the more current is drawn. The relationship is almost a straight line for small currents, and so it is useful to know the rate the voltage drops as the current increases, or dV/dI. Now of course the units of volts/amps is ohms.\n\nWhere this relationship is linear (small currents) you can also usefully include the internal resistance as a fictitious resitance in a circuit - it will do just as well as accounting for the change in voltage (can you see why?).
107334
This question needs a little work. By convention, current flows from + to - although in real life, current is electrons which go from - to +. But only physics types do it that way. Electrical engineers say + -> -.\n\nNow, that being said, if you have a non-zero resistance across a voltage source, you can find the current flow by measuing the voltage across the load (resistance). The lower  the resistance, the lower your voltmeter has to read. The sign of the voltage would indicate the current flow. \n\nSo say you have a 10V battery and two 5 ohm resistors in series across it. The total current will be 1A (ohm's law). Now if I measure the voltage across one resistor, I should read 5V (kirchoff's law by the way) and the sign (+ or - 5V) will tell me the current direction. Using conventional + to - flow, if the voltage is + the current is flowing from the red lead to the black lead.
250746
This is the best answer you'll get -\nhttp://electronics.howstuffworks.com/battery.html
757379
Resistance is connected to conductance as resistivity is related to conductivity.\n  By ohm's law\n   J= E * condutivity\nor \n  J= E/resistivity\n Resistance=( resistivity X length) /  Cross Sectional Area
879048
Resistance = Voltage / Current\n= 1.4/0.8\n= 1.75 ohms.
233940
A battery is a method of either STORING or GENERATING an electrical charge chemically. In generator type of battery usually (2) disimiliar metals provide the anode (+) and Cathode (-) of the battery an electrolyte solution causes a chemical reaction which destroys the anode and cathode as it produce an electrical charge. Examples of this type of battery are the carbon zinc battery, alkaline batteries etc. \nA storage battery stores an electric charge in the form of ions on plates in the battery this type of battery may use the same material for the anode and cathode and also uses an electrolyte material but it must be charged before it will "produce" an electrical charge. Examples or the lead acid, Nickel Cadmium, Lithium ion etc.
36655
potential difference\ndue to which the crrent flows  through the ends of the battery due 2 which appliances work on cell or a battery
25244
What type of resistance--thermal, electral, or what?\n\nBoth are generally determined by the bandwidth of free electrons if any are present.  In electrical, it can also deal with impurities in the material.  In thermal, it can have to do with the irregularity of the molecular structure, the density of the solid, and the heat capacity of the molecules.
891056
An ideal voltmeter has infinite resistance: It is an open circuit. Although it is impossible to make a physical voltmeter with infinite resistance, a well designed voltmeter exhibits a very large internal input resistance. In some experiments, it is important to take into account the finite, non-ideal, internal resistance. To determine the internal resistance of the voltmeter, set up the circuit shown in figure . The voltmeter reads the voltage across itself, which includes its internal resistance. Since the circuit has only a single branch, the current flowing through the resistor also flows through the voltmeter.\n\nFor the required formulas and schematics, check out the reference.
489342
A VRLA battery is a basic lead acid battery and an internal between one or a number of cells will render it useless.  An internal short will prevent that cell from working (will produce no EMF) and so the output voltage of the battery will be reduced by the amount that would be given by that cell.\nBattery no good I'm afraid and no chance of recovering it.\nP.
609858
Its is the amount of resistance.  Same reason the wire in a lightbulb grows.  There is some level of electrical resistance, even in conductive materials.  The resistance creates heat and eventual combustion.
244228
Hi! You haven't mentioned which of the resistances are in series or parallel. Too bad we can't provide images in Yahoo! Answers.\n:(\n\nDo this: Remove V2 and V3 from the circuit, and replace them by a simple wire. Then find the equivalent resistance of the circuit taking V1 as the battery source. Then find the voltage drops across each of the resistances R1, R2, R3, R4 (you can do this by finding current through them using simple Ohm's Law and then multiplying by the R value).\n\nTAKE CARE that you note the DIRECTIONS of the voltage drops. The direction of voltage drop across R is the direction of current through R. So you must see if its positive or negative.\n\nOnce you have done this, repeat the procedure by removing V1, V3 and taking only V2 in circuit. This time, again specifically note the direction (+ or -). Then do it with V3.\n\nFinally, add the drops across each R, and this is when you use the signs (directions) of the voltage drops. For instance (just an EXAMPLE. You will get diff. values), if you get VR1 with V1 battery as -6V, with V2 as 4V, and with V3 as -2V, then VR1 will be:\n\nVR1 = (-6)+(4)+(-2) volts = -4 volts\n\nYou can use the same procedure to compute IR1: find current each time with V1, V2, V3 resp.\n\nHope this helps!
225362
Atoms electrons neutrons protrons and neucleic acids (or some type of acid) are probally inside a battery
953385
With electronics, they're typically built for a specific voltage.  Always stick with the same voltage rating.\n\nThe mA-hr rating refers to how long the battery will last.  It's the product of the current (how much power the battery is providing) and the total time at that current.  For example, a 250 mAh battery will last for one hour when providing 250 mA.  It will last 0.5 hrs when providing 500 mA.\n\nThe bigger the number, the longer it will last.
260531
The battery has two terminals, one positive and one negative, with two different types of metals.  Floating inside the battery is also a supply of acid, which scientists would call an electrolyte.  This acid keeps the electrons flowing between the positive and negative terminals all the time.\n\nWhen you set up a wire loop, one end is connected to the positive terminal and the other is connected to the negative terminal.  The free electrons inside the battery like to flow from the positive down to the negative terminal.  Now they have a way out of the battery through the wire.\n\nThe electrons flow through the wire and anything else connected to it, like a small lightbulb.  Some of the electrons are used to make the filament inside the lightbulb glow red hot.  The rest of the electrons continue around the loop and end up at the positive terminal of the battery again.  This process keeps going until either almost all of the electrons are used up in the lightbulb (the battery is 'dead') or the circuit is broken.
402553
Your circuit diagram looks strange - some resistance are denoted in Volts. Hmm, anyway, otherwise it's rather simple. You got to use your formulaes for calculating resistance in a series and ohm's law V=IR. I hope I'm right - I last did physics in 1999.\n------------------------------------------------\n\nThe resistors are all in series, therefore:\nTotal Resistance = 33 + 23 + 2/I + 1/I + 43\n                 = 99 + 3/I Ohms\n\nUsing Ohms law, V = IR. Therefore:\n=> 100 = I . Total Resistance\n=> 100 = I (99 + 3/I)\n=> 100 = 99 I + 3\n=> I = 97/99 Amperes  --- ans\n\nSub. value for I into Total Resistance:\n=> Total Resistance = 99 + 3/(97/99)\n                    = 9900/97 Ohms (or 102.06 Ohms) --- ans \n\nUsing V=IR\n\nV(3) = potential difference from 100 V to when current I passes through the 33 Ohm resistor\n     = 100 - [(97/99 x 33)] \n     = 202/3 V or 67.33 V (The reading of V (3)'s galvanometer) --- ans.\n\nV(2) = potential difference from V(3) to when current I passes through the 23 Ohm resistor\n     = 202/3 - [(97/99 x 23)] \n     = 4435/99 V or 44.80 V (The reading of V (2)'s galvanometer) --- ans.\n\nV(1) = potential difference from V(2) to when current I passes through the 2V resistor\n     = 4435/99 - 2\n     = 4237/99 V or 42.80 V (The reading of V (1) galvanometer) --- ans.
598213
On a battery, you have two terminals- a positive and a negative.  As a result, the net charge is 0.  Also, a battery is the source of 'power' in an electrical circuit, it is direct current (DC) instead of alternate(ing) current (AC).  The battery produces a certain number of Volts and from that, you can find the current at different points in the circuit, as well as the number of amps.  (I hope this is what you are talking about)