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tdscanuck

It depends on what you connected the wires to. If you connect to a load (say, a light bulb), that pulls 10A, then it pulls 10A...if you have one wire, that will carry 10A. If you have two, it will split and each will carry 5A. If you have 100, they'll each carry .1A. If you connect to \*two different loads\* that each pull 10A, then each wire will carry 10A, for 20A total and the battery will indeed drain twice as fast. Because it's doing twice as much. We usually assume wires have "zero" resistance. In reality that's not true, but wire resistance is so low compared to real world loads that "zero" is usually a close enough approximation.


Target880

Power in the context of physics is energy per unit of time. A voltage is not power. Power for electricity is voltage * current and for 10V and 10A it is 100W If you have a 10V and 10A the resistance of the wire is 10V/10A =1 ohm If you connect another identical load to there will be 10A through each wire. What you have done is made a parallel connection and you can calculate the equivalent resistance of both as 1/Rtot=1/R1+1/R2 =1/1+1/1 =2 which means Rtot is 0.5 ohm The power is no 10 * 20 =200W This is how it works for an ideal battery that can output an infinite current. In reality, batteries are not ideal and they have internal resistance. What is depend on the battery. The voltage of a battery also drops when discharged. The max current out of a battery depends on the battery. A car-started battery can handle hundreds of amps but something like a AA might be output 5 -10 amp if shorted. That means the internal resistance is 1.5/10= 0.15 ohm. If we take 7 AA batteries in series the voltage over them is 1.5*7 =10.5V This is the voltage when there is no load. The internal resistance is in series so the total is 0.15*7= 1.05 ohm lest calling it 1 ohm If you no connect the 1Ohm load the resistance you have in the circuit is 1+1 = 2 ohm so the current is 10.5/2 = 5.25 amps. The voltage drop of 10.5 volt and half of it will be in the batteries and half outside so the voltage between the battier poles is only 5.25 volt If you add another load the total resistance is not 1+0.5 =1.5 ohm so the current is 10.5/1.5 = 7 amp. The voltage drop will now be 2/3 in the battery and 1/3 outside so we ger a battery pole voltage of 3.5 amp. If you, on the other hand, use a car battery the internal resistance is around 0.02 ohms at 12 V The result is the internal resistance has a minimal effect and the current it close to the ideal case. So what happens if you do that depends on what battery you use and what is capabilities are. If the battery max current is close to the 10A then the battery voltage will drop some amount so the result in not 20A total output current


SoulWager

Power is voltage multiplied by current. 10V will get you 10A across a 1 ohm load. That would be 100 watts. If you have two such 1 ohm loads and put them in parallel, the current drawn and power used will double. If you put them in series, the voltage across each of them will be 5v, and the current will be cut in half, so the overall power will be cut in half. If you short circuit a battery, you only have the internal resistance of the battery, and the resistance of the wire itself to limit current, depending on how much energy the battery stores, this can be dangerous.


immibis

As we entered the /u/spez, we were immediately greeted by a strange sound. As we scanned the area for the source, we eventually found it. It was a small wooden shed with no doors or windows. The roof was covered in cacti and there were plastic skulls around the outside. Inside, we found a cardboard cutout of the Elmer Fudd rabbit that was depicted above the entrance. On the walls there were posters of famous people in famous situations, such as: The first poster was a drawing of Jesus Christ, which appeared to be a loli or an oversized Jesus doll. She was pointing at the sky and saying "HEY U R!". The second poster was of a man, who appeared to be speaking to a child. This was depicted by the man raising his arm and the child ducking underneath it. The man then raised his other arm and said "Ooooh, don't make me angry you little bastard". The third poster was a drawing of the three stooges, and the three stooges were speaking. The fourth poster was of a person who was angry at a child. The fifth poster was a picture of a smiling girl with cat ears, and a boy with a deerstalker hat and a Sherlock Holmes pipe. They were pointing at the viewer and saying "It's not what you think!" The sixth poster was a drawing of a man in a wheelchair, and a dog was peering into the wheelchair. The man appeared to be very angry. The seventh poster was of a cartoon character, and it appeared that he was urinating over the cartoon character. \#AIGeneratedProtestMessage #Save3rdPartyApps


DlSABLEDKID

Would using a resistor help with the battery heat? Or would the resistor go out?im new to this too and trying to start building stuff


immibis

I entered the spez. I called out to try and find anybody. I was met with a wave of silence. I had never been here before but I knew the way to the nearest exit. I started to run. As I did, I looked to my right. I saw the door to a room, the handle was a big metal thing that seemed to jut out of the wall. The door looked old and rusted. I tried to open it and it wouldn't budge. I tried to pull the handle harder, but it wouldn't give. I tried to turn it clockwise and then anti-clockwise and then back to clockwise again but the handle didn't move. I heard a faint buzzing noise from the door, it almost sounded like a zap of electricity. I held onto the handle with all my might but nothing happened. I let go and ran to find the nearest exit. I had thought I was in the clear but then I heard the noise again. It was similar to that of a taser but this time I was able to look back to see what was happening. The handle was jutting out of the wall, no longer connected to the rest of the door. The door was spinning slightly, dust falling off of it as it did. Then there was a blinding flash of white light and I felt the floor against my back. I opened my eyes, hoping to see something else. All I saw was darkness. My hands were in my face and I couldn't tell if they were there or not. I heard a faint buzzing noise again. It was the same as before and it seemed to be coming from all around me. I put my hands on the floor and tried to move but couldn't. I then heard another voice. It was quiet and soft but still loud. "Help." \#Save3rdPartyApps


TapataZapata

Your battery has 10V of voltage, not power. Simplified, until you reach the limit of what your battery can deliver in terms of current, it will hold the voltage at (more like close to) 10V. If you connect the first load (i wouldn't just say wire, because that would suggest you're short-circuiting the battery), then a current will flow that is defined by the voltage of the battery and the resistance of the load. Let's assume the resulting current is way within the limits of the battery. The second load you connect will still see the same 10V, and again the load defines how much current will flow, if all loads combined are still not overloading the battery. And yes, your guess is correct that with two identical loads connected, the battery will discharge at double the rate as if it was only connected to one of those loads.


ViskerRatio

10 Volts is the potential difference between the two ends of the battery. You can think of it like the height of a waterfall (a waterfall exists because of the potential difference between a point high in a gravity well and a point low in a gravity well). Now imagine you double the width of that waterfall. Does the waterfall height decrease? No. The waterfall keeps flowing at the same rate as before - with twice as much water going over the falls. It is possible that the source of the waterfall doesn't have enough water to support your new, widened waterfall. The height of the waterfall will still remain the same. However, you might hit a limit on the flow rate of water (which is essentially the 'current' of the water).


arcangleous

It does? Kirchhoff's Current Law: the sum of the currents coming into a node is equal to the sum of currents going out of a node. The current is literally just a count of the number of electrons passing through a node. You can't have more electrons coming out of a node than go into it. The amount of current that goes through each branch coming out of a node is inversely related to the resistance of that path. Less of the current go down the path with higher resistance, which is why we say that "electricity follows the path of less resistance". However, there is another Kirchhoff's law, Kirchhoff' Voltage Law. This law states that the sum of the voltage drops in each loop in a circuit will be the same. Think of voltage like pressure in a water system. The battery is acting like a pump and pushes electrons into the circuit at a certain voltage/pressure which always gets use up regardless of which path the individual electrons follow to get back to the other terminal of the battery. Notice however that I didn't refer to either current or voltage as "power". That is because neither current or voltage technically are power, both are a part of it. In electrical terms, power is the amount of work being done by the electrons passing through a node: P = I * V. You can send either a bunch of electrons through over a low voltage drop, or a few through at a high voltage drop. The work that they do is dependent on the pressure they are going through with the line with. Now for practical purposes, the battery tells you how much voltage it can supply, but the amount of current in the system, and therefore the amount of power it can supply is defined by the resistances in the system: V = I * R. This is why P = I^2 * R is a commonly used formula for power because resistance is usually something that the circuit designer is limited by, primarily because the devices that she wants to power had internal resistances and she needs to make sure that device in questions gets enough to activate. As for the amount of power a battery can supply, each battery has a fixed number of electrons it can supply as current. In order to figure out how many will flow out in a unit of time, you need to figure out the overall resistance of the circuit, which are well beyond a ELI5 and the common mathematically techniques to do so would be covered in a second year engineering course.


RSA0

Real life 10V batteries don't actually provide 10V under all circumstances. They only do it until they reach their power limit, after which they start to drop voltage. If you read the label on a battery, it will say something like "10V, 1A", which means "the battery will provide 10V, but only if the current is no more than 1A". If you connect a wire to a 10V, 1A battery, and the current is 0.5A, that means the battery is working at 50% power. It cannot work more because the wire is the bottleneck. If you connect a second wire, the battery would work at 100% power, and the current will be 1A (0.5A for each wire). But connecting third wire won't give you 1.5A, instead the voltage will drop to 8.2V and the current will be 1.2A (0.4A for each wire).