# Why electric current doesn’t decrease for each wire i connect to a battery?

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For example, if i have a battery with 10V of power, and then i connect a wire to the battery it will have 10A of current in it.

If i connect 2 wires, why current doesn’t split between the two wires so that i have 5A in each one?

I have no knowledge in this area, the only logical conclusion i can come to with the little information i have is that both wires will have 10A of current because both will “work at full potential”, but then, the battery will run out of power at double the speed compared to having only 1 wire connected.

Is my guess correct or the solution is another one?

In: 1 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. 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. If you overload the battery and it can’t provide more current then it will absolutely split the current.

Usually we do things at relatively low currents, where the battery is able to provide more current than we need, so we don’t encounter this scenario. Batteries get hot when you use too much current so it’s not good.

It’s possible to make a constant current power supply circuit (instead of the usual constant voltage), and it works freakily backwards: to power several things you wire them in series, because if you wire them in parallel then the current splits when you don’t want it to, and to turn something off you short-circuit it instead of disconnecting it. 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 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. 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). 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. 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).