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?
Thanks in advance.
In: 1
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
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