Question about dependence of current in a circuit.

Firstly, sorry for my poor English and I know that this is a idiot question in electronic. I am a fresh beginner.

1st question, Imagine that I have a portable battery which can output 5v/2A. What does it means? Is it means this battery will force itself to output 5v with 2A? Or just means this battery will output 5 volt and can provide 2 amp at maximum?

2nd question, for those AA battery in the market state that they can output 1.5v but they doesn't stated the output current. Then what is the current they output? Assume I have a LED which consume 3v to light up, then I connect 2 1.5v battery to become 3v and supply power for the LED. Then what is the current in the circuit? Is the resistance of the LED constant? Why the LED won't blew up? << (I have tried, the LED work perfectly)

3rd question, I want to buy a motor that consume 1 - 6v and the reference current is 0.35-0.4A. What is the meaning of reference current? Is it means that the motor will regulate the current pass through itself to 0.35-0.4A? If I supply the power to the motor by the portable battery stated in 1st question (output 5V/2A), will the motor blew up because of the current supply (2A) is much larger than the reference working current stated by the motor (0.4A)?

4th question, I have a device that should be powered by 5v/1.2A via micro USB. Will I damage the device if I use 1st question portable battery as power source? If no, what is the current output by the battery? If the current is 1.2A then is it means the device can regulate the circuit current to 1.2A even the power source can output 2A? If the answer is YES, how about I change another power source that output 10A? Will the device be damaged? (because I have tried to supply power to a LED by a laboratory DC power supply, I keep increasing the supply current, the LED blew up finally.)

Sorry for those long questions, I am a newbie. I know the basic theory of KCL, KVL and ohms law, it will be appreciated if anyone can explain the answer to me by these laws.
Thousand Thanks!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!:thumbsup:

So many questions!!

1) You are right; the current rating is how much the battery is CAPABLE of supplying. It will only supply as much as the load requires. The answer to this this question is summed up in Ohm's Law : V=IxR.

2) You were fortunate you didn't damage the LED. LEDs (and diodes in general) have a "forward voltage", that is, when they are conducting, the voltage across the device remains relatively constant, regardless of the current through them. That's why usually a resistor is placed in series with the LED to limit the current. In your case, your LED forward voltage was close to the supply (3V) voltage, so you didn't pass excessive current through the LED. If you had used, say 9V, you would be asking why your LED was damaged.

3) I have never heard the term "reference current" for motors. Maybe that's a problem in translation. Your motor will draw higher current at higher voltage. But if it's specified as a 5Volt motor, it shouldn't be damaged hooking it up to 5 volts.

4) Again, the battery will essentially maintain a constant voltage and will provide whatever current the load demands. Look at Ohm's law again.

Here's a thought experiment. My house gets 120 Volts from the power company, 200 Amps maximum. If I only plug in a 60 Watt light, I don't pass 200 amps through it, I pass 0.5 Amps.

barry said:

So many questions!!

1) You are right; the current rating is how much the battery is CAPABLE of supplying. It will only supply as much as the load requires. The answer to this this question is summed up in Ohm's Law : V=IxR.

2) You were fortunate you didn't damage the LED. LEDs (and diodes in general) have a "forward voltage", that is, when they are conducting, the voltage across the device remains relatively constant, regardless of the current through them. That's why usually a resistor is placed in series with the LED to limit the current. In your case, your LED forward voltage was close to the supply (3V) voltage, so you didn't pass excessive current through the LED. If you had used, say 9V, you would be asking why your LED was damaged.

3) I have never heard the term "reference current" for motors. Maybe that's a problem in translation. Your motor will draw higher current at higher voltage. But if it's specified as a 5Volt motor, it shouldn't be damaged hooking it up to 5 volts.

4) Again, the battery will essentially maintain a constant voltage and will provide whatever current the load demands. Look at Ohm's law again.

Here's a thought experiment. My house gets 120 Volts from the power company, 200 Amps maximum. If I only plug in a 60 Watt light, I don't pass 200 amps through it, I pass 0.5 Amps.

Click to expand...

Thank you Barry! thousand thanks!!!
I am still confusing that what is the demanding current of LED? as much as it burn out? If I supply correct voltage, it keeps drawing "excessive" current untill it burn out. (assuming I have a unlimited current supply source.)

A motor uses a fairly low current when it is running with no load. But when it has a heavy load its current is much higher. When the motor starts running then its current is maximum like it is simply a piece of wire because it is stalled when it starts running or when something mechanically stops it.

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An AA battery cell can be a cheap old carbon-zinc type that is usually dead and is leaking by the time it is sold so its maximum output current is very low.
Energizer and Duracell make AA alkaline cells that have a maximum output current of about 1A for maybe 10 minutes. The voltage continues to drop as it discharges.

mannokhkhkhk2207 said:

Thank you Barry! thousand thanks!!!
I am still confusing that what is the demanding current of LED? as much as it burn out? If I supply correct voltage, it keeps drawing "excessive" current untill it burn out. (assuming I have a unlimited current supply source.)

Click to expand...

If you look at the data sheet for an led, there's a spec for nominal current and maximum current. You need to size your resistor based on those values and the supply voltage.

An LED is not a resistor where the current doubles when the voltage is doubled. Instead an LED is a diode where the current increases A LOT when the voltage fed to it is increased only a little.
Therefore a resistor is added in series with the LED to limit its current.

mannokhkhkhk2207 said:

Thank you Barry! thousand thanks!!!
I am still confusing that what is the demanding current of LED? as much as it burn out? If I supply correct voltage, it keeps drawing "excessive" current untill it burn out. (assuming I have a unlimited current supply source.)

Click to expand...

Hello. This is only about above question of yours,for rest I think you have got fairly good idea.
You please observe that LEDs being diodes, have completely different functioning than resistors. They have constant voltage drops across them up to very high currents(see I-V characteristics of diodes). For understanding purpose you first visualize as we have a simple circuit having a 5V Dc supply in series with a resistor(100ohm) and an LED. Now as said, our LED will have constant voltage drop if forward biased(for specific values look into datasheet of LED) assume that to be 2V. The remaining voltage drop(i.e difference between voltage provided by supply and that appearing across LED) which is 3V in our analysis, will appear across a series resistor (or simply across wire,if no series resistor is present and LED is directly connected to supply). Now that resistor will decide how much current it wants to be flowing through circuit, according to Ohm's law.

I = (Voltage across resistor)/(resistance of resistor) = (Vcc - Vled)/(100) = (5 - 2)/100 = 0.03 = 30mA.

In this way the series resistor decides the current flowing through circuit. The higher the value of series resistance the lower the current, that it will allow to flow and vice versa . Now if specifications of our LED tells it can withstand only up to 20mA current, Off course we will have to connect a higher resistance(how exactly that you can find by ohms law ) otherwise 100ohm resistor, allowing 30mA will definitely blow our LED. If no resistor is connected, we will have only our wire(having very low resistance) in series with our LED. This wire will allow the highest current your supply can provide, the higher the supply voltage the sooner your LED will blow. You can even try this . I hope this explanation would help you.