How to boost current??
- Thread starter krishh9
- Start date
Daniel Payne
Moderator
That 3.3v battery will produce .250A for one hour, .500A for 30 minutes, 1A for 15 minutes, and 2A for about 7.5 minutes. If you need longer life, then you will have to place multiple batteries in parallel.
Remember that P = I*V, so any increase in current means a corresponding decrease in V at the same power output. What are your designing that draws 2A of current?
Remember that P = I*V, so any increase in current means a corresponding decrease in V at the same power output. What are your designing that draws 2A of current?
"so which circuit can boost up to 2amp current keeping same voltage??"
Something something laws of thermodynamics
Batteries can supply a range of currents. 250mah capacity is an energy metric, not a current metric. Daniel did the math for you on how long that battery could support 2 amps, provided its internal resistance is low enough, and provided it doesn't overheat on you.
Something something laws of thermodynamics
Batteries can supply a range of currents. 250mah capacity is an energy metric, not a current metric. Daniel did the math for you on how long that battery could support 2 amps, provided its internal resistance is low enough, and provided it doesn't overheat on you.
Daniel Payne
Moderator
tty2,
Thanks for the follow up.
This reminds me of an interview question posed to me in college: "I have two black boxes, each with two terminals. Inside one is a Thevenin circuit, and the other a Norton circuit. How can you tell the difference between the two black boxes?"
The question drove me nuts, and I answered, "Well, because they are equivalent circuits, there would be no measurable difference on the two output pins."
Wrong answer.
Thanks for the follow up.
This reminds me of an interview question posed to me in college: "I have two black boxes, each with two terminals. Inside one is a Thevenin circuit, and the other a Norton circuit. How can you tell the difference between the two black boxes?"
The question drove me nuts, and I answered, "Well, because they are equivalent circuits, there would be no measurable difference on the two output pins."
Wrong answer.
Maybe you shouldn't have been thinking...
Outside the box
https://www.youtube.com/watch?v=6YMPAH67f4o (thank you, CSI Miami!)
Outside the box
https://www.youtube.com/watch?v=6YMPAH67f4o (thank you, CSI Miami!)
Daniel Payne
Moderator
The correct answer to the question is that the Norton equivalent circuit has a current source connected to a resistor and therefore draws power, thus it would be warmer to the touch than the Thevenin circuit which has a voltage source connected to a resistor in series, drawing no current and consuming no power.