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From what I've come to understand, a higher voltage source leads to a higher current from source to the destination.

Example: On a 1.5V battery let's say I'm getting 50mA load from my device rated for 25mA to 100mA.

Question: I can use a resistor to lower the current but how can I make the current higher? How can I figure out the limits of current for any given power source?

Jacksonkr
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    your question is a bit unclear. Recommended current draw from batteries are usually on the datasheet (for example: you can draw more but lifetime is not optimal). Short circuit current is limited mainly by internal resistances. Or.. you want to find out how to drive your load with more current? In which case we'll need to know what your load is. – Wesley Lee Aug 01 '16 at 15:18
  • @WesleyLee There's no project involved, I'm merely trying to make sense of the world of electricity. My question is more about proper ways to manipulate current from a battery (or any) power source. – Jacksonkr Aug 01 '16 at 15:21
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    In your example do you mean the 1.5 volt battery is connected to a load which is drawing 50mA? You apparently know that if you were to place a resistor in series between the battery and the load that the current will drop below 50 mA. (By the way, generally true, but not strictly true in all instances.) And your question is how would you get the current higher in the same example with the same load connected to the battery. Or do you mean in general how do you get more current from a voltage supply? – FiddyOhm Aug 01 '16 at 15:25
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    @Jacksonkr -- then I'm afraid that the answer will be "It depends, look at the datasheet". This one will give nice examples of lifetime and behavior with different loads: http://data.energizer.com/PDFs/cr2032.pdf – Wesley Lee Aug 01 '16 at 15:30
  • Otherwise its impossible to know without characterization of the battery. And/or some ballpark guesses by the type/size/quality of battery you have. – Wesley Lee Aug 01 '16 at 15:31
  • @FiddyOhm More in the general way. I will eventually want to understand the 50mA example and the general example so the more the merrier. I'll reword my question a bit to make it more sensical too. – Jacksonkr Aug 01 '16 at 15:31
  • Not sure it this is the answer you want but.. once I was driving LEDs for lightpainting on the street with batteries. The current was limited by battery internal resistance. So I put more batteries in parallel to get more current. This may or may not be applicable in your 50mA example. – Wesley Lee Aug 01 '16 at 15:35
  • 1- as long as your circuit total resistance is Constant, increasing the voltage will increase the amount of current taken from the source but take in mind that your power source has a maximum amount of current that can be supplied [Maximum power] exceeding this value will heat up the power supply and destroy it 2-Buy putting a series resistance in your circuit : YES you decrease the total current flowing in the light bulb BUT=> your light bulb is not having the same voltage across its terminals; less than the power supply voltage – Elbehery Aug 01 '16 at 15:39
  • Possible duplicate of [What does a resistor do?](http://electronics.stackexchange.com/questions/136377/what-does-a-resistor-do) – Ignacio Vazquez-Abrams Aug 01 '16 at 15:40
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    Jackson, there is a lot of confusion about the relationship of voltage, current and load resistance amongst newcomers to this topic. You need to study Ohm's Law in a meaningful way to get a correct insight into the relationship between V,I & R. IMHO most U-tube videos on this topic will probably confuse you more than enlighten you. Try to find the U-tube video by Dan Sullivan about Ohm's Law, it is the least confusing of the lot and lays it out in a logical manner for most beginners. Spend some quality time with a DMM, a battery and an assortment of resistors to get meaningful insight into OL – FiddyOhm Aug 01 '16 at 15:43
  • @FiddyOhm I am definitely new and am just looking to get pointers. Thank you for seeing that and addressing it in a productive manner. I will certainly watch the videos you mentioned. That being said, transistor gave a solid answer for what I was asking. Thanks for your pointers! – Jacksonkr Aug 01 '16 at 15:46

3 Answers3

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I'll assume you're talking in the most general terms. If that's what you want, you should always talk about a "voltage source" rather than a battery, since batteries have some specific characteristics which will get in the way of your question.

So, what you seem to be referring to is a circuit like this

schematic

simulate this circuit – Schematic created using CircuitLab

Since this is an ideal circuit, the ammeter has no resistance, and drops no voltage when it measures the current. Real ammeters do have resistance, and do drop voltage, but these are kept as small as possible.

Also, you'll note that I've tied the negative lead of the voltage source to ground. This not absolutely necessary, but is custormary when discussing the operation of a circuit. It gives a reference point, which makes discussions clearer.

Now, the current is .05 amps, and the voltage is 1.5 volts. What is the resistance? That's simple (for simple resistors). $$V = iR$$ That is, voltage equals current times resistance. This can be rewritten as $$R = V/i = 1.5/.05 = 30$$ Another way to rewrite the relationship is $$i = V/R$$ and this addresses your question.

There are only two variables affecting current, V and R. So if you want to increase current, you have 3 choices:

1) You can increase V. If you double the voltage, you'll double the current.

2) You can decrease R. If you halve the resistance, you'll double the current.

3) You can do both. And you can do the math.

WhatRoughBeast
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I can use a resistor to lower the current but how can I make the current higher?

For a fixed battery voltage, V, and a fixed load resistance, R, the current is given by Ohm's law:

$$ I = \frac {V}{R} $$

If you want to make I larger you have two choices:

  1. Increase \$V\$.
  2. Decrease \$R\$.

Increasing V is possible by increasing adding more cells in series or by using a voltage boosting circuit but you need to be sure the load (which you haven't specified) is rated for the increased voltage.

Generally we can't reduce the resistance of the load unless we have the ability to modify the load. Again details missing in the question.

How can I figure out the limits of current for any given power source?

That's a bit more tricky but here are a few pointers:

  • Mains supplies: check the ratings of the supply transformer, the incoming fuse rating, the cable rating, the socket rating, etc.
  • AC/DC supplies: rating plate, datasheet, etc.
  • Batteries: label, datasheet.

Improve the question and the answers will improve.

Transistor
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  • This is what I was looking for. Thanks for being patient with my limited-knowledge question. @FiddyOhm gave some great pointers too. – Jacksonkr Aug 01 '16 at 15:48
  • @Jacksonkr: Thanks for accepting the answer but I'd be happier if you un-accept for a day or two to encourage others to answer. That way you will get several viewpoints which will probably enhance your understanding - or lead to total confusion! ;^) – Transistor Aug 01 '16 at 15:50
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For any source, its voltage and current limits will be specified in it datasheet, and then there's Ohm's law,

$$ E = IR, $$

which states how voltage, current, and resistance relate to one another, and where

  • E is voltage in volts,
  • I is current in amperes, and
  • R is resistance in ohms

Using your example, and rearranging the formula to solve for the resistance required to draw 50 milliamperes from a 1.5 volt source yields:

$$R =\frac{E}{I} = \frac{1.5V}{0.05A} = 30\text{ ohms} $$

In order to answer your question we can again rearrange to solve for I, like this:

$$ I = \frac{E}{R}$$

Note that if you want to decrease \$I\$ you can either decrease \$E\$ or increase \$R\$, or do both, and if you want to increase \$I\$ you can either increase \$E\$ or decrease \$R\$, or do both.

EM Fields
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