so I have a 500 ohm resistor with a known voltage across it. a known voltage input. and a load across the output with a known voltage on the output
how do I calculate the current flowing into the load by the voltage drop across the resistor??
I cant find the right google search terms to search "this" specific situation.
and I dont want to cut wires and use my multimeter in current setting in series to figure it out.
8.1 to 8.5v on output across the load
input is about 21v
voltage drop across resistor 6.1v approximately.
how do I calculate the current flowing into the load by the voltage drop across the resistor??
I cant find the right google search terms to search "this" specific situation.
and I dont want to cut wires and use my multimeter in current setting in series to figure it out.
8.1 to 8.5v on output across the load
input is about 21v
voltage drop across resistor 6.1v approximately.
current = voltage/resistance = 6.1V / (500 Ohm) = 0.012 A (or 12mA)
You have to search for Ohm's law and Kirchhoff's law.
One thing in your numbers does not fit: you wrote that supply is 21V, voltage drop across 500 Ohm resistor is 6.1V (I assume it is in series with the load), voltage drop across the load is 8.1 to 8.5V. If all is presented correctly then voltage drop across load plus voltage drop across series resistance has to be equal to supply voltage. In your case it does not fit: 6.1V + 8.1 V is not equal to 21V supply. I would suggest to post schematic to better understand your setup.
You have to search for Ohm's law and Kirchhoff's law.
One thing in your numbers does not fit: you wrote that supply is 21V, voltage drop across 500 Ohm resistor is 6.1V (I assume it is in series with the load), voltage drop across the load is 8.1 to 8.5V. If all is presented correctly then voltage drop across load plus voltage drop across series resistance has to be equal to supply voltage. In your case it does not fit: 6.1V + 8.1 V is not equal to 21V supply. I would suggest to post schematic to better understand your setup.
21v power supply driving an LED through a resistor (adjustable power supply)
with 6.1v-ish across the resistor. and 8.1 to 8.5v-ish across the LED (the LED is 3 in series as far as I can tell through the die of the chip)
can you guestimate approximately what the current might be.
I think it's around 30 to 60mA but im not 100% sure
with 6.1v-ish across the resistor. and 8.1 to 8.5v-ish across the LED (the LED is 3 in series as far as I can tell through the die of the chip)
can you guestimate approximately what the current might be.
I think it's around 30 to 60mA but im not 100% sure
Your numbers do not add up.
Voltage across series resistor (6.1V) plus voltage across load (8.5V) MUST add up to source voltage (21V) and they do not.
Of course, your "source" may have a hidden internal resistor , so we expect more details about your circuit to go on.
In any case, LED and series resistor are in series, so current through both is the same, so current is: 6.1V/500 ohm=0.012A=12mA
Math is right, but data is doubtful, so recheck it.
I couldn't figure out the equation because google wasn't returning the search results that I wanted
i was getting instead how to convert the current through a resistor into the voltage across it
rather than converting the voltage across it into current
It was giving me the opposite search results that I wanted
and It wasn't showing me the formula to figure it out myself.
and I dont know the formula. so I was trying to find a calculator that would do it for me and i still couldn't find that
Try searchin for it yourself using various search terms. its hard to find this specific calculation
I dont want to convert the current through the resistor and the voltage across the output into the voltage across the resistor. i already know the voltage across the resistor
I need to find the CURRENT through the resistor. by USING the voltage across the input and output and across the resistor.
but google wouldn't return this in the results and gives me the opposite scenario instead
I decreased the voltage boost circuit so that the measured current across the LED was 20mA and its not very bright.
It has 3 tiny LED's in series.
I dont think it can handle much more than half a watt or 1/4 a watt or less since it doesnt have any significant form of heatsinking
the front of the LED still gets pretty searing hot with only 20mA of current.. so it must be three high voltage LED's in series
with 46mA of current. each LED would be getting 46mA. and since the voltage is about 8.5 to 9v across the 3 LED's in series so about half a watt of power in the LED's already
at 100mA it would probably be near 1 watt in total for the LED chip which is probably too much. at near 50mA it gets pretty hot. would burn your finger if you held your finger on for more than a second at a time.
so at 100mA it would probably overheat if it doesnt have some form of heatsinking.
Is there some kind of solderable miniature heatsink strip?? Like a flat finned blade of copper that I could solder at an angle?
aluminum foil doesnt seem to take solder. and just shrinks and shrivels up
and the LED is on a PCB with solder joins and wire leads coming out the back. so I cant just remove it.
I dont think the solder leads on the LED can heatsink enough heat away anyways.
so I will only power it with 20 to 50mA max.
I wanted to use it as a decently powerful flashlight or such.
if I have it at 20mA I could use it as a nightlight really nicely running from a 5v USB portable power bank and be pretty useful in emergencies
we're going all out here where im at with solar power and emergency supplies
i was getting instead how to convert the current through a resistor into the voltage across it
rather than converting the voltage across it into current
It was giving me the opposite search results that I wanted
and It wasn't showing me the formula to figure it out myself.
and I dont know the formula. so I was trying to find a calculator that would do it for me and i still couldn't find that
Try searchin for it yourself using various search terms. its hard to find this specific calculation
I dont want to convert the current through the resistor and the voltage across the output into the voltage across the resistor. i already know the voltage across the resistor
I need to find the CURRENT through the resistor. by USING the voltage across the input and output and across the resistor.
but google wouldn't return this in the results and gives me the opposite scenario instead
I decreased the voltage boost circuit so that the measured current across the LED was 20mA and its not very bright.
It has 3 tiny LED's in series.
I dont think it can handle much more than half a watt or 1/4 a watt or less since it doesnt have any significant form of heatsinking
the front of the LED still gets pretty searing hot with only 20mA of current.. so it must be three high voltage LED's in series
with 46mA of current. each LED would be getting 46mA. and since the voltage is about 8.5 to 9v across the 3 LED's in series so about half a watt of power in the LED's already
at 100mA it would probably be near 1 watt in total for the LED chip which is probably too much. at near 50mA it gets pretty hot. would burn your finger if you held your finger on for more than a second at a time.
so at 100mA it would probably overheat if it doesnt have some form of heatsinking.
Is there some kind of solderable miniature heatsink strip?? Like a flat finned blade of copper that I could solder at an angle?
aluminum foil doesnt seem to take solder. and just shrinks and shrivels up
and the LED is on a PCB with solder joins and wire leads coming out the back. so I cant just remove it.
I dont think the solder leads on the LED can heatsink enough heat away anyways.
so I will only power it with 20 to 50mA max.
I wanted to use it as a decently powerful flashlight or such.
if I have it at 20mA I could use it as a nightlight really nicely running from a 5v USB portable power bank and be pretty useful in emergencies
we're going all out here where im at with solar power and emergency supplies
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What you need is given by the Ohm's law. I suggested you to google for it.
It states: I=U/R,
where I is the current through the resistance R and U is the voltage across this resistance.
Also google for Kirchhoff's laws.
I think Ohm's and Kirchhoff's laws are absolute basic knowledge anyone dealing with electronics shall have. I am even more surprised after seeing your more or less technical discussion in your 5V 1A linear regulator.
EDIT: and yes, I searched using google before suggesting it to you. It gives the right links as the very first hit in the search result.
It states: I=U/R,
where I is the current through the resistance R and U is the voltage across this resistance.
Also google for Kirchhoff's laws.
I think Ohm's and Kirchhoff's laws are absolute basic knowledge anyone dealing with electronics shall have. I am even more surprised after seeing your more or less technical discussion in your 5V 1A linear regulator.
EDIT: and yes, I searched using google before suggesting it to you. It gives the right links as the very first hit in the search result.
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all of those symbols are just scribbles to me. I don't speak mathematical functions and scientific notation.
i just know the basics of how to add subtract divide and multiply.
is there a way to do it without having to know all of those symbols?
Or some easy way to learn how to use those symbols and what they all mean?
Also I was trying to search for strings like "how to divide resistor voltage input and output voltage and get current across output" and nothing useful was coming up
I couldn't figure out the right search string to use
i just know the basics of how to add subtract divide and multiply.
is there a way to do it without having to know all of those symbols?
Or some easy way to learn how to use those symbols and what they all mean?
Also I was trying to search for strings like "how to divide resistor voltage input and output voltage and get current across output" and nothing useful was coming up
I couldn't figure out the right search string to use
. The 5 basic maths symbols really cover the majority of what you need, most of the others, like squares and square-roots are just shorthand of divisions and multiplications
Well, if you understand add subtract divide multiply, that only leaves the two others, and if they don't come up then you should have it covered.
My suggestion is not to decide you don't know it in advance, just because it has "math". To me it is just arithmetic, I do it all on a cheap little calculator.
Ohm's LAw is basic. Volts is ohms times amps (V=RxI). Amps is volts divided by ohms(I=E/R). Ohms is volts divided by amps (R=E/I). Those are just three ways to say the same relationship. If you know two of the three things, you can always calculate the third.
The only ones that require square roots or squaring are the ones involving power.
My suggestion is not to decide you don't know it in advance, just because it has "math". To me it is just arithmetic, I do it all on a cheap little calculator.
Ohm's LAw is basic. Volts is ohms times amps (V=RxI). Amps is volts divided by ohms(I=E/R). Ohms is volts divided by amps (R=E/I). Those are just three ways to say the same relationship. If you know two of the three things, you can always calculate the third.
The only ones that require square roots or squaring are the ones involving power.
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