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# How to calculate value of a Resistor for a Circuit ?

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#### ThinkHimanshu

##### New Member
Simple LED circuit doubt.

Assume : I have a LED with forward voltage of 3.0 V and forward Current of 350mA.

With 9V battery, i would be calculating the required Resistance with the Ohms law :

R = ( Source Voltage - Forward Voltage) / Forward Current.

Here we dont consider the Current Ouput of Battery ....

Doubt : Will it make any difference for R when i have

1) 9V battery with 1A output.
2) 9V battery with 2A output.

Kindly explain the behavior.

The current is determined by the voltage across the resistor. The amount of current the battery can supply doesn't matter as long as it's enough to provide the calculated current. The LED and resistor only draw the calculated current and no more.

The second part of your post looks almost textbook. A battery can be viewed as a voltage source in series with a resistor. Hence comes the term equivalent series resistance (esr).

As JonSea noted, that usually doesn't make a difference. However, in your two examples, the resistance of the battery is halved from about 9 ohm to 4.5 ohm. Using your equation, the calculated R is 17 ohm. That would be the total resistance of the series circuit. So, the 1-A battery would need a smaller series resistor than the 2-A battery for the required current.

Two more things to consider in a practical circuit:
1) If 350mA is the rated maximum current of any component, it is unwise to drive the component at that current if you want the component to have a long life.
2) A '9V battery', if of the PP3 type/size, holds a relatively small amount of energy so will soon die if asked to provide 350mA.

jpanhalt what you suggesting is When i have a battery with 9V 1A then i need (17-9) ohm. And when i have 9V 2A then i need (17 - 4.5) ohms ??

Kindly revert. (i know this may b very silly doubt)

I was reluctant to give the exact calculated values, as they would apply only to the textbook situation. In that case, your calculation is correct. However, as alec-t suggests, one generally does not face that situation in real life. Also a tiny battery will warm and its internal resistance will change if subjected to a 350 mA discharge for long.

And finally, 17-4.5=12.5 ohm. Not even the E192 series (https://www.logwell.com/tech/components/resistor_values.html ) has a 12.5 ohm resistor. So, unless one plans on using very expensive resistors or multiples in parallel to get the unusual calculated values, you will have to use something else (probably the next higher value).

Simple LED circuit doubt.

Assume : I have a LED with forward voltage of 3.0 V and forward Current of 350mA.

With 9V battery, i would be calculating the required Resistance with the Ohms law :

R = ( Source Voltage - Forward Voltage) / Forward Current.

Here we dont consider the Current Ouput of Battery ....

Doubt : Will it make any difference for R when i have

1) 9V battery with 1A output.
2) 9V battery with 2A output.

Kindly explain the behavior.

Hi,

Really the only way to get this right is to try a resistor value and then modify that value after you measure the current. So you adjust the resistor value to suite.

The other problem though is the battery run down. The voltage will change as the battery looses energy. The LED will dim little by little. The only way to make up for that is with a regulating circuit or adjustable brightness circuit.

Thank you all. I got my answer.

Battery dont have "A" component, it have "mAh" from where the current is depleted.

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