Voltage Divider Question.

shaneshane1 said:

Hi just wondering if my calculations are correct on this subject?

I've made a voltage divider (in theory) with the output 2.1V from 12V

I used a 248R and a 52R with the output current at 20mA

is this equation correct?

12V/0.040A = 300R and 480mW

And

2.1V/0.040A = 52R and 84mW

300R - 52R = 248

R1 = 248R

R2 = 52R

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ericgibbs said:

Hi,
That should give about 2.08V at the junction.

Whats the 2.1V going to be connected too.?

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bryan1 said:

Hiya Shane,
Eh mate checkout this link https://electronics2000.co.uk/download.php#assistant

I've been using that calculator for years now and I'm sure you'll find it very handy too and the question you posted can be answered in no time by yourself.

Cheers Bryan

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Nothing, its just some theory!
but i would say a 20mA LED with a forward voltage of 2.1V

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ericgibbs said:

hi shane.


That could be a problem, normally you would not connect a resistor in parallel with a LED. Work out the currents in the LED and 52R..?

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shaneshane1 said:

This is what i am having trouble with!

The reason i put 40mA in the equation for R1 and R2 is because i though the current would halve at the center of the resistor's equaling to 20mA?

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shaneshane1 said:

The reason i put 40mA in the equation for R1 and R2 is because i though the current would halve at the center of the resistor's equaling to 20mA?

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crutschow said:

The current does not change at the junction of the two resistors if there is no load. They will both carry 40mA. If there is a load at the junction then it will depend upon the resistance of the load in relation to the equivalent resistance at the resistor junction (look up Thevenin's Theorem).

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