how to calculate resistance

[misterT]

ok, then tell me how to use voltage devider rule in circuit posted in this thread

Just looking at the reguired voltages I can easily see that R1 must be (2/3)*R2. So I would choose 15k for R2 and 10k for R1. Both from E6 set.

I don't know what rule you are referring to. Ohm's law?

One handy equation is:

V1 / V2 = R1 / R2

Which means that the voltage across the resistors will divide with the same ratio as the resistor values.

 

[Misterbenn]

misterT is correct. If you really want to go down the resistor path then do the above.

But i still maintain that if you then put a significant load, lets say .5A on the output then you'll totaly unballance your resistor divider. You can only really use the resistor solution if you know your desired load current and take this into account during the calcuation.

 

[misterT]

There is this one time I needed to calculate a voltage divider to drive a LED. But that was a special case for bicolor LED (and I wouldn't call it an application of voltage divider):

**broken link removed**

Notice how the voltage at the divider junction (right picture) is 2V, but the resistor values are 160 ohms and 3k3 ohms. The LED changes the problem dramatically.
Original post: https://www.electro-tech-online.com/threads/challenge-of-the-day-dual-color-led.120564/#post993022

What are you trying to drive with the voltage divider? What is the practical application you need a solution for? We can't give you any rules to a problem that does not exist. If you don't have a practical application, make something up as an example.

 

[Misterbenn]

misterT, not sure why you needed R1 in that circuit. The led voltage drop would of regulated the voltage for you, all you needed to do was limit the current - this is why R1 is usualy called a current limiting resistor.

In any case your final comments are right, we need more info to solve this question.

 

[misterT]

misterT, not sure why you needed R1 in that circuit.

Read the original thread and you will understand. It is about driving a two-leg Dual color LED with two microcontroller pins so that both LEDs are driven with same current. If anyone has questions about that circuit, post the question to the original thread.
https://www.electro-tech-online.com/threads/challenge-of-the-day-dual-color-led.120564/#post993022

 

[Misterbenn]

ah i understand, very good

 

[neptune]

ok now i understand , if i connect some circuit in parallel to R1 which requires 2V to operate, then the current division will take place , which will change the voltage drop from 2V to something less, and so resistor value of R1 would have to be changed to get 2V back !

 

[Misterbenn]

That's it
Also if your load changes then the output will change. If your load is fixed and resistive then your ok. Else use better regulation.

 

[neptune]

i calculated current in RC series network to be 2mA , but spice is giving me current to be 42mA
can anybody verify

 

[ericgibbs]

i calculated current in RC series network to be 2mA , but spice is giving me current to be 42mA
can anybody verify

At 50Hz the Zimp of the series RC is 10500R, so 220Vrms approx 325Vpk gives ~30mA according to my calculator.??

EDIT:
The Directive in your Vsrc says 230V not 220V

 

[neptune]

that was 42mA peak to peak , so rms current is 29.702mA

How did you calculated Xc reactance ? i calculated it by 1/2.π.f.c , which is 0.0314/c (f=50)
taking C= 1 x 10^-6
Xc= 31400 ohms

 

[ericgibbs]

that was 42mA peak to peak , so rms current is 29.702mA

How did you calculated Xc reactance ? i calculated it by 1/2.π.f.c , which is 0.0314/c (f=50)
taking C= 1 x 10^-6
Xc= 31400 ohms

The total impedance of a Res and Cap in Series is:
Zimp= √( R^2+Xc^2)

Where Xc= 1/(2Π*f*c)

 

[neptune]

yes thats what i said
so Z= √ 10k^2 + 31400^2
then Z= 32953 ohms
then current I= V/Z .... 220/32953 = .00667 Amps
or 6 mA but LT spice is giving me 40mA peak to peak which is 30mA RMS. what is going wrong ?

 

[ericgibbs]

yes thats what i said
so Z= √ 10k^2 + 31400^2
then Z= 32953 ohms
then current I= V/Z .... 220/32953 = .00667 Amps
or 6 mA but LT spice is giving me 40mA peak to peak which is 30mA RMS. what is going wrong ?

hi,
Xc= 1/( 2 * pi * F * C)

1/(2 * 3.142 *50 * 1 * 10^-6)

1/0.000314

= 3184R not 31400R

so Z = √ (10,000)^2 + (3184)^ 2 = 10,494R

so Irms = 230/10,494 = 21.9mA RMS

 

[neptune]

so Irms = 230/10,494 = 21.9mA RMS

which is not 30mA RMS given by LTspice , why dont you try this circuit on your simulator

 

[ericgibbs]

which is not 30mA RMS given by LTspice , why dont you try this circuit on your simulator

I have already pointed out that you are actual using 230V not 220V as your SIM.

The point you are missing is that LTspice, in the Directive expects Vpk, so for 230Vrms you need the directive to state (0 325 50)

I have tried your sim with the correct Voltage