how to calculate resistance

[neptune]

hi,
can anybody tell me how to calculate resistance required R1 an R2 to generate below mentioned voltages,
how do you calculate it ?

 

[hexreader]

Make it easy for yourself by choosing resistors that add up to 5k Ohms. R1+R2=5000.

Calculate the current around the circuit using Ohms law. I = V/R. 0.001 Amps (1 milliamp).

So I is 0.001. Now use Ohms law again. R=V/I. to calculate the two resistances.

There are quicker formulae to get the result, Google voltage divider, but the long way gives better understanding.

 

[neptune]

so we have to assume current flowing through this network before calculating anything, i will assume 50 mA of current as it is required to light up LED.
how will voltage devider rule apply to it ?

 

[hexreader]

Just use Ohms law V=IR using 0.05 as your value of I.

 

[neptune]

Just use Ohms law V=IR using 0.05 as your value of I.

but why cant i take any other values of I , 0.05 is too low to drive anything !

 

[hexreader]

but why cant i take any other values of I , 0.05 is too low to drive anything !

It was you that wanted to assume 50mA as your current. 50mA is 0.05 Amps.

You can pick any current you want. If you choose 1 Amp, the maths gets really easy.

V=IR Assumes units of Volts, Amps and Ohms.

 

[Jugurtha]

Ohm's Law : V=Z.I . It is one equation with three variables. You got to have at least two to calculate the third. Makes sens, right ?

So, for any given V and I, you'll have your Z. For any given V and Z, you'll have your I... etc, for any given PAIR of known values, you'll have the third.

What do you have here ? The V0= 5volts. V1=2 volts. V2=3volts. V0=V1+V2. Anyway :

V1=R1.I ... 1

V2=R2.I ... 2

Either:

a- You got V1 so you SET arbitrarly R1 (to a normalized value if it's a board you entend to really make), you could then get the I, inject it in 2 and get R2.
(Same thing can be used in 2, you set R2, get I and inject it in 1)

b- You set arbitrarly I, and inject it in both 1 and 2 and get R1 and R2.



But the point being that, you always look at what you have, at the equations. V=Z.I here. It takes TWO to tango.

 

[misterT]

hi,
can anybody tell me how to calculate resistance required R1 an R2 to generate below mentioned voltages,
how do you calculate it ?

Give R1 any value you like and then R2 = (3*R1)/2

 

[neptune]

Give R1 any value you like and then R2 = (3*R1)/2

your method is not working. it doesnt give me freedom to choose value of I

 

[neptune]

Ohm's Law : V=Z.I . It is one equation with three variables. You got to have at least two to calculate the third. Makes sens, right ?

So, for any given V and I, you'll have your Z. For any given V and Z, you'll have your I... etc, for any given PAIR of known values, you'll have the third.

What do you have here ? The V0= 5volts. V1=2 volts. V2=3volts. V0=V1+V2. Anyway :

V1=R1.I ... 1

V2=R2.I ... 2

Either:

a- You got V1 so you SET arbitrarly R1 (to a normalized value if it's a board you entend to really make), you could then get the I, inject it in 2 and get R2.
(Same thing can be used in 2, you set R2, get I and inject it in 1)

b- You set arbitrarly I, and inject it in both 1 and 2 and get R1 and R2.



But the point being that, you always look at what you have, at the equations. V=Z.I here. It takes TWO to tango.

b method is prefferable , it gives me independence to choose value of current in circuit.

 

[Misterbenn]

Seriously, this is a real question??

You should be aware that if your current value differs from the one you use to select the resistances then you wont get the voltage drop you want

 

[neptune]

Seriously, this is a real question??

You should be aware that if your current value differs from the one you use to select the resistances then you wont get the voltage drop you want

we choose current first and voltage drop is known so resistance value can be calculated

 

[misterT]

You should be aware that if your current value differs from the one you use to select the resistances...

It never does, not when the input voltage is constant like in the example.

we choose current first and voltage drop is known so resistance value can be calculated

I have never chosen a specific current in a problem like the example voltage divider.. why would I do that? I usually want the current to be "low" and then I choose the other resistor from E24.. and then I calculate the value of the other resistor.

If I choose some fixed current to work with, then I can easily end up with two resistor values that are inconvenient in practice.

 

[neptune]

I have never chosen a specific current in a problem like the example voltage divider.. why would I do that? I usually want the current to be "low" and then I choose the other resistor from E24.. and then I calculate the value of the other resistor.

in one statement you are saying that you dont choose current and in another statement you are saying i usually want current to be low .. that means you are choosing low current value !!! :-|

If I choose some fixed current to work with, then I can easily end up with two resistor values that are inconvenient in practice

if i choose 30 mV current to run in above circuit.. then R1 and R2 comes out to be 60 ohm and 100 ohm .. which are easily available in market.
if it R1 and R2 comes some different value we can make resistor network by joining resistance sereis and parallel to obtain required equilent resistance

 

[misterT]

in one statement you are saying that you dont choose current and in another statement you are saying i usually want current to be low .. that means you are choosing low current value !!! :-|

I said that I have never chosen a specific current (value) for a voltage divider. I never use current to calculate resistor values in a voltage divider. I do calculate the current after I have chosen the resistors, just to check if it's "low" enough.

For example, if I need a voltage divider that divides 5 volts in half, I would just put two 10k resistors in series and never calculate the current. And if you want to know why I chose 10k resistors.. It's because I have a s*it load of them in stock. Can't say that about 60.4 ohm and 100 ohm resistors.

 

[Misterbenn]

misterT,
I'm afraid your wrong

It never does, not when the input voltage is constant like in the example.

If there is a load on the voltage divider output then you are changing the circuti resistance, thus you end up changing the voltage divider output.

 

[neptune]

For example, if I need a voltage divider that divides 5 volts in half, I would just put two 10k resistors in series and never calculate the current. And if you want to know why I chose 10k resistors.. It's because I have a s*it load of them in stock. Can't say that about 60.4 ohm and 100 ohm resistors.

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

 

[Misterbenn]

dont use a resistor divider, use a zener regulator which will give you a much more stable voltage source.

or even better use a bipolar voltage regulator

both are found here
https://en.wikipedia.org/wiki/Linear_regulator

 

[neptune]

oh common. this is a stable DC spply ... not a rectifier output
zenner cannot provide a desired voltage drop. its voltage is fixed. ..

 

[Misterbenn]

I dont care if its a stable dc supply. Whats the load like? is it stable and resistive? maybe you should tell us what your trying to do with the voltage refrence.

Also a zenner can defo produce the voltage drop you want easily, you just need to select the right one!