Voltage drop resistance

Status
Not open for further replies.
In general, 253 ohms/amp but somehow I get the feeling that is not going to be the answer you need.

What are you trying to do?
 
oscar47 said:
I am sorry! What sive resistors do I need to divide 277vac and 24vac?
Click to expand...
Your questions don't make a lot of sense.
Maybe you could explain what you are trying to achieve.
Do you want a voltage divider that will get you 24V from a 277V supply? If so, what will the load be on the 24V?
Without knowing that, there is no way the resistance values and power rating can be calculated.
Have you considered the possibility of using a 277V to 24V transformer?
 
OP understand that your question doesn't give enough information. It would be like me asking you how many mile is it from where I am to New York? Since you don't know where I am, you can't answer. Secondly, using a resistor to drop the voltage from 277 to 24 is an unusual request hence our confusion.
 
pete m. said:
1028 miles? :)

Pete
Click to expand...

No, because I was in Gainesville when I asked the question. Pretty slick of you though! At first I though you made it up, since that is an exact binary equivalent, but went to Google and found the exact same figure.
 
120413-0800 EDT

oscar47:

Suppose you just want a voltage divider made from two resistors, and there will be no current drawn by whatever is placed across the resistor with the lower voltage, and assume that the resistor with the large voltage is R1 and the low voltage one is R2,

then Vout / Vin = R2 / (R1 + R2) = 24 / 277 = 0.0866 .

Next assume R1 = 100,000 ohms. This will need to be a resistor with a power rating greater than P = V2 / R = 2772 / 100,000 = 76,729 / 100,000 = 0.8 watts. By comparison very little power is dissipated in R2.

If you changed R1 to 1000 ohms, then how much greater power would be dissipated in R1. Note: it is much greater.


Change the problem to one where you have a known fixed resistive load load at 24 V. Suppose it is 1 W, or 24 * 24 = 576 ohms. Then (R1 + 576) / 576 = 277 / 24 = 11.542 . Thus, R1 = (11.542 - 1) * 576 = 10.542 * 576 = 6072 ohms. Power dissipation in R1 is (277-24)2 / 6072 = 64009 / 6072 = 10.6 W.

Your efficiency is a little less than 10% when comparing output power to input power. Output power and voltage will not be very stable as the load resistance (R2) varies.

By comparison if you use a small transformer to do the same voltage change, then voltage stability will be much better with changes in load resistance. Also efficiency will be more in the 90% range.

Somebody check my math. I have great problems proofreading my own work.

.
 
gar said:
120413-0800 EDT

oscar47:

Suppose you just want a voltage divider made from two resistors, and there will be no current drawn by whatever is placed across the resistor with the lower voltage, and assume that the resistor with the large voltage is R1 and the low voltage one is R2,

then Vout / Vin = R2 / (R1 + R2) = 24 / 277 = 0.0866 .

Next assume R1 = 100,000 ohms. This will need to be a resistor with a power rating greater than P = V2 / R = 2772 / 100,000 = 76,729 / 100,000 = 0.8 watts. By comparison very little power is dissipated in R2.

If you changed R1 to 1000 ohms, then how much greater power would be dissipated in R1. Note: it is much greater.


Change the problem to one where you have a known fixed resistive load load at 24 V. Suppose it is 1 W, or 24 * 24 = 576 ohms. Then (R1 + 576) / 576 = 277 / 24 = 11.542 . Thus, R1 = (11.542 - 1) * 576 = 10.542 * 576 = 6072 ohms. Power dissipation in R1 is (277-24)2 / 6072 = 64009 / 6072 = 10.6 W.

Your efficiency is a little less than 10% when comparing output power to input power. Output power and voltage will not be very stable as the load resistance (R2) varies.

By comparison if you use a small transformer to do the same voltage change, then voltage stability will be much better with changes in load resistance. Also efficiency will be more in the 90% range.

Somebody check my math. I have great problems proofreading my own work.
Click to expand...

Gar

I think Oscar47 might just be a one post wonder.
As for your maths, the only problem I saw was the dissipation in R1 when it was 100,000 ohms. If R2 has 24V across it then R1 will have 253V across it giving a dissipation of about 0.64W.
Nothing of any practical significance.
 
Status
Not open for further replies.
Top