[SOLVED] selective about resistances

hi,
what kind of resistance i have to use if i want to put a resistance in series with the load in ac circuit?
can resistances upto 2w be used ,can they sustain??

I think you need to be more specific.
- What is your source Voltage
- What frequency
- How much power draws you load
- Why you want to insert a resistor, to limmit power ? If so, how much current do you want in the circuit,.

It is a big difference if you want to insert 10 Ohm in a 2kV 1 MHz line or 0.1 Ohm in a 12V/ 50 Hz line

i am actually going to put that resistor in series with the load in ac circuit (50hz frequency).
i am designing a ammeter kind of device ...
The current how much ever a device draws in ac circuit has to be measured by the resistor ..

Rse = Vo-max / Imax

Vo = (max voltage of sample needed)
Imax = (maximum load current)

Vo-max is .5V is more than enough and select it for your load current...

Also think about P = ( U x U ) / R for that shunt. U is the Voltdrop over the resistor R and P is power in Watt. Be carefull, 230V AC can be lethal.
How you gonna measure this AC Voltdrop ?

yes i am going to measure it .
I know the formula of finding the resistance value ,but only thing is i am confused on what kind of resistor i have to select so that it can be used in ac circuit in series with the load..,like we have power ratings for the resistors, so what has to be power rating of the resistor ?...because if i use .25w resistor in ac circuit in series with any load then certainly it will burn so then what i have to place??????????????

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u are telling me pertaining to particular load what resistor i have to insert ....but, the project i am doing here is nothing but the digital meter which will measure any fluctuations in the current(i.e from smallest to many times larger)....so i want to know whether an ordinary resistor is ok for this or have i to go for some other kind??

;-) you can use your 0.25w resistor, No problem if it is a tiny load.. ;-)

That is why I gave you the formula. You say you know that formula but I think dio do not know it. Because that formula is the answer to your last question.. I gave the formula to calculate the power rating, or in other words, how many Watt the resistor needs to be. And you can rearange that to get the unkown.

Example: If your load draws 1A and the resistor has a voltdrop of 5V then it will dissipate 5W. So to be safe you take a 10W resistor.
And you can calculate the value too. 5V / 1A = 5 Ohm (U=IxR, Ohms law)

So if you want it universal and, lets say, up to a load of 1000W. Then we know 230V. The current will be 1000 / 230 = 4.35A. We take 5A as max for easw of math.
Then you have to think about resolution and voltdrop. If the voltdrop is 20V the load will get not enough. So lets say we find a use a 5V drop as the max.
U/I=R so 5/5= 1 Ohm Then P will be (5x5)/1= 25W. So you can use a 25W powerresistor but if you use it for 1000W you need to cool it very good. Better take a 50W resistor.

But the fact you are thinking about 0.25W and 2W makes me think you have no clue about what you are doing. So be very caerefull. Do not use some cheap multimetyer to measure mains. They are often not enough protected for that. (cat rating and max voltage rating, isolation on pcb, fuses, overvoltage protection etc. And to get good resolution you need a True RMS meter (for most for things like inductive or capacitive loads that polute the mains.

perhaps u can suggest me rather than opting for power resistor of 50w or something else it would better to use just a current transformer ...it can handle any amount of current & can be easily measured by micro-controller & even no need to add additional circuitry or to do any arrangement on heat dissipation simple as that ...

You can use a current transformer. They are for sale but the output of that is AC. So you will have to rectify it. You can use an opamp as precision rectifier so you have loss because of the voltdrop over the diodebridge and to scale it to meet the voltage needed by the microprocessor. Also add a zener diode on the input of the uP and input of opamp so the input will be protected agains high voltage caused by surge currents.

Did you want to use that microprocessor to measure the voltdrop over that resistor without diodes and without galvanic isolation ?
If so I would advise to forget this project for a while and buy a good book about electronics and do projects that does not involve 230V AC. You probably kill the microprocessor and even worse, your self.

First calculate the value of the resistance you need using the basic equations.

Calculate the power of the resistor. Then find the near value of the resistance that is available at the market.

Tips: always try to use a little bit extra power for resistors. Also keep in mind that do not loss much power cross the resistance because it will create heat.

hey i never gave u description about the circuit ... i just was commenting onto the heat evolved by placing resistor and the smae when placed currnt x'mer that's it ....I never would rather give direct ac supply like hat to microc ..i was commenting on your statemnt as u have mentioned abt using power resistor & it's cooling system...

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hey i never gave u description about the circuit ... i just was commenting onto the heat evolved by placing resistor and the smae when placed currnt x'mer that's it ....I never would rather give direct ac supply like hat to microc ..i was commenting on your statemnt as u have mentioned abt using power resistor & it's cooling system...

Then it would be a good start to give more information like the circuit so we do not have to guess. And it could help if you did not used those shortcuts. That way we avoid miscommunication. Like u , currnt, x'mer (?) , smae (?) microc (microprocessor), abt (?) , hat (?) because I now do not understand what you are writing in the second line.

we are dealing here with electronic circuits & having talk about electronic related components so let me tell u what exactly i wanted u to know about (usually x'mer means transformer this we usually use, any way that is not the part of discussion )::"" i never gave u description about the circuit ... i just was commenting onto the heat evolved by placing power resistor and the same when placed current transformer that's it ....I never would rather give direct ac supply like that to micro-controller(uc) ..""i am commenting on your statemnt as u have mentioned about using power resistor & it's cooling system..."".
surely i will certainly give detailed description of the circuit ....

OK, if you like local words that can be done but I think you will not understand in that case what I'm talking about
You now say:

I never would rather give direct ac supply like that to micro-controller

Click to expand...

But my comment was on what you said here:

current transformer ...it can handle any amount of current & can be easily measured by micro-controller & even no need to add additional circuitry

Click to expand...

No additional circuitry is in my opinion connecting the transformer direct to you microprocessor.

But you are not very clear, you do not answer questions and do not read the answer.
I and Mithun tell you how to calculate the resistor (power and resistance) and after that you ask about the power needed.
I asked, how are you gonna measure the current, your answer: yes I;, gonna measure it (???)

This way we get nowhere.

You need to think about :
- the minimum and maximum current (that is not the same as power) you want to measure. 1 mA to 10A will need a very good many bita ADC
- the maximum possible voltdrop over the resistor
- the powerrating of the resistor
- the range you microprocessor needs for good resolution
- how you make a galvanic isolation between uP and mains
- how you are going to rectify the AC to DC ?, a diode, fine, but then you lose 0.35 to 1V depending on the Diode, so better a opamp precision rectifier circuit if you need resolution.

If you want a current transformer most of the above questions have to be answered too.
The advantage is you have isolation, no voltdrop in the mains line.

Anwer these questions and use the formulas and tips given in this topic and you are done.

Isolation is also an advantage..

i will give up a fresh start and tell u what i want to implement:::
i want to make an anti power theft system to a digital meter,for that purpose i am first trying to make digital meter to measure how much power is being used in our house
to make an energy meter i will require to know
1)voltage being supplied on the mains.
2)current in the line being drawn by the appliances connected at home.
so considering domestic load(i.e resistive load) i am doing this project:
*line voltage supplied in here is about 220v-240v
-so considering this to be constant the voltage across the line will always be 220v.
*now line current has to be found out as we know to find out power i.e power=line voltage* line current (in watts)..Voltage is already known(i.e 220v)
now coming to the line current-so here i want to place a power resistor or current transformer..So as we have discussed lot about resistors let me consider a resistor of 10ohm & 30w resistor placed in series with the load....So inorder to find current in load i will have to find voltage difference across the resistor & then will have to find the current by ohm's law as i know the value of resistor.This way i will get the line current or the current flowing in the line.

so if i have to calculate the p.d. across the resistor so that i can find line current then this can be done as follows:
i want to use differential op-amp circuit which could measure exactly the voltage difference across the resistor/ current transformer and could be fed to the adc.. But, the input is ac if i opt for rectification & various other procedures then there will be lot of stuff to be included and the signal will attenuate.. so can any one suggest whether that idea of using differential opamp is right??...and how should signal be maintained?

Sorry, just forget it and start with something less dangerous and easy. You still do not understand it or I do not understand your text.

A pd = powerrdrop ? And then calculate Voltage drop ? This is nonsens, read what we wrote about that and the formulas.

You can only measure an AC voltdrop over a resistor. You will always need some kind of diff amp to measure that voltdrop. And to read it with an ADC you need to rectify that AC voltage or use a TRMS converter. In both cases you end upo with DC. You will need circuits for protection, signal condition, signal scaling, galvanic separation, a powersupply for the circuits with several voltage rails .

An other problem is I now understand you want to measure the total current of your house. You still do not understand the principle. A 10 Ohm 30W resistor. Are you kidding ? Do the math, how much current your house draws ? Lets say 10A in peaks. 10 Ohm x 10A = 100V Voltdrop. 100V x 10A is 1000W. And your TV etc sees 220-100=120V so they stop working if it has a classic powersupply.
With a 10 Ohm and 30W: P=I^2 x R So 30W / 10 Ohm is the sqaure root of 3 = 1,7A. That will give a 17V Voltdrop. After rectification you have 24V. To much Volt, a to high voltdrop for your house and not enough current to be usefull. An other problem for a good power meter is you need to measure the cpomplex power because inductive and capacitive loads change the cos-pi.

Try building something with a 555, a sinewave generator, audio amp or blinking led instead of something that will kill you and burn down your house.

i really don't understand in what language should i talk with u well because if you don't know what is p.d which means potential difference & which also means voltage drop & not power drop. I don't know from which place are u graduate....
anyway this is a question & answer site where in some questions are been posed and perhaps may be we try to give ideas and may be they are sometimes wrong and so ask for a rectified answer ....
And perhaps it's a foolish thing if i directly start working on the hardware. Of-course i will trace my circuit on multisim software & then go for hardware implementation...U might have directly implemented circuits on ac that is why u are advising me ..
Thank you for your comments

Just try English without those fantasy abriviations, that you think are international used. Try writing without leaving out words, use kapitals, comma's etc.
I did not care about that because it is not your native language, nor is it mine. Until I could not understand it anymore. I then asked if you would stop the abriviations, but I still tried to help you.

But if you refuse to use normal English, do not read/study the answers and on top of that, tell me that I miss some education you cross the line.

I wish you good luck with your project and hope you and your relatives survive this experiment.