A 5400 ohm resistor is rated at 1.5watts.what is the maximum safe working current?

Question

anyone please help.is maximum safe working current is simply it's current

Answer 1

P=I^2R
1.5=I^2(5400)
I=0.017 A

Answer 2

solid question. The wattage is the max potential the resistor can tolerate without burning. The formulation for potential is going: P = V x I, or P = v^2 / R, or P = I^2 x R. Given: max P = .5 watt, and R = 2400 ohms use formulation v^2 = P x R then v^2 = .5 x 2400 = 1200 then take sq. root v = 34.641 volts

Answer 3

P=IE E=IR - P=Watts I=Amps E=voltage R=Ohms

We know P=1.5W and R=5400 What we don't know is the voltage, and the amperage will be dependent on this.
This is why resisters are rated in Watts and not Amps.
A 1.5 Watt resister, for instance, with a 10 volt source will handle a peak amperage of 1.5=10*I or 150mA. (0.15A)
At 100 volts, it will only handle 1.5=100*I or 15mA (0.015A)

Answer 4

Current is equal to the square root of power (1.5watts) divided by the resistance (5400 ohms). sqrt of (1.5/5400) => sqrt of 1.5/5.4x10^3 => sqrt of {(1.5/5.4)*10^-3} =>sqrt of .000277 => .01643 amps which is 16milliamps or 16ma. Oh and the voltage across it is 90volts


From Ohm's law which; you could have found by typing in (ohm's law) in a search engine and that would have taken you to this site.Which is the first of many on the search results page.

Answer 5

Using Aldrin's answer of 17 mA for the current, a good rule of thumb is to design in a margin of 1.5 so for Aldrin's answer of 17mA you would do the following

17mA/1.5 = 11.3 mA
This would give you a margin of 1.5

It is generally not a good idea to use any component at its maximum specified value if you can avoid it

Answer 6

About 17ma for a voltage drop of about 91V