Calculating Voltage, Current, Resistance, and Power Using Ohm's Law and Power Supply Specs

hi, I hope ur good and happy and happier... I a little confused in ohms law...for 2amp 20volt we need 40w resistor???!!!! (p=vi)

and one more question...if my i=3A and r=1000ohm then my voltage had to be 3000V?!!is this example another way to solve??

to this law power supply or battery have only voltage!and in reality, I have for example dc power 24v 3 Amp so that amp how effect in ohm law? how I can consider ps current to ohm law?

 

Why do you post the same question twice?

Hi David, there appears to have been a major glitch in introducing a new interface, I have had problems logging in, and a number of posts disappeared for some time.
Cheers,
Richard

If 1.0 volt DC is measure across a component and 1.0 amp DC is flowing through that component then 1.0 watt will be dissipated in that component
This is the basics of Watts formula: P(watts) = E(volts) X I(amps)
In your example of 20volts and 2amps for a resistor the power dissipated in that resistor will be: 20V X 2A = 40W
But a resistor of 40 watts will be operating at maximum temperature therefore choose a 50W resistor to create some operating margin (20%).

Question #2
Referring to the formula E(volts) = I(amps) X R(ohms) [if you have a known I & R then E(volts) can be calculated.]
E = I * R . . . . . . E = 3A X 1000ohms = 3000 volts.
Yes, it will require 3000 volts to push 3 amps through a 1000 ohm resistor.
....and the power dissipated in that 1000 ohm resistor will be: P = E * I
P = 3000V X 3A = 9000 watts [this applies to steady state DC and nor pulsed]
ie:
If you apply 5 volts across a 1000 ohm resistor only 5 milliamps will flow (I = E/R)....or 5V/1000ohm = 0.005 amps.

If 1.0 volt DC is measure across a component and 1.0 amp DC is flowing through that component then 1.0 watt will be dissipated in that component
This is the basics of Watts formula: P(watts) = E(volts) X I(amps)

Hope this information is helpful to you.