A resistor in parallel with a voltages source will not cause the voltage to drop, theoretically.
To get a 20 volt drop you need a resistance in series, and the number of ohms is 20 divided by the current in amps.
If the current is unknown or variable, the voltage can't be dropped by using a resistor.
the physical size tells how much power it can dissipate (watts)
The voltage drop across a resistor is equal to the resistance times the current flowing through it. If you do not have a constant current load you cannot use a resistor and must use an active solution. Please read up on ohms law before continuing with your project.
To calculate the amperage, you need to use the formula: Amps = Watts / Volts. Assuming a standard voltage of 120V, you can calculate the amperage by dividing 600 watts by 120 volts, which equals 5 amps. Therefore, 600 watts would require 5 amps of current at 120 volts.
Fuses are rated in Amps. Although the physical size of a fuse is to do with volts; the further the terminals are apart the less likelihood there is of 'sparkover' between them.
7.5kVA is a value derived from the voltage multiplied by the current. Without knowing one, you cannot know the other. For example, think of the number 10kVA This could be made by :10kV * 1A or 5kV * 2A or 1kV * 10A or 2kV * 5A. You couldn't possibly know which is the initial pair.
Its dependent what will by rated power of the device (current).
A: LEDS are devices that needs a certain voltage and current typically 1.8 and 20 ma
If there is nothing else in the circuit, then the voltage drop across the resistor will be the full supply voltage of 5 volts. The size of the resistor does not matter in this case - it will always be 5 volts.
Assuming that you're talking about 12V DC you would use a 4 ohm resistor. If you mean AC then you would need a step-down transformer with a 4:1 ratio.
When a resistor is added the current goes down, that is expressed in the equation current= voltage/ resistance
This question cannot be answered because you did not specify the current.
There is no relation between the resistor's ohms value and its size. The power of the resistor can be seen by its size. If the power is too small, the resistor can be destroyed.
Reducing 6 volts to 4 volts at 500ma trickleThe formula is voltage / current6v-4v=2v / .5a = 4 ohmspower = current squared x resistance or voltage x currentpower of R = .5a x .5a x 4 ohm = 1 watt minimumor 2 volts x .5 amps = 1 watt minimum
You will need to take the resistance of the load into account if you are going to design a voltage divider. The resistance of the load can completely change the voltage ratio of a voltage divider if not factored into the calculation. you can measure or read R(load), then R(needed) = 0.8 R(load)
there might be ways to get the power rating by measuring the size of the resistor. but as the physical size of the resistor increases, its power rating also increases..
You will need to know the amount of current flowing through the coil when 220 volts is applied across it. A resistor in series with the coil will limit the current so that the coil only sees 220 volts. The resistor will need to drop 57 volts. So, 57 volts divided by the current in amps will give you your required resistance. You will need a resistor with a high power dissipation rating with 57 volts across it. Your resistor will probaly need to dissipate several watts. For example: A 220 volt coil with 300 milliamps (.3 amps) will require a resistor of 733 ohms. The power dissipation of the resistor would need to be 17.1 Watts! You might try using a light bulb as a series resistor. Ensure that it can handle 57 volts. To complicate matters, is that AC or DC you are using? AC relays have inductance build in, in order to increase the specific "ac resistance", thus the same coil could use as little as 0,001A so you will need a very low value resistor. Anyway, if any 220V relay uses as much as 300mA, I doubt if you will be able to pick it up with one hand! Such a relay coil will draw about 66W of power! I have a 16A rated contact 230V relay. Its current is 0,0015A that is equivelant to 0.33W at 220V!
A potentiometer.