Electrical Engineering

It is 230V single phase and 440V in 3 phase system at 50 Hz.

If the single-phase voltage is 230 V, then the three-phase voltage must be 400 V, not 440 V. The line voltage is 1.732 times the phase voltage.

resistive load

If the current is driving a motor, then the load is resistive-inductive.

Alternating Current can be converted to Direct Current by using a DC Converter which contain a Bridge Circuit , a Capacitor and if needed a Transformer.

A 'DC converter' is normally called a rectifier.

Ohm's Law

It's not known as anything; it's simply an equation. Ohm's Law describes constant proportionality between current and voltage for certain, but not all, materials (linear or ohmic) -it has nothing to do with this equation.

In a parallel circuit, more power is provided to the lights. Power = V2/R . The resultant resistance of the circuit is lower, and the potential difference is not divided as in a circuit in series. Thus, lights in parallel burn brighter.

The switch will not increase the current. The only thing that will increase the current is adding a larger load.

Another Answer

To maximize current flow in a circuit with 2 Flashlightbatteries, 2 Flashlight bulbs, and a switch, set aside the bulbs, put the batteries in series, connect the switch between the end poles of the battery set, and close the switch. The batteries will get hot and probably leak electrolyte, and be drained in seconds.

If you want to maximize current through the circuit withthe bulbs in it, connect the bulbs in parallel to each other, and in series with the switch.

The formula for a simple DC voltage drop across a cable is:

VDrop = Vmeasured at the input of cable - Vmeasured at the output of cable

This formula may seem simplistic however keep in mind that the sensitivity, accuracy and resolution of measuring instrument instrument is what is really important.

If you know the DC current flowing through the cable and the impedance of the cable then you can use Ohm's law. Vdrop = I R where Vdrop is the voltage drop across the cable in volts I is the current flowing through the cable in amperes R is the resistance of the cable in ohms.

ADDITIONAL

Calculation of the voltage drop is given by the formula below

Voltage Drop (Volt) = [(mV/Am)/1000] * I * l

Where mV/Am = millivolt drop per meter per ampere of the cable

(this information is given in the table based on IEE Wiring Regulation)

I = Current in the cable (in Ampere)

l = Distance of cable (in Metre)

See related links below

You question does not say what size DC motor you are wanting to control, so this answer will have to be generic.

First of all, you DON'T drop the voltage "OF" the DC motor, you drop the voltage "TO" the motor.

The simple answer is to place a "RESISTANCE," or "RESISTOR" in series with the motor. A resistor limits the amount of current flowing through it, and thus the voltage.

If you are talking about the very small DC hobby type motors [designed for 1 to 3 volt inputs], one way would be to use a variable resistor, like a large potentiometer ["pot"], scavanged out of a discarded "old" TV or "old" radio [both "pre" solid state era].

Another trick I've used [requires a lot of trial and error] is to use an assortment of old [but working] incandescent/filiment type light bulbs [small, like from flashlights, Christmas tree bulbs, dial lights from "old" radios and TVs, etc.] as resistors.

You can experiment around with "stacking" these in series to add more resistance, with each addition slowing the motor more. And if even the largest of them slows the motor TOO MUCH, then they can be wired in PARALLEL to decrease resistance, allowing more current flow. As said above, this can require much trial and error, but it can be made to work.

Another method [if using batteries] would be to use fewer batteries in series. This only works for larger voltage motors, like out of 24/18/12/9/7.2 volt motors [like out of cordless drills for example].

As an example, consider using an 18 volt motor. To provide the "normal" 18 volts would require 12 1.5 volt alkaline batteries in series. If you used only 6 batteries, you would have reduced the voltage by half. By varying the number of batteries, you would be able to vary the motor speed. Remember however, that there is a limit to how much you can "starve" the motor and still have it run or do any work.

If your aplication allows you to use an AC power source, then another method [much more expensive] would be to use a benchtop AC powered, variable power supply.

Another common method is pulse width modulation (PWM). The full available voltage is switched on and short time later it is turned off again. This cycle repeats continuously at a relatively high frequency (hundreds, to tens of thousands of times per second). The ratio of the on-time to the off-time determines the apparent percentage of the DC supply that the motor sees.

The circuits to drive motors this way can be relatively simple to extremely complex, depending on the size of the motor, available DC voltage, and other controls needed (such as controlling speed instead of voltage supplied to the motor).

ANSWER: motor are not voltage related but rather power related so the answer is not to reduce the voltage but to reduce the power going to it. REDUCE THE POWER WILL REDUCE THE SPEED. either reducing current and /or voltage will do the trick.