The rheostat makes the current high by decreasing the resistance.....and as v=ir.. if r has been decreased and v remains constant than i will increase..
yes
Rheo
We can treat the rheostat as a resistor. There is a formula "V squared over R" that gives the power lost. In this case, that would be 400/2.5 or 160 Watts. You could also find the current using Ohm's law: V=IR so I=V/R=8 Amps. You can then find power as P=VI = 20V x 8A = 160 Watts.
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Yes, However not all fluorescent lights are capable of doing so. Check your local store to see which ones are usable with a rheostat and which ones aren't
A rheostat is a variable resistor and these control the flow of current.
yes
A rheostat is simply a variable resistor used to control current. It does not have a positive or a negative terminal.
It is a variable resistance device which control the flow of current
It isn't. It is only kept at maximum resistance when the motor is not running. That is done to limit the starting current.
Yes.
Series motors have the highest starting torque. The torque is proportional to the square of the current, and the starting current is so high it has to be limited by a resistor called a rheostat. Series motos are used mainly on trams and trolley buses.
Rheostat
For a shunt dc motor the rheostat would vary from zero ohms to a value that produces the required minimum field current, but rheostats are not common with shunt motors except as a crude way of controlling the speed. Series dc motors normally use rheostats for starting, especially in trams etc. because the starting torque and current is very high. The rheostat resistance starts at a value equal to the supply voltage divided by the maximum allowable current drawn, and is reduced as the speed builds up.
The heat released by the rheostat with double the voltage will quadruple. When voltage is tripled, the power loss is 32 or 9 times that before. A rheostat is a kind of variable resistor. Since E = IR (voltage equals current times resistance), then I = E/R (current equals voltage divided by resistance). If the voltage is doubled and the resistance stays the same, then--you can see by the formula--the current would double. Now, power dissipated by a resistor is related to the product of the current and voltage (P = IE). But since a doubling of voltage produces also a doubling of current, double the current results in 2X2=4 times the power (heat) loss.
try RHEOSTAT
the back emf increases so that high currents doesn't pass through the field windings