Two SCR's are connected in an inverse-parallel arrangement to one leg of a single phase circuit. This arrangement allows one SCR to control one half of the sine wave, and the other SCR controls the other. A special circuit is used, sometimes called a "trigger circuit". This circuit sends a pulsed signal to each gate of the two SCR's. The SCR's are so fast to open close the circuit, that they can be "fired" open, then "shut" off in increments far smaller than 1/60th of a second, the typical AC line frquency. So, if you were to tell the trigger circuit to fire each SCR for only 1/240th of a second, (1/2 of 1/2 of a complete cycle) 60 times a second, then each SCR would only be "on" or letting electric current through for 1/2 of each +/- cycle of the sine wave. Therefore output power would be 50% of maximum, but complete in the sense that the 60Hz cycle would remain intact. To put it another way you would still have an alternating current 60 times per second, but the upper and lower portion of the sine wave would be cut off early, and at the same points respectively. This is called Phase-Angle Control, and is the most common form of solid state power control. Resistive loads are the easiest to control with solid state techniques, though power factor is increased with the use of solid state switching. There are ways around that. The advantages of this type of power control is infinitley variable power output, with no mechanical contactors to wear out. SCR's can switch 100's of amps, hundreds of times per second, and as long as you keep them cool, theoretically they will never wear out. If you would like to know more, there is a site: http://www.payneb5.com , for Payne Engineering. They have a couple of animations and schematics depicting what i just explained, but with pictures.
The PF will increase
An electric current through a resistive circuit can be increased by decreasing the resistive load or increasing the voltage of the circuit.
resistive loadAnswerIf the current is driving a motor, then the load is resistive-inductive.
It's the product of the resistance of that resistive load and the current passing through it.
The power will be the product of the square of the current and the resistance of the load. The fact that the circuit is a parallel circuit is irrelevant to this question.P = I2R = 0.032 x 1000 =0.9 W
The PF will increase
ratio between true power and apparent power is called the power factor for a circuit Power factor =true power/apparent power also we conclude PF=power dissipated / actual power in pure resistive circuit if total resistance is made zero power factor will be zero
In a resistive load circuit, the power = multiplication of voltage and Current. By increasing the voltage power will not be increased. Power is defined by the load as per its design. If the voltage is higher the load current will reduce. However running a load at double the rated voltage is not good for the device. Insulation may fail.
An electric current through a resistive circuit can be increased by decreasing the resistive load or increasing the voltage of the circuit.
Apparent power is VA. Real power is W reactive power is VAR. Under an inductive+resistive load the VA is higher than W
when a resistive load is applied there is no phase angle difference between voltage and current. when a inductive load is applied there is phase difference between voltage and current. current lags voltage by an angle of 90 degrees for pure inductive load
The voltage before it is hooked up to a resistive load.
For power control of inductive load.They can be successfully used for resistive load as well.
With a pure resistive load the Power Factor should be 1.
A pure resistive load always has a power factor of one. This is because the current and voltage waveforms are in phase in an AC circuit.
these two types of circuit loads are the purely capacitive loads and purely inductive loadsAnother AnswerApparent power will be larger than true, or active, power in ANY circuit, other than a purely-resistive circuit or an R-L-C circuit at resonance.
It is resistive much load