Recent content by Dr.kW

Also... v(t) = Vmax/ exp[jwt] i(t) = Imax exp[j(wt + phi)] Veff (phasor) = v(t)/√2 Ieff (phasor) = i(t)/√2 I*eff = Imax/√2 exp[-j(wt + phi)] (note minus sign) so (product of phasors) Veff x I*eff = Vmax/√2 exp[jwt] x Imax/√2 exp[-j(wt + phi)] Seff = Vmax Imax/2 exp...

This is not an easy task, you are in for a lot of study and thinking. The foundation you need is IEEE Standard 115, "IEEE Guide: Test Procedures for Synchronous Machines," particularly: Part II-Test Procedures and Parameter Determination for Dynamic Analysis This will get you to the...

a little more... (dc ckts) Neutral is midpoint between a positive and a negative. Relative to ground, the potential differnce is usually zero, but that is not necessarily true. In power electronics, it may be true only on average, such as in some switched-mode power supplies where the...

Sizing the motor has two issues--acceleration time (depends on inertia) and load torque. Yes, your system has load torque. In this case it is the friction and air-resistance of all rotating parts--bearings, fans, all rotating surfaces. Once runing, the motor will probably run very close to...

Going down, your machine is acting as a generator. It will have retarding toque in proportion to the load seen at the electrical terminals. This means you need either a 4-quadrant drive for regenerative braking, or a dynamic brake circuit which essentially puts a resistor across the terminals...