Electrical Equations ( Alternating Current ) AC - Engineers Edge

Instrumentation, Electrical, Control and Sensing Devices

Electrical circuits and alternating current AC equations anf formulae.

AC  Equations
DESIRED DATA	ALTERNATING CURRENT			DIRECT CURRENT
	Single-Phase	Two-Phase* Four-Wire	Three-Phase
Amperes when kva is shown	kva x 1000 E	kva x 1000 2 x E	kva x 1000 1.73 x E	kva x 1000 E
Amperes when kilowatts are shown	kw x 1000 E x pf	kw x 1000 2 x E x pf	kw x 1000  1.73 x E x pf	kw x 1000 E
Amperes when horsepower is shown	hp x 746   E x %Eff x pf	hp x 746     2 x E x %Eff x pf	hp x 746      1.73 x E x %Eff x pf	hp x 746 E x %Eff
Kilovolt-Amperes	I x E 1000	I x E x 2 1000	I x E x 1.73 1000	I x E 1000
Kilowatts	I x E x pf 1000	I x E x 2 pf 1000	I x E x 1.73 x pf 1000	I x E 1000
Horsepower	I x E x %Eff x pf 746	I x E x 2 x %Eff x pf 746	I x E x 1.73 x %Eff x pf 746	I x E x %Eff 746

OHMS LAW

 Volts (E) = Amps (I) x Ohms (R)

Amps (I) = Volts (E) / Ohms (R)

Ohms (R) = Volts (E) / Amps (I) R=Ohms, E=Volts, I=Amperes

POWER - AC CIRCUITS

Efficiency = 746 x Output HP / Input Watts

3 KW = Volts x Amps x PF x 1.732 / 1000

3 Amps = 746 x HP / 1.732 x Eff. x PF

3 Eff. = 746 x HP / 1.732 x Volts x Amps x PF

3 PF = Input Watts / Volts x Amps x 1.732

1 KW = Volts x Amps x PF / 1000

1 Amps = 746 x HP / Volts x Eff. x PF

1 Eff. = 746 x HP / Volts x Amps x PF

1 PF = Input Watts / Volts x Amps

HP (3) = Volts x Amps x 1.732 x Eff. x PF / 746

HP (1) = Volts x Amps x Eff. x PF / 746

1 KW = 1000 Watts

Eff. = Efficiency, PF = Power Factor

KW = Kilowatts

HP = Horsepower