### Electrical Equations ( Alternating Current ) AC - Engineers Edge

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