Electrical Equations ( Alternating Current ) AC - Engineers Edge
Instrumentation, Electrical, Control and Sensing Devices
Electrical circuits and alternating current AC equations anf formulae.
AC Equations |
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DESIRED |
ALTERNATING CURRENT |
DIRECT |
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Single-Phase |
Two-Phase* |
Three-Phase |
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Amperes when |
kva x 1000 |
kva x 1000 |
kva x 1000 1.73 x E |
kva x 1000 E |
Amperes when |
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 |
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
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