1 5A 200A 6600V HOW TO DETERMINE LIE WATTAGE... WATT The standard voltage rating is 110V, the standard current rating is 5A, the transducer range is 1000W for 100% input with the standard specifications for 3phase watt transducers. Since the transducer input, 110V and 5A,is already transformed by PT and CT one the power line, the actual line wattage must be multiplied by the rate of transformation. W = PT Ratio x CT Ratio x 1000W (INPUT RANGE) [example] PT ratio:6600V/110V = 60 CT ratio:200A/5A = 40 W = 60 x 40 x 1000W = 2400kW The line wattage rating is then 0 to 2400kW. WATT RANGE _ W =  3 EI cos  where E = 110V (rating) I = 5A (rating) When the cos  equals 1, W will be given as Considefing the operational function of E value, the watt range is 0 to OUTPUT LOAD A 110V SOURCE WATT TRANSDUCER 1000W PT CT ? 110v WATTHOUR A standard 3-phase watt transducer accepts the input 110V, 5A and 1000W range.When the watthour is 1pulse/1Wh, mode is open collector or Dry-contact or Voltage pulse.  Wh pulse range = WATT standard range pulse/1hour [example] 3-phase INPUT range 110V,5A and 1000W range, pulse range = 1000pulse/1hour WATTHOUR PULSE RATIO 1 pulse ratio = PT ratio x CT ratio x Wh [example] PT ratio:3300V/110V = 30 CT ratio:200A/5A = 40 1 pulse = 30 (PT ratio) x 40 (CT ratio) x Wh = 1200Wh TERMINOLOGY  APPARENT POWER(VA) REACTIVE POWER (Var) ACTIVE POWER (W)  = Phase angle between current and voltage POWER The general term “ power ” means the “ active ” power, and in more strict terms, the “ power ” includes also the “ apparent ” power and the “ reactive ” power. APPARENT POWER The power value calculated by simple multiplication of the rated voltage by the rated current. ACTIVE POWER The power value to be converted into any energy form at the load. Unit is W (watt). The W is given with the equation : W = El cos  (3-phase) The cos  equals the power factor, cosine of the phase angle between voltage and current, therefore the equation is also given as: W = Voltage  Current  Power Factor WATT RATIO A standard 3-phase watt transducer accepts the input 110V, 5A and 1000W range.When the watt ratio with these conditions is given as 1, the acceptable Watt Ratio will be 0.5 to 1.2. W = PT Ratio x CT Ratio x 500 to 1200W [example] PT ratio:6600V/110V = 60 CT ratio:200A/5A = 40

2 RIPPLE AC factor existing in a transmitter output. Represented as peak-to-peak value or as RMS value. REACTIVE POWER The power value which is not converted into energy. When the load has a greater inductive factor, more current is required for creating the same active power value. Unit is Var (volt-ampere reactive). The Var is given with the equation: Var = El sin  INPUT BURDEN The power value consumed at transformers at the primany side. Unit = VA = voltage rating  current rating OVERLOAD CAPACITY The rush current value, generated at start-up of a motor, allowable to the transmitter. ■ HOW TO REPRESENT ALTERNATING POWER... An instantaneous value I alternating current/voltage varies according to time. In the other part, a peak value is not able to include effects by waveform distortion, while a simple average value calculation equals zero. The following are examples for re presenting alternating current/voltage. PEAK VALUE The maximum value in the waveform. PEAK-TO-PEAK VALUE The absolute value between the positive peak value and the negative peak value. RMS(root-mean-square) VALUE The square root of the average of the squares of the instantan  eous value. This AC power are able to generate the same heat (energy) as the same DC power value. AVERAGE VALUE (in alternating current) The average of absolute value. Ripple is represented with “peak-to-peak” value or “r.m.s” value  Ripple DC Factor AverageRMS Peak-to-Peak Peak WAVEFORM TYPERMSMADRMS/MADCREST FACTOR Sinusoidal Waveform Square Waveform or DC Triangular Waveform Vm Vm Vm 2  Vm Vm  =  = = = RMS SENSING & AVERAGE SENSING Vm The crest factor indicated in the above in the above table is given from the peak value divided by RMS value. It means the r.m.s value is stable in any waveforms. The ratio value [RMS/MAD] in sinusoidal waveform equals 1.111, Therefore by multiplying the average value by 1.111, an accu  rate output is given in average sensing, while there will be a significant difference in other type of waveforms such like a thyristor output.