ऐम्पीयर एवं वोल्टेज से किलोवाट कैसे निकालें
The power P in kilowatts (kW) is equal to the current I in amps (A), times the voltage V in volts (V) divided by 1000:
I=current(A)=ampereP(kW) = I(A)× V(V) / 1000
The power P in kilowatts (kW) is equal to the power factor PF times the phase current I in amps (A), times the RMS voltage V in volts (V) divided by 1000:
P(kW) = PF × I(A)× V(V) / 1000
The power P in kilowatts (kW) is equal to square root of 3 times the power factor PF times the phase current I in amps (A), times the line to line RMS voltage VL-L in volts (V) divided by 1000:
P(kW) = √3 × PF × I(A)× VL-L (V) / 1000
The power P in kilowatts (kW) is equal to 3 times the power factor PF times the phase current I in amps (A), times the line to neutral RMS voltage VL-N in volts (V) divided by 1000:
P(kW) = 3 × PF × I(A)× VL-N (V) / 1000
Do not use typical power factor values for accurate calculations.
Device | Typical power factor |
---|---|
Resistive load | 1 |
Fluorescent lamp | 0.95 |
Incandescent lamp | 1 |
Induction motor full load | 0.85 |
Induction motor no load | 0.35 |
Resistive oven | 1 |
Synchronous motor | 0.9 |
300 Amp mai 132000 Watts hote hai
573 kWh ka bill kitna hua
202 किलो वाट पर कितना एंपियर लोड चला सकते हैं
Ia=30 amp
Ib=40 amp
Ic=30amp
V=440volts
Pf=95
Kw=?
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288 Amp.=kW ??