Why does the Power-Mate show a lower power reading than the multiplication of volts and current?
The Power-Mate displays True Power – ie what the electricity company bills you for. If you wish to know the apparent power (VA), simply multiply the RMS Voltage by the RMS current readings. For a purely resistive load the results between true and apparent powers should be identical. For a capacitive or inductive load the apparent power will be higher than the real power – because of the phase difference between voltage and current. To calculate power factor, simply divide the true power by the apparent power. You can also then calculate the phase angle if you wish. When resistive loads are on the mains, I x E x pf = W (assume E = 240V, I = 10A, power factor pf = 1), therefore W=2400 When reactive loads are present, pf is not equal to one, Irms x Erms = VA (Apparent Power) assume E=240V, I = 20A, VA=4800 BUT we must take Power-Factor into account so although the apparent power is 4.8kW, due to say a pf = 0.5 the true power consumed would be actually half or 2400W, which is the same watt
Related Questions
- I have a 130wpc receiver and speakers rated at 120 watts max. Should I use speakers that have a lower max power output rating than my receiver?
- How is the FNX14701 able to operate at significantly lower power than existing solutions?
- Can I post a chart to track homework, reading, multiplication tables, etc.?
