Return to Home



KVAR Box The KVAR unit 

A KVAR unit can save you 8% to 15% of the typical bill.  It will cost $350 for the typical 1400 sq. ft. house with A/C, so, if your power bill runs $325.00 Jun. - Sep. and about $200 a month for the rest of the year, the 1st years savings alone would pay for the unit.

The KVAR unit
is made with sealed capacitors and safety bleed down circuits included to discharge any residual high voltages and an 'on' indicator.

These capacitors are used to put capacitative loads on AC circuits to correct typical power factors toward a '1.0' power factor from the typical customer's .85 kvar power factor. (.85 is a utility industry accepted typical value)  If a you do NOT have such a capacitor, you will always use additional wattage to overcome this .85 power factor.  This is wattage you have to pay the utility for, but it does no real work for you. In the typical .85 power factor example it will be 15% of the $ costs shown on your bill.
**  If you are in the world of Pacific Gas & Electric's surcharges for high use, it can exceed the 15% of the $ cost by as much as an additional 25% for a total of 40 % of the bill. 
OUCH!! This is what got me started investigating power factor correction,  as my 2002 PG&E bills totaled $4,400 with 4 months over $750.  I was a design engineer working as a systems analyst at the time, and had done power factor correction work for mainframe computer power supplies on mainframe  computers for NCR, so I was aware of the concept of power factor correction, unlike most people, even electricians and architects.

What causes this
.85 power factor? All inductive devices create a lagging power factor load ( +0.25) because of when wire is wound around a core, current changes will induce a magnetic field, but not all of your power loads are inductive loads, so yours will vary between 1.00 and .75 usually.  Pure resistive loads have a power factor of 1.00, so they pull the combined power factors toward the ideal 1.00. However, capacitative loads (like the KVAR unit) create a offsetting power factor load ( -0.25), so they can be used to cancel out equalalent inductive loads.

I do recommend replacing as many incandescent bulbs as practical with their lumen equivalent fluorescent bulbs.  Although these equivalent fluorescent bulbs do create an inductive load, this load is very small, typically about 5 watts for each bulb and would be easily compensated for with this KVAR unit, thereby reducing the effective wattage of the 60 W bulbs replaced to only 18 W.
This means if you replaced 10 of  your 60 W. bulbs with equivalent fluorescent bulbs and if they were turned on 1.5 hours a day, you would save 70% of the annual power cost. This is either $206.96 @ $0.09 / kWh (national average)  or for suffering Pacific Gas & Electric customers (like me) $347.00 @ $0.1509 / kWh. (my 2002 average rate)

What are some inductive devices:

1.   Any motor in
air conditioners, refrigerators, freezers, washers, dryers, any pumps, fans transformers,  fluorescent light ballast's, power supplies, fans and microwave ovens.

What are
NOT an inductive devices: (These do not cause power factor losses but may quite inefficient)
1.   Regular incandescent light bulbs, Stove burners & regular ovens, portable heaters, electric water heaters, baseboard heaters, electric furnaces and electric clothes & hair dryer heating elements.  Sorry, I can't help you here!  I did put my electric water heaters (2 - 50 gal.) on timers so they don't heat during the peak load period.