Choosing the right generator size for your home is not a trivial task because there are many variables to consider. The simplest and quickest way to determine the required wattage is to match the rating of your existing electric service. For example, if you have a 70A 120/240V service, it would yield 70240=16,800 watt. If you are not sure about your amperage rating, you can find this number on the main breakers. Of course, in practice people rarely use all available power. You can save money by selecting a smaller generator, which may still be sufficient to feed the entire house. The question is- how to determine your actual home's power consumption?

The typical wattages of various appliances provided in numerous online sizing guides and electric load calculators are not very helpful for two reasons.

120/240V electric wiring diagram Firstly, different models obviously consume different amount of power (see for example the consumption of a refrigerator). Secondly, most guides don't take into account imbalance of the load in your home electrical system. I'll show you how to properly determine the required wattage, but first I need to explain the basic configuration of the wiring. Most residential houses in U.S. have single-phase three-wire service (see wiring diagram). In essence, you have two separate 120V lines connected to two busses in your main panel with common neutral. Large appliances such as a central a/c or a dryer run off 240V and therefore draw equal currents from both lines. However, all other devices draw current only from one of two lines. As the result, two busses in most cases are imbalanced. A home generator likewise has two 120V outputs with common neutral, each of which can supply not more than half of its total rated wattage. That's why it is not enough to know your lump power consumption- you need to know how it is split between L1 and L2. Let me illustrate it with the following example. Suppose your house consumes P=8kW. If this load splits equally between L1 and L2, in theory you could use an 8kW genset. However, if you have for example P1=6 kW and P2=2 kW, you would need a 12kW system to be able to supply 6kW on the first bus. Of course, it is always recommended to have at least 20% derating, but we will get to that later.


Measuring AC power To properly size a whole-house generator, the first thing to do is to measure electric current on each of two lines of your home. If you have a proper electrical training and know how to work safely with AC voltage you can do it by yourself. You will need special rubber gloves and a clamp meter preferably with peak reading function. The cover from the main distribution panel has be removed in order to reach L1 and L2. If you found a significant imbalance between them, ask your electrician to swap appropriate breakers in the circuit panel to better balance the loads. Once you know final I1 and I2, the minimum required wattage of your genset is the larger of these two currents times 240 plus 20% margin:

GENSET RATING(watts)= max(I1,I2)2401.2

For example, if I1=40A, I2=30A, you would need 402401.2=11,520 watt system. If you have a central a/c or another high-power motor-driven appliance, you may have to increase the size of your system to handle high startup currents. To determine the requirement you would need to measure initial surge current of your device by using peak function of the clamp meter and choose a generator with corresponding surge capability (read about sizing a generator to start a/c or motors). Unfortunately, generator manufacturers rarely specify their surge characteristics for standby models. As a rule of thumb, most fixed models can briefly provide at least 50% more current compaired to their continuous rating.