Da, sigur e o solutie.

Sa facem calculele:

Energia solara la nivelul pamāntului la latitudinea de 45 de grade ajunge la 1000Wati /ora / metru2. Vara.
Eficienta conversiei energiei solare in energie electrica este de 13% cel mult. In ultimii 30 de ani aceasta eficienta a crescut de la 9%  la 13%.
Asta inseamna ca pe fiecare metru2 cultivat cu celule fotovoltaice obtinem 130 wati pe ora. La 12 volti.
Pe ansamblul unui an numarul mediu de ore insorite la latitudinea 45 este de 3,6 pe zi. Asta inseamna 3,6*130W = 468W /zi /m2. Adica consumul normal pe o zi intr-o casa foarte, foarte saraca, fara niciunul din aparatele de mai jos. Doar cāteva becuri chioare.(de 25 de Wati)
Consumul cātorva aparate casnice: (Wati/ora)
Frigider 350-600
Televizor 120-180
Masina de spalat 700-1.000
Calculator ( PC) 250-350
Aspirator 700-1500
Fier de calcat 750-1600
Ca sa putem sa ne uitam la televizor doar, ne mai trebuiesc vreo 2 metri2 de celule solare.
Uitati de boiler. Aici discutam de 2-3.000 de Wati. Adica 20 de metri2 de fotovoltaice.
Si sa nu uitam, discutam de 12 volti sau 24 volti, nu de 220.
Pretul ?
Panourile fotovoltaice propiu-zis nu costa enorm. Metrul2  instalat ajunge pe la 300-500 de euro. Dar sistemul de stocare al energiei electrice e alta poveste. Pentru ca discutam de perioade lungi in care soarele nu apare deloc. Chiar si vara se intāmpla sa fie inourat mai multe zile la rānd. Una din firmele care vinde in Romania celule olandeze. http://www.kerychip.dk/EnergieSolara.htm

Solar energy. This is a favorite possible source of future energy for many people, comforted by the thought that it is unlimited. But, quite the contrary is true. The Sun will exist for a long time, but at any given place on the Earth's surface the amount of sunlight received is limited--only so much is received. And at night, or with overcast skies, or in high latitudes where winter days are short and for months there may be no daylight at all, or available in small and low intensity quantities. Direct conversion of sunlight to electricity by solar cells is a promising technology, and already locally useful, but the amount of electricity which can be generated by that method is not great compared with demand. Because it is a low grade energy, with a low conversion efficiency (about 15%) capturing solar energy in quantity requires huge installations--many square miles. About 8 percent of the cells must be replaced each year. But the big problem is how to store significant amounts of electricity when the Sun is not available to produce it (Trainer, 1995), for example, at night. The problem remains unsolved. Because of this, solar energy cannot be used as a dependable base load. And, the immediate end product is electricity, a very limited replacement for oil. Also, adding in all the energy costs of the production and maintenance of PV (photovoltaic) installations, the net energy recovery is low (Trainer, 1995).

Energy is essential to life, and energy from the sun is the ultimate basis for all life.

> Well,
> As one in the solar biz, here are the calculations used in the
> I live in New England and we use insolation tables that show
> sun hours (1 sun hour is equal to 1000w/m squared). At 13%
> per module. That is 130 watts per sq meter of electric output for 1
> sun hour. The average annual daily sun hours here is 3.6 per day.
> So 130 watts * 3.6 * 365 gives the answer.  170,820 watt hours per
> year per sq meter of module. 170 kWh / year per sq meter (rpughly
> sq feet...100 sq feet is 1 kw of PV installed).
>Also note that the most beneficial use of PV is in the south for
peak aircon load. Measured insolation in Arizona and much of Texas is
in excess of 2200 peak hours per year. Module efficiencies of 18% are
now realistic, giving conversion to a/c of 14%. With even a 20 sun
concentrator these values go to about 22% and 17% respectively. So in
much of the SW we can expect 370kWh/yr/sq. m. a/c out. Murray

In Australia  on average there are 5.5 hrs of usable sunlight per day, x 365
days = 2000 hrs (round figures) of sunlight.

2000hrs x 1000W/m2 = 2MWhrs/m2/yr

assume 12% efficiency = 240kWh/m2/