If there is full light then the output power will be 48W (doh!). However, this condition requires that the optimal volt/current is set at the load (or the input of a special power converter, to maintain the optimal condition).
Since W=IxV, then the current at full power at 70V would be 48/70 = 680mA. This is at full power but your solar panel may have a different optimal power point (see above) so is likely to be less than this.
it depends on your solar panels, i can tell you if you know the amperage and the volts, you can find the amount of watts it produce, the equation is watts = amps x volts. hate to give you an equation as an answer
The voltage of the solar panel is less important than the total amperage. Proper wire size is determined by amperage that will be going through the wire. Wire INSULATION determines the voltage that a wire can carry.
Watts = Volts x Amps, if you use your algebra you will find that it's approx 14 Amps.
Check the battery voltage on your tester. The voltage on the panel is the same throughout the whole electrical system.
If it is a 12 volt panel it will light a 12 volt bulb. Most likely it is not a 12 volt panel, it is some other voltage, so you then need equipment like an inverter to convert the energy to 12-volt energy.
500 volts!!
A solar panel develops volts of EMF across its terminals when it's illuminated. If a conducting path is provided between its terminals, then amps of current flow. Just like any other battery or generator.
Most likely not. In order to charge the battery to its nominal rated 4.8 volts, youreally need a source capable of more than 4.8 volts open-circuit.You need to take the solar panel and a voltmeter, and measure the output voltageof the solar panel with no load connected to it. If it's more than 4.8 volts, then itwill charge your battery.But . . .That's not saying anything about how long it will take. 0.4 watt is not an awful lotof power, and your solar panel will not even deliver that much before its outputvoltage sags to 4.8 . So I would think that this solar panel will not be an acceptablecharger for that battery.
Solar panels are rated in watts output. To find the amperage use this equation, Amps = Watts/Volts. The wattage will be on the nameplate of the solar panel.
it depends on your solar panels, i can tell you if you know the amperage and the volts, you can find the amount of watts it produce, the equation is watts = amps x volts. hate to give you an equation as an answer
The voltage of the solar panel is less important than the total amperage. Proper wire size is determined by amperage that will be going through the wire. Wire INSULATION determines the voltage that a wire can carry.
A three phase panel will not give you 110 and 220 volts. A three phase four wire panel will, but not at these voltages. The nearest voltages will be 120 and 208 volts. The 120 volt is the wye voltage of 208 volts. 208/1.73 = 120 volts. A single phase three wire panel will give you 110 and 220 volts.
u multiply your voltage times max current (amps) and this gives u max wattage.
Try the manufacture.OR Wattage=volts x amps. the brighter the sun the more they put out, up to a limit. Use a 17volt light bulb(good luck), as high a wattage you can find, in series with an ampmeter and connect them to the solar panel. Place a volt meter across the solar panal leads.Then place them in bright sunlight to read the ampmeter. Make sure you do not exceed to ampmeter limit by observing the meter as the solar panel is slowly exposed to sunlight. This should work. 17 volts is low voltage and probably cannot hurt you. Multiply the volt meter reading by the amp meter reading to get your wattage reading in watts.
Yes but you will need vast area of space to put solar panels and they only work during the day. But with technology come by with more efficiently in collecting solar radiation, this is possible to be done on the roof of your house in the future.
Watts = Volts x Amps, if you use your algebra you will find that it's approx 14 Amps.
You would have to add panels in parallel to get more amps and then add a variable resistance in one leg to drop the voltage to your device.