None by itself. Watts are measurement of power, and power is volts times amperes.
Secondly, the 6 volts must be in an electrical circuit (loop) to create current flow (Amps).
Thirdly, the resistance of whatever the 6 volts is powering determines current flow (Amps). So, without knowing the current flow, you cannot calculate power.
Fourth, power (Watts, horsepower, ergs, chevals vapeur, etc) is a measurement taken over a certain time - it is not instantly measureable with an inexpensive meter like a voltmeter or an ammeter (Amp-meter).
It generates 12V unless its an older 6V model the correct question is probably how many Watts does it produce
The peak-to-peak (P-P) value of a 6V square wave is the difference between its maximum and minimum voltage levels. For a square wave that oscillates between +3V and -3V, the peak-to-peak voltage would be 6V (3V - (-3V) = 6V). If the square wave oscillates between 0V and 6V, the peak-to-peak voltage would also be 6V (6V - 0V = 6V). Therefore, regardless of the specific levels, a 6V square wave has a peak-to-peak voltage of 6V.
It is: -10v+6v = -4v
Power = V x A = 6 x 0.5 = 3 watts = 3 joules per seconddissipated by the lamp into the surrounding environmentin the form of heat and light.
yes, you can.
0
6 and 18v, respectively.
The power used by the battery can be calculated using the formula P = V x I, where P is power, V is voltage (6V), and I is current (0.5A). Substituting the values, the power used by the battery is P = 6V x 0.5A = 3 Watts.
2u^2-6v-uv
5(6v + 7)(6v - 7)
6v + 7 = -296v = -36v = -6
100wats