P sun=180000/3600=50W
30W/50W=0.6=60% (input 50, output 30, therefore utilization rate 60%)
I=P/ U=30/12=2.5A
W=0.05kW*8h=0.4kW.h
t=W/P=0.4/ 0.03=13.3h
< h3>Urgent help in case of physical problems! ! ! ! ! ! !In complete darkness conditions, when the solar cell is forward biased, the point voltages are measured in nine groups at voltages of 1'5.2'0.2'3.2'5.2'6.2'7.2' 8.2'9 volts. about. Under lighting conditions, when the solar cell is not polarized, the resistance is 90.60.30.10.8.7'5.7'0.6'6.6'5.6'4.6'3.6'2.6'0.5'9.5'8.5'7.5'6.5 '5.5'0.4'. 24 groups were measured under the conditions of 5.4'0.3'0.1'0 kiloohms.
Solution: The area of the sphere formed by the sun's ray to the earth is S = 4π (r^2), and the area of the solar panel is s.
The total radiated power of the solar light before reaching the panel is 0.7P,
The power radiated towards the panelwater is (0.7P/S)*s=0.7Ps/S.
The power actually converted into electrical energy is 0.15*0.7Ps/S=0.105Ps/S.
The power of the motor is P'=UI=300V*50A=15000W, therefore 0.105Ps/S=P'
That is to say 0.105Ps/[4π (r^2) ]= P', deformed to obtain s=[P'*4π(r^2)]/0.105P
Setting to P'=15000W, r=1.5×10^11m , P=4× 10^26W Obtaining s=100m^2
The minimum solar panel area required is 100m^2