Respuesta :
Answer:
A. Power incident on panel = 3000W
B. Electrical power generated = 600W
Explanation:
Actual solar intensity I = 750W/m2
A) if area A of panel = 4m2, then,
Power incident on panel = I x A
P = 750 x 4 = 3000W
B) if panel efficiency is 20%,
Then,
Power generates = 0.2 x 3000 = 600W of electricity.
Answer:
a) 3000W
b) 600W
Explanation:
A) To solve this problem we have to use
[tex]P=IA[/tex]
where I is the intensity of light and A is the area of our solar panels. By replacing we have
[tex]P=(750\frac{W}{m^2})(4m^2)=3000W[/tex]
B) However we know that the panel only absorb 20% of the total power. Hence we have that the real power is
[tex]P_r=0.2IA=0.2(3000W)=600W[/tex]
HOPE THIS HELPS!!