#1.) A solar photovoltaic panel has an area of 3 m^2. The panel's maximum power output is 595 W @ 1100 W/m^2.
a) What is its efficiency?
b) If the panel receives 1300 W/m^2 of irradiance, what is its new power output?
#2.) A solar voltaic module has a maximum output of 200 W @ 1100 W/m^2 when connected to a 2 ohm resistor. What is the maximum voltage and current?
#1.) Efficiency of a solar photovoltaic panel is defined as the ratio of the actual power output to the maximum power output it could potentially produce. To calculate the efficiency, we can use the following formula:
Efficiency = (Actual Power Output / Maximum Power Output) * 100
a) To find the efficiency of the panel, we need to know the actual power output. In this case, the maximum power output is given as 595 W @ 1100 W/m^2, but we are not given the actual power output. Therefore, we cannot calculate the efficiency without that information.
b) To find the new power output of the panel when it receives 1300 W/m^2 of irradiance, we can use the concept of proportional relationships. We know that the maximum power output is 595 W @ 1100 W/m^2. So, we can set up the following proportion:
(595 W / 1100 W/m^2) = (New Power Output / 1300 W/m^2)
To solve for the new power output, we can cross-multiply and solve for it:
New Power Output = (595 W / 1100 W/m^2) * 1300 W/m^2
#2.) The maximum voltage and current of a solar voltaic module can be determined using Ohm's Law, which states that voltage (V) is equal to the current (I) multiplied by the resistance (R).
Given the maximum output of 200 W @ 1100 W/m^2 and the connected 2 ohm resistor, we can find the current (I) by rearranging the formula:
I = V / R
To find the maximum voltage (V), we can rearrange the formula again:
V = I * R
We can substitute the values we are given:
I = 200 W / 2 ohm
I = 100 A
Now, we can find the maximum voltage (V):
V = 100 A * 2 ohm
V = 200 V
So, the maximum voltage is 200 V and the maximum current is 100 A.