Answer :
Answer:
The answer is "13% and 83%"
Explanation:
The highest power is generated for:
[tex]\to \frac{dP}{dV_a}=0= I_s(e^{\frac{V_m}{V_t}} -1)-I_{Ph} +\frac{V_m}{V_t} I_s e^{\frac{V_m}{V_t}}[/tex]
Where voltage, [tex]V_m[/tex], was its calculated maximum power wattage. This same following transcend national stable coins the power output:
[tex]\to V_m=V_t \ In \frac{1+\frac{I_{ph}}{I_s}}{1+\frac{V_{m}}{V_t}}[/tex]
The accompanying consecutive values of [tex]V_m[/tex] are done utilizing iteration and an initial value of 0.5 V:
[tex]\to V_m = 0.5,\ 0.542,\ 0.540 \ V[/tex]
Efficacy is equivalent to:
[tex]\eta = |\frac{V_{m} I_{m}}{p_{in}}| =\frac{0.54 \times 0.024}{0.1}=1.3\%[/tex]
Currently, [tex]I_m[/tex] was calculated with the formula of voltage [tex]V_m[/tex] (4.6.1) as well as the sun is presumed to produce [tex]100 \frac{mW}{cm^2}[/tex] of power.
The fulfillment factor is equal to:
fill factor[tex]= \frac{V_{m} I_{m}}{V_{oc}I_{sc}}= \frac{0.54 \times 0.024}{0.62 \times 0.025}=83\%[/tex]
where open circuit tension of equations (4.6.1) and I = 0. is calculated. The current is equal to the total current of the photo.