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A Si Solar cell has short-circuited current of 100 mA and open-circuit voltage of 0.7 V under full illumination. If the fill factor is 0.71 then the Maximum power delivered (in mW) to load by this cell is
Correct answer is '49.7'. Can you explain this answer?
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A Si Solar cell has short-circuited current of 100 mA and open-circui...
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A Si Solar cell has short-circuited current of 100 mA and open-circui...
Given data:
- Short-circuited current (Isc) = 100 mA
- Open-circuit voltage (Voc) = 0.7 V
- Fill factor (FF) = 0.71
Calculating the maximum power:
To calculate the maximum power delivered to the load by the solar cell, we need to find the maximum power point (MPP) on the current-voltage (I-V) curve.
Step 1: Find the maximum power point voltage (Vmpp):
The maximum power point voltage (Vmpp) is the voltage at which the power output is maximum.
We can estimate Vmpp as the average of the Voc and the short-circuited voltage (Vsc):
Vmpp = (Voc + Vsc) / 2
Given that Voc = 0.7 V, we need to find Vsc.
Step 2: Find the short-circuited voltage (Vsc):
The short-circuited voltage (Vsc) is the voltage across the solar cell when it is short-circuited, i.e., the current is maximum (Isc) and the voltage is zero.
Given that Isc = 100 mA, we can calculate Vsc using Ohm's Law:
Vsc = Isc * Rs
Here, Rs is the series resistance of the solar cell. However, the value of Rs is not provided in the given data, so we cannot calculate Vsc accurately. Therefore, we assume Vsc to be zero for simplicity.
Hence, Vsc = 0 V.
Step 3: Calculate Vmpp using the estimated values:
Vmpp = (Voc + Vsc) / 2
= (0.7 + 0) / 2
= 0.35 V
Step 4: Calculate the maximum power (Pmpp):
The maximum power (Pmpp) is given by the product of Vmpp and the corresponding current (Impp).
Since the fill factor (FF) is given, we can calculate Impp as a fraction of Isc:
Impp = FF * Isc
= 0.71 * 100 mA
= 71 mA
Pmpp = Vmpp * Impp
= 0.35 V * 71 mA
= 24.85 mW
Therefore, the maximum power delivered to the load by this solar cell is 24.85 mW.
- Short-circuited current (Isc) = 100 mA
- Open-circuit voltage (Voc) = 0.7 V
- Fill factor (FF) = 0.71
Calculating the maximum power:
To calculate the maximum power delivered to the load by the solar cell, we need to find the maximum power point (MPP) on the current-voltage (I-V) curve.
Step 1: Find the maximum power point voltage (Vmpp):
The maximum power point voltage (Vmpp) is the voltage at which the power output is maximum.
We can estimate Vmpp as the average of the Voc and the short-circuited voltage (Vsc):
Vmpp = (Voc + Vsc) / 2
Given that Voc = 0.7 V, we need to find Vsc.
Step 2: Find the short-circuited voltage (Vsc):
The short-circuited voltage (Vsc) is the voltage across the solar cell when it is short-circuited, i.e., the current is maximum (Isc) and the voltage is zero.
Given that Isc = 100 mA, we can calculate Vsc using Ohm's Law:
Vsc = Isc * Rs
Here, Rs is the series resistance of the solar cell. However, the value of Rs is not provided in the given data, so we cannot calculate Vsc accurately. Therefore, we assume Vsc to be zero for simplicity.
Hence, Vsc = 0 V.
Step 3: Calculate Vmpp using the estimated values:
Vmpp = (Voc + Vsc) / 2
= (0.7 + 0) / 2
= 0.35 V
Step 4: Calculate the maximum power (Pmpp):
The maximum power (Pmpp) is given by the product of Vmpp and the corresponding current (Impp).
Since the fill factor (FF) is given, we can calculate Impp as a fraction of Isc:
Impp = FF * Isc
= 0.71 * 100 mA
= 71 mA
Pmpp = Vmpp * Impp
= 0.35 V * 71 mA
= 24.85 mW
Therefore, the maximum power delivered to the load by this solar cell is 24.85 mW.
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A Si Solar cell has short-circuited current of 100 mA and open-circuit voltage of 0.7 V under full illumination. If the fill factor is 0.71 then the Maximum power delivered (in mW) to load by this cell isCorrect answer is '49.7'. Can you explain this answer?
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