A solar cell, 3.0 cm square, has an output of 350 mA at 0.80 V when exposed to full sunlight. A solar panel that delivers close
to 1.0 A of current at an emf of 120 V to an external load is needed. How many cells will you need to create the panel? How big a panel will you need, and how should you connect the cells to one another? How can you optimize the output of your solar panel?
1 answer:
Answer:
You will need 450 cells (3 cm each) to meet the voltage/current requirement.
The panel must be 3 cells in one side, by 150 cell in another side. 1350 cm^2 or 0.135 m^2. They must be connected 3 in row in parallel (to add current), then each of the former group must be connected in series to meet the voltage, so it would be 150 rows of connected in series.
The panel can be optimized using a voltage inverter, to convert current to voltage. In this way, less cells can be used achieving the same output specs.
Explanation:
To meet the voltage:
120 [v] required voltage
0.8 [v] voltage of each cell
So we need 150 cells in series for the voltage.
To meet the current
1.0 [A] Required current
350[mA]=0.35[A] cell current
1/0.35=3 cell So we need 3 cells in parallel to add the currents and meet the requirement.
See the attached figure
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(a) The speed of the target proton after the collision is:, and (b) the speed of the projectile proton after the collision is:
.
Explanation:
We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle:, and y axle:
. Now replacing the value given as:
,
for the projectile proton and according to the problem
are perpendicular so
, and assuming that
, we get for x axle:
and y axle:
, then solving for
, we get:
and replacing at the first equation we get:
, now solving for
, we can find the speed of the projectile proton after the collision as:
and
, that is the speed of the target proton after the collision.