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mote1985 [20]
3 years ago
8

Why is it not possible to make use of solar cells to meet all of our energy needs?why​

Physics
2 answers:
Nady [450]3 years ago
5 0

Answer:

In the solar cells, the energy is obtained only during the day, when the Sun shines. ... So, it increases the cost of using solar panels as the source of energy. So, the solar cell is not used to meet all our energy needs

Explanation:

n the solar cells, the solar panel convert solar energy into electricity, which is stored in storage battery. The storage battery give the direct current but all the appliances are working by the alternating current, so first of all direct current is converted into alternating current by any suitable appliances before it can be used to run various devices. So it increases the cost of using solar panels as the source of energy. 

aleksley [76]3 years ago
4 0

Answer:

In the solar cells, the energy is obtained only during the day, when the Sun shines. .So, it increases the cost of using solar panels as the source of energy. So, the solar cell is not used to meet all our energy needs

hope this helps

have a good day :)

Explanation:

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The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
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Answer:

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Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

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