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JEE Main 2017 Solved Paper Solved Paper
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© examsnet.com
Question : 29
Total: 90
A capacitance of 2µF is required in an electrical circuit across a potential difference of 1.0 kV.A large number of 1µF capacitors are available which can withstand a potential difference of not more than 300 V.The minimum number of capacitors required to achieve this is :
24
32
2
16
Validate
Solution:
To hold 1 KV potential difference minimum four capacitors are required in series
⇒
C
1
=
1
4
for one series
So for Ceq to be 2μF, 8 parallel combinations are required.
⇒ Minimum no. of capacitors = 8 × 4 = 32
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