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Question

For a Physics project, a student requires a 4 μF capacitor in a circuit across a potential difference of 1 kV. Unfortunately, 4 μF capacitors are out of stock in the lab. However, 2 μF capacitors which can withstand a potential difference of 400 V are available in plenty. Find the minimum number of capacitors required.

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Solution


1000 V=n1×400V
n1=2.5
3 capacitors are required in each row (Equivalent capacitance in each row = 23 μF)
Now, Ceq=4
n2.23=4n2=6
Total capacitors =6×3=18


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