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Question

A capacitance of $$2\ \mu F$$ is required in an electrical circuit across a potential difference of $$1.0\ kV$$. A large number of $$1\ \mu F$$ 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 :


A
32
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B
2
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C
16
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D
24
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Solution

The correct option is B $$32$$
$$m\times 300 > 1000$$

$$m > \dfrac{10}{3}$$

or $$m = 4$$

For $$m = 4$$, capacitance of 1 branch is $$\dfrac{1\mu F}{4}=0.25 \mu F$$

For overall capacitance of $$2\mu F$$, $$8$$ such branches or equivalent

is $$C_{eq}=8\times 0.25 = 2\mu F$$

Total capacitors required= $$8\times 4 = 32$$


Physics
NCERT
Standard XII

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