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
B
2
C
16
D
24

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 equivalentis $$C_{eq}=8\times 0.25 = 2\mu F$$Total capacitors required= $$8\times 4 = 32$$PhysicsNCERTStandard XII

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