Electrostatics 5 Question 17
19. A capacitance of $2 \mu \mathbf{F}$ is required in an electrical circuit across a potential difference of $1 kV$. A large number of $1 \mu \mathbf{F}$ 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
(2017 Main)
(a) 16
(b) 24
(c) 32
(d) 2
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Solution:
- Let there are $n$ capacitors in a row with $m$ such rows in parallel.
As voltage not to exceed $300 V$
$$ \therefore \quad n \times 300>1000 $$
[a voltage greater than $1 kV$ to be withstand]
$$ \begin{array}{lll} \Rightarrow & n>\frac{10}{3} \Rightarrow n=4 & \text { (or 3.33) } \\ \text { Also, } & C _{\mathbf{E q}}=\frac{m C}{n}=2 \mu \mathbf{F} & \\ \Rightarrow & \frac{m}{n}=2 \Rightarrow m=8 & {[\because C=1 \mu \mathbf{F}]} \end{array} $$
So, total number of capacitors required
$$ =m \times n=8 \times 4=32 $$
