# 2.23 An electrical technician requires a capacitance of $2\mu F$ in a circuit across a potential difference of 1 kV. A large number of $1\mu F$ capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors

Let's assume n capacitor connected in series and m number of such rows,

Now,

As given

The total voltage of the circuit = 1000V

and the total voltage a capacitor can withstand  = 400

From here the total number of the capacitor in series

$n=\frac{1000}{400}=2.5$

Since the number of capacitors can never be a fraction, we take n = 3.

Now,

Total capacitance required = $2\mu F$

Number of rows we need

$m=2*n=2*3=6$

Hence Capacitors should be connected in 6 parallel rows where each row contains 3 capacitors in series.

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