Question

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

Answer 1

Step 1: Find the minimum number of capacitors in series

Given, Required Capacitance, $C=2\mu F$
Potential difference, $V=1kV=1000V$
Capacitance of each capacitor , $C^{\prime }=1\mu F$
Potential difference that the capacitors can withstand, $V^{\prime }=400V$

Suppose $m$ number od capacitors are connected in series.
Then, across each capacitor the potential difference is given by,
$\frac{V}{m}=V^{\prime }$
$m=\frac{1000}{400}=2.5\approx 3$
Therefore, the number of capacitors connected in series is three (in one row).

Step 2: Find the effective capacitance of one row

So, the effective capacitance of one row is,

$\frac{1}{C^{\prime }}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=3$
$C^{\prime }=\frac{1}{3}\mu F$ ....$\left(i\right)$

Step 3: Find the number of rows required in parallel

Let there be $n$ parallel rows and we know each of these rows will have $3$ capacitors.

Therefore, the equivalent capacitance of the circuit is given as

$C^{\prime \prime }=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}....\frac{n}{3}$ ...$\left(ii\right)$

Since, the required capacitance of the circuit is $2\mu F$

Therefore, from equation $\left(ii\right)$

$2=\frac{n}{3}$

$n=2\times 3=6$

Hence there are $6$ rows of three capacitors in the circuit.

Step 4: Find the number of capacitors for required arrangement
Then, a minimum of capacitors required for the arrangements is,
$N=m\times n=3\times 6$
$N=18$
Therefore $18$ capacitors are required for the possible arrangement.

Final Answer: $18$ capacitors are required for the possible arrangement.