Question

A number of condensers, each of capacitance 1μF and each one of which gets punctured if a potential difference just exceeding 500 V is applied, are provided. An arrangement suitable for giving capacitance of 2μF across which 3000 volts may be applied requires at least.
A. 6 component capacitors
B. 12 component capacitors
C. 72 component capacitors
D. 2 component capacitors

Answer 1

Hint:We have capacitors and we know that capacitors can be grouped in series and parallel combination. Capacitor stores charge and when it becomes fully charged no current flows through it. Here the maximum potential difference that could be applied without burning the capacitor is 500 V. We can try different arrangements to arrive at a meaningful solution.

Complete step by step answer:
Maximum voltage which we need= 3000 V
Also, maximum voltage that s single capacitor can withstand= 500 V
So, the number of capacitors is given by: \[n=\dfrac{3000}{500}=6\]
So, we are putting 6 capacitors in one row and that in a series combination, so net capacitance is given by:
$\Rightarrow \dfrac{1}{C}=1+1+1+1+1+1 \\
\therefore C=\dfrac{1}{6}\mu F \\$
Now let there be n such rows, then according to the condition given in the question we get
$\dfrac{n}{6}=2 \\
\therefore n=12$
Since, the number of capacitors in one row were 6, \[\therefore 12\times 6=72\]

So, 72 component capacitors are required and the correct option is C.

Note:The capacitors can be arranged broadly in two combinations: one is series combination and the other is parallel combination. Also, the formulas used are just the opposite that we use for series and parallel combination of resistors. Potential is due to electric charge and we always measure the potential difference, while doing the calculation here we had assumed that potential at one point to be zero would not affect the measurement because we always measure the potential difference and not the potential.