Question

n capacitors of 2 uF each are connected in
parallel and a p.d. of 200 V is applied to the
combination., the total a charge on them was
1C. Then n is equal to
1) 2500
2) 3000
3) 3333
4) 25​

Answer 1

Answer : (d) 12nCV2

Explanation : Each capacitor has voltage difference = V

Number of capacitors = n

Total energy = n× energy across one capacitor

=12nCV2.

Answer 2

Concept: When two or more resistances are connected in parallel between two places, each has a separate current direction, they are said to be connected in parallel. As the branches converge at a common point in such circuits, the current is branched out and recombined.

Given: n capacitors of 2μF connected in parallel

           potential difference = 200V

           charge = 1C

To find: number of capacitors

Solution:

In this problem, we must determine the number of capacitors linked in parallel. To calculate the number of capacitors, we'll apply the formula that relates the capacitance of the capacitors, the total charge, and the potential difference.

$C = \frac{Q}{V}$

As a result of the data, we have the following information.

Two potential differences are applied to the capacitors in combination, V = 200 V.

Q = 1 C is the total charge on the capacitors when they are combined.

C = 2F is the capacitance of each capacitor.

The total number of parallel capacitors connected is n =?

A circuit schematic depicting the parallel connection of n capacitors.

Where C is the capacitance of the parallel/series linked capacitors, Q is the net charge on the capacitors, and V is the potential difference applied across the capacitor plates.

The basic formula for calculating the capacitance of the capacitor can be changed while maintaining the original form.

The 'n' number of capacitors is given to us.

The capacitance of any number of capacitors can be represented by the formula, depending on how we utilise it.

As a result, below is the amended formula:

$nC = \frac{Q}{V}$

$n = \frac{Q}{CV}$

Replace the given values in the equation above.

$n = \frac{1}{(2*10^{-6})(200) }$

n = 2500

Option (A) is valid since the number of capacitors linked in parallel is 2500.

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