You have many 2.0μFcapacitors, each capable of withstanding 200 Vwithout undergoing electrical breakdown (in which they conduct charge instead of storing it). How would you assemble a combination having an equivalent capacitance of (a)0.40μFand (b)1.2μF, each combination capable of withstanding 1000 V?

Short Answer

Expert verified
  1. Five capacitors in series are connected to get capacitanceof0.40μF.
  2. Three arrays of the capacitors in parallel having five capacitorsin series are connected to get 1.2μF.

Step by step solution

01

The given data

  1. Value of each capacitor, C=2.0μF
  2. Potential that each capacitor can withstand, V = 200 V
  3. Equivalent capacitance,C'eq=0.40μF
  4. Equivalent capacitance of withstanding potential of is C'eq=1.2μF.
02

Understanding the concept of the equivalent capacitance

We will use the idea of the combination of capacitors in series and parallel to get the required capacitance.

Formulae:

The equivalent capacitance of a series connection of capacitors,

1Cequivalent=1Ci …(i).

The equivalent capacitance of a parallel connection of capacitors,

1Cequivalent=Ci …(ii)

03

(a) Calculation of the number of capacitors

If we connect N capacitors in series to get capacitance40μF, then the equivalent capacitance can be given using equation (i) as follows: (as each capacitance value is given to be same)

1Ceq=i=1n1Ci=NC

Now, using the given values, we have

N=CCeq=2.0μF0.40μF=5.0

So, we need to connect 5 capacitors in series.

04

(b) Calculation of number of capacitor in arrays

Above combination can withstand 1000 V

If same sequences of capacitors connected in parallel, having five capacitors connected in series are used, then the number of equivalent capacitance can be given using equation (ii) as:

C''eq=N×C'eq1.2μF=N×C'eqN=1.2μF0.40μF=3.0

So, we must connect 3 arrays of capacitors in parallel, each having 5 capacitors in series.

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