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The resultant capacitance of four capacitors connected in series is ------------ the smallest individual capacitance

a)less than b)greater than c)equals to d) four times

User Shena
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1 Answer

3 votes

Answer:

a)less than

Step-by-step explanation:

The resultant capacitance of N capacitors connected in series is given by


(1)/(C)=(1)/(C_1)+(1)/(C_2)+....+(1)/(C_N)

In this problem, since we have 4 capacitances in series, the resultant capacitance is


(1)/(C)=(1)/(C_1)+(1)/(C_2)+(1)/(C_3)+(1)/(C_4)

Each of the single term on the right-side of the equation is a positive term: this means that the term
(1)/(C) will be larger than the individual terms
(1)/(C_i).

Therefore, this means that the reciprocal, C, will be smaller than each of the individual capacitances
C_i.

In particular, if the 4 capacitances are identical,


C_1=C_2=C_3=C_4=C_0

So the resultant capacitance will be:


(1)/(C)=4(1)/(C_0)\\\rightarrow C=(C_0)/(4)

So the correct answer is

a) less than

User Malganis
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