Three capacitors each of 4 µF are to be connected in such a way that the effective capacitance is 6 µF. This can be done by connecting them :

  • Option 1)

     all in series

  • Option 2)

     two in series and one in parallel

  • Option 3)

    all in parallel

  • Option 4)

    two in parallel and one in series

     

 

Answers (2)

As we learnt in

Series Grouping -

\frac{1}{C_{eq}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\cdots

- wherein

 

 

Parallel Grouping -

C_{eq}=C_{1}+C_{2}+\cdots

- wherein

 

 a: \frac{1}{C_{a}}= \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}=\frac{3}{4}\Rightarrow {C_{a}} = \frac{4}{3}\mu\, f

b: {C_{b}}=4+4+4= 12 \mu\, f

c: {C_{c}}=\frac{(4+4)\times 4}{{(4+4)+ 4}}\, = \frac{8}{3}\, \mu\, f

d: {C_{d}}=\frac{(4\times 4)}{{(4+4)}}+4\, = 6\, \mu\, f

 


Option 1)

 all in series

Incorrect Option

Option 2)

 two in series and one in parallel

Correct Option

Option 3)

all in parallel

Incorrect Option

Option 4)

two in parallel and one in series

 

Incorrect Option

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