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The equivalent capacitance between A and
B in the circuit given below, is :

  • Option 1)

    2.4 \mu F

     

     

     

  • Option 2)

    4.9 \mu F

  • Option 3)

    3.6 \mu F

  • Option 4)

    5.4 \mu F

 

Answers (2)

As we learned

 

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

 

 \frac{1}{c_{eq}} = \frac{1}{6\mu F}+\frac{1}{12\mu F} + \frac{1}{6\mu F} = \frac{2+1+2}{12\mu F}

c_{eq} = 2.4\mu F


Option 1)

2.4 \mu F

 

 

 

Option 2)

4.9 \mu F

Option 3)

3.6 \mu F

Option 4)

5.4 \mu F

Posted by

Vakul

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