For the arrangement of capacitors as shown in the circuit , the effective capacitance between the points A and B is
(capacitance of each capacitor is `4 muF` )

Options

A. `4 muF`
B. `2 muF`
C. `1 muF`
D. `8 muF`

Answer

A

Explanation

The given circuit can be redrawn as shown in figure (ii).

As `(4muF)/(4muF)=4/4(muF)/(muF)`
Therefore , the given circuit is a balanced Wheatstone bridge and the capacitance in arm CD is ineffective.
Thus, it reduces to the equivalent circuit as shown in figure (iii) .
As, `4muF and 4 muF` are in series in the upper branch , so their equivalent capacitance `C_1` is
`C_1=((4muF)(4muF))/(4muF+4muF)=2muF`
Similarly , `4muF and 4muF` are in series in the lower branch , so the their equivalent capacitance `C_2` is
`C_2=((4muF)(4muF))/(4muF+4muF)=2muF`
And the corresponding equivalent circuit is shown in figure (iv). Since `C_1 and C_2` are in parallel, the effective capacitance between the points A and B is
`C_(AB)=C_1 +C_2= 2 muF + 2muF= 4muF`