Question

Find equivalent capacitance across $\text{AB}$ (all capacitance are in $\mu \text{F})$.

• A.
  $\frac{20}{3}\mu \text{F}$
• B.
  $9\mu \text{F}$
• C.
  $48\mu \text{F}$
• D.
  $36\mu \text{F}$

Answer 1

The correct option is B

$9\mu \text{F}$
The given circuit can be rearranged by applying point potential technique. Let us assign the same potential to all the points present on the common wire.


Hence, we have four different potential points i.e. $1,2,3$ & $4$$\left(\text{near B in diagram}\right)$. Now redrawing the circuit as shown below;


Now, we get


We can see that,

$C_{12}{C}_{34}=C_{13}{C}_{24}=60\mu F$

Hence, it is a balanced Wheatstone bridge and the $5\mu F$ capacitor connected across points $2$ & $3$ will be non-functioning.


Now for the above series branches, the equivalent capacitance,

$C_{eq}{)}_1=\frac{30\times 10}{30+10}=\frac{30}{4}\mu \text{F}$ and

$(C_{eq}{)}_2=\frac{6\times 2}{6+2}=\frac{12}{8}\mu \text{F}$


So, the final equivalent capacitance across $\text{AB}$,

$(C_{AB}{)}_{Final}=\frac{30}{4}+\frac{12}{8}=\frac{72}{8}=9\mu \text{F}$

Hence, option (b) is the correct answer.

Why this Question? Always try to apply point potential technique to simplify complex circuits. If we obtain ${}^{\prime }{4}^{\prime }$ points with different potentials, then place these points as vertices of the square and redraw the circuit.