Four capacitors are connected as shown in figure. If $C_1 =2 μF, C_2 =3 μF, C_3=5 μF$ and $C_4=10 μF$, the equivalent capacitance between point A & B will be:

5 μF

The correct answer is Option (2) → 5 μF

$C_{AC}=C_1+C_2+C_3$ [Capacitors in Parallel]

$=(2+3+5)μF$

$=10μF$

Now,

$\frac{1}{C_{AB}}=\frac{1}{C_{AC}}+\frac{1}{C_{BC}}$ [Capacitor in series]

$\frac{1}{C_{AB}}=\frac{1}{10}+\frac{1}{10}⇒C_{AB}=\frac{10}{2}μF$

$⇒C_{AB}=5μF$