A network of four capacitors of 5 µF is connected to a 500 V supply as shown in figure. Find the equivalent capacitance of the network.

6.67 µF

The correct answer is Option (1) → 6.67 µF

According to the diagram, 3 capacitor are connected in series-

$∴\frac{1}{C_{eff}}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}$

$C_{eff}=\frac{3}{5}μF$

Now,

∴ Both are connected in parallel,

$C_{result}=\frac{5}{3}+5$

$=6.67µF$