Two capacitors of 8µF are connected in series then the equivalent capacitance is ______?

¼ µF

2 µF

3 µF

4 µF

When two capacitors of equal capacitance (C) are connected in series, the equivalent capacitance (Ceq) is half the value of one capacitor:

Ceq = (C × C) / (C + C) = (C × C) / 2C = C/2

In this case, C = 8 µF, so:

Ceq = 8 µF / 2 = 4 µF

Therefore, the equivalent capacitance of the two 8 µF capacitors connected in series is 4 µF.