The equivalent capacitance between points A and B in the below-shown figure will be ________ μF.

CONCEPT:

• The capacitance of series - The capacitance series are in series we have;

            \(\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} +....\)

            Here we have C_s are equivalent is series, C₁, C₂, C₃,....  is the capacitance in series in 1, 2,3... capacitor.

• The capacitance of parallel - The capacitance series are in series we have;

            Cp = C1 +C2 +C3 +...

           Here we have CP are equivalent is series, C1, C2, C3,....  is the capacitance in parallel in 1, 2,3...capacitor.

CALCULATION:

Note: The two capacitors connected with the same wire without junction will vanish from the circuit. Therefore the revised circuit is;

As we can see in the above image that the three capacitances are parallel to each other than according to the parallel capacitance we have;

C_p = C₁ + C₂ + C₃

⇒ C_p = 8 + 8 + 8 

⇒ C_p = 24 μF

In the above figure capacitance, C_p, and 8 μF are in series with each other.

 ∴ \(\frac{1}{C_s} =\frac{1}{C_p}+\frac{1}{8}\)

⇒ \(\frac{1}{C_s} =\frac{1}{24}+\frac{1}{8}\)

⇒ \(\frac{1}{C_s} =\frac{1+3}{24}\)

⇒ \(\frac{1}{C_s} =\frac{4}{24}\)

⇒ C_s = 6 μF

Hence,  Cs = 6 μF