From the circuit given below, the capacitance between terminals A and B shown in the circuit is ______ $$\mu$$F.
(take $$C_1 = C_2 = C_3 = 1$$ $$\mu$$F and $$C_4 = 2$$ $$\mu$$F.)

The three capacitors $$C_1 , C_2 , C_3$$ are connected end-to-end between the terminals A and B, so they are in series.
Their series combination is

$$\frac{1}{C_{\,s}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$

$$\frac{1}{C_{\,s}} = \frac{1}{1\,\mu F} + \frac{1}{1\,\mu F} + \frac{1}{1\,\mu F} = 3$$

$$\Rightarrow\; C_{\,s} = \frac{1}{3}\,\mu F$$

The capacitor $$C_4 = 2\,\mu F$$ is connected directly across A and B, hence it is in parallel with the above series combination.

For parallel capacitors, capacitances simply add:
$$C_{\text{eq}} = C_{\,s} + C_4 = \frac{1}{3}\,\mu F + 2\,\mu F = \frac{1 + 6}{3}\,\mu F = \frac{7}{3}\,\mu F$$

Therefore, the equivalent capacitance between A and B is $$\displaystyle \frac{7}{3}\,\mu F$$.

Option C which is: $$\frac{7}{3}\,\mu F$$