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.)

Let the node between $$C_1$$ and $$C_2$$ be P. 
The npode between $$c_2$$ and $$C_3$$ is A (since that is shorted)
The two capacitors $$C_1 , C_2 $$ are connected end-to-end between the terminals A and P, so they are in paralell.
Their  combination is

$$C_{\,p}$$ = $$C_1$$ + $$C_2$$

$$C_{\,p}$$ = \frac{1}{1\,\mu F} + \frac{1}{1\,\mu F}  = 2$$

$$\Rightarrow\; C_{\,p} = \frac{1}{2}\,\mu F$$
Now this is connected in series with $$C_3$

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: $$2,\mu F$$