Hi there!
Recall the following.
Capacitors in series:
[tex]C_T = \frac{1}{C_1} + \frac{1}{C_2} +...+ \frac{1}{C_n}[/tex]
Capacitors in parallel:
[tex]C_T = C_1 + C_2 + ... + C_n[/tex]
Begin by solving for the resulting capacitance of both paths.
Path on the left:
[tex]\frac{1}{C_T} = \frac{1}{2C} + \frac{1}{C} = \frac{3}{2C}\\C_T = \frac{2C}{3}[/tex]
Path on the right:
[tex]\frac{1}{C_T} = \frac{1}{C} + \frac{1}{3C} = \frac{4}{3C}\\\\C_T = \frac{3C}{4}[/tex]
Now, since we ADD capacitors in parallel, we can add the resulting capacitances together:
[tex]C_T = \frac{2C}{3} + \frac{3C}{4}[/tex]
Substitute in 12 F for C and solve.
[tex]C_T = \frac{2(12)}{3} + \frac{3(12)}{4} = \frac{24}{3} + \frac{36}{4} = 8 + 9 = \boxed{17F}[/tex]
