(a) $2\,\mu\mathrm{F}$ and $4\,\mu\mathrm{F}$ capacitors are to be connected in series. So, the effective capacitor will be $C_{\text{eff}} = \frac{2 \times 4}{2+4} = \frac{4}{3}\,\mu\mathrm{F}$ With $C_{\text{eff}}$, the $3\,\mu\mathrm{F}$ capacitor is to be connected in parallel. So, the equivalent capacitor will be $C_{\text{eqv}} = \frac{4}{3} + 3 = \frac{13}{3}\,\mu\mathrm{F}$ (b) Maximum value of capacitance may be obtained by connecting the capacitors in parallel. In that case, the equivalent capacitance will be, $(2+3+4)=$ $10\,\mu\mathrm{F}$. Minimum value of capacitance may be obtained by connecting the capacitors in series. In that case, the equivalent capacitance will be $C_{S}$. $\frac{1}{C_S} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{13}{12}$ $\therefore C_S = \frac{12}{13}\,\mu\mathrm{F}$