Explanation: When capacitors are connected in series, their total capacitance is calculated differently than resistors in series or capacitors in parallel. In a series connection, the effect is similar to increasing the distance between the "plates" of a single capacitor, which *reduces* the overall capacitance. The formula for total capacitance ($C_T$) of capacitors in series is: $1/C_T = 1/C_1 + 1/C_2 + 1/C_3 + ...$ For the given values of 3 $\mu F$, 5 $\mu F$, and 7 $\mu F$: $1/C_T = 1/3 + 1/5 + 1/7$ To sum these fractions, find a common denominator, which is $3 \times 5 \times 7 = 105$: $1/C_T = (35/105) + (21/105) + (15/105)$ $1/C_T = (35 + 21 + 15) / 105$ $1/C_T = 71 / 105$ Now, invert to find $C_T$: $C_T = 105 / 71 \approx 1.47887$ microfarads. Rounding to three decimal places gives 1.479 microfarads. This makes (B) the correct answer. Options (A) and (C) are incorrect because they represent a capacitance greater than the individual capacitors, which only occurs when capacitors are connected in parallel. For capacitors in series, the total capacitance will always be less than the smallest individual capacitor.