Three capacitors each of capacitance \( 3\,\mu\mathrm{F} \) are provided. These cannot be combined to provide the resultant capacitance of:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 6\,\mu\mathrm{F} \)
List every combination that uses all three identical capacitors of \( 3\,\mu\mathrm{F} \). Series gives \( C/n \) and parallel gives \( nC \).
\( \displaystyle \text{All three in series: } \frac{3}{3} = 1\,\mu\mathrm{F} \qquad \text{All three in parallel: } 3 \times 3 = 9\,\mu\mathrm{F} \)
\( \displaystyle \text{Two in parallel } (6\,\mu\mathrm{F}) \text{ in series with one: } \frac{6 \times 3}{6 + 3} = 2\,\mu\mathrm{F} \)
\( \displaystyle \text{Two in series } (1.5\,\mu\mathrm{F}) \text{ in parallel with one: } 1.5 + 3 = 4.5\,\mu\mathrm{F} \)
These four are the only arrangements of the three capacitors, so \( 6\,\mu\mathrm{F} \) cannot be obtained.
Option (1) \( 1\,\mu\mathrm{F} \) — Possible. All three in series.
Option (2) \( 2\,\mu\mathrm{F} \) — Possible. Two in parallel, in series with the third.
Option (3) \( 4.5\,\mu\mathrm{F} \) — Possible. Two in series, in parallel with the third.
Option (4) \( 6\,\mu\mathrm{F} \) — Not possible. With all three used the available values are \( 1,\ 2,\ 4.5 \) and \( 9\,\mu\mathrm{F} \), and \( 6\,\mu\mathrm{F} \) is not among them.