Three capacitors each of capacitance 3 μF 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 μF
Given: Three capacitors, each $C = 3 \,\mu F$.
Possible combinations:
• All in series: $C_{eq} = \frac{C}{3} = \frac{3}{3} = 1 \,\mu F$
• All in parallel: $C_{eq} = 3C = 9 \,\mu F$
• Two in series, one in parallel:
$C_{series} = \frac{C}{2} = 1.5 \,\mu F$
$C_{eq} = 1.5 + 3 = 4.5 \,\mu F$
• Two in parallel, one in series:
$C_{parallel} = 6 \,\mu F$
$C_{eq} = \frac{6 \cdot 3}{6 + 3} = \frac{18}{9} = 2 \,\mu F$
Thus, achievable capacitances: $1, 2, 4.5, 9 \,\mu F$
From options given: $1 \,\mu F$ ✔ $2 \,\mu F$ ✔ $4.5 \,\mu F$ ✔ $6 \,\mu F$ ❌ (not possible)
Answer: 6 μF