Question:
Three capacitances, each of \(3\ \mu\mathrm{F}\), are provided. These cannot be combined to provide the resultant capacitance of :
Three capacitances, each of \(3\ \mu\mathrm{F}\), are provided. These cannot be combined to provide the resultant capacitance of :
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List the networks that use all three capacitors: 9, 1, 2 and 4.5 \(\mu\)F.
Updated On: Nov 11, 2026
- \(2\ \mu\mathrm{F}\)
- \(6\ \mu\mathrm{F}\)
- \(1.5\ \mu\mathrm{F}\)
- \(1\ \mu\mathrm{F}\)
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The Correct Option is B
Solution and Explanation
Using all three capacitors, the possible networks are:
All in parallel: \(9\ \mu\)F. All in series: \(1\ \mu\)F.
Two in parallel (\(6\ \mu\)F) in series with the third: \(\dfrac{6\times3}{6+3}=2\ \mu\)F.
Two in series (\(1.5\ \mu\)F) in parallel with the third: \(1.5+3=4.5\ \mu\)F.
So 2 and 1 \(\mu\)F can be obtained, while 6 \(\mu\)F (and 1.5 \(\mu\)F) cannot come from a network using all three; 6 \(\mu\)F is taken as the intended value that cannot be obtained.
Answer: (2) \(6\ \mu\mathrm{F}\)
All in parallel: \(9\ \mu\)F. All in series: \(1\ \mu\)F.
Two in parallel (\(6\ \mu\)F) in series with the third: \(\dfrac{6\times3}{6+3}=2\ \mu\)F.
Two in series (\(1.5\ \mu\)F) in parallel with the third: \(1.5+3=4.5\ \mu\)F.
So 2 and 1 \(\mu\)F can be obtained, while 6 \(\mu\)F (and 1.5 \(\mu\)F) cannot come from a network using all three; 6 \(\mu\)F is taken as the intended value that cannot be obtained.
Answer: (2) \(6\ \mu\mathrm{F}\)
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