Question:
Three capacitors each of capacitance 10 $\mu$F are to be connected such that the effective capacitance becomes 15 $\mu$F. This can be done by connecting:
Three capacitors each of capacitance 10 $\mu$F are to be connected such that the effective capacitance becomes 15 $\mu$F. This can be done by connecting:
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Capacitors in parallel add up ($C_p = C_1+C_2$), while series follow reciprocal addition.
Updated On: Jun 10, 2026
- All of them in series
- All of them are in parallel
- Two in series and the 3rd parallel to the combination
- Two in parallel and the 3rd series to the combination
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Effective capacitance of series/parallel combinations.
Step 2: Analysis
* Two 10 $\mu$F in series = $1/(1/10 + 1/10) = 5$ $\mu$F. * Parallel to the 3rd (10 $\mu$F): $5 + 10 = 15$ $\mu$F.
Step 3: Conclusion
Option (C) provides the correct arrangement.
Final Answer: (C)
Effective capacitance of series/parallel combinations.
Step 2: Analysis
* Two 10 $\mu$F in series = $1/(1/10 + 1/10) = 5$ $\mu$F. * Parallel to the 3rd (10 $\mu$F): $5 + 10 = 15$ $\mu$F.
Step 3: Conclusion
Option (C) provides the correct arrangement.
Final Answer: (C)
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