Question:

Two capacitors when connected in series have a capacitance of 3 μF, and when connected in parallel have a capacitance of 16 μF. Their individual capacities are

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When capacitors are in series, their total capacitance decreases, and when they are in parallel, their total capacitance increases.
Updated On: Jan 12, 2026
  • 12 μF, 2 μF
  • 6 μF, 2 μF
  • 4 μF, 12 μF
  • 3 μF, 2 μF
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The Correct Option is A

Solution and Explanation

Step 1: Use the formulas for capacitance in series and parallel.
For capacitors in series, the total capacitance \( C_s \) is given by: \[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \] For capacitors in parallel, the total capacitance \( C_p \) is: \[ C_p = C_1 + C_2 \]
Step 2: Solve the system of equations.
From the given information, we have the following equations: \[ \frac{1}{3} = \frac{1}{C_1} + \frac{1}{C_2} \quad \text{and} \quad C_1 + C_2 = 16 \] Solving these equations gives \( C_1 = 12 \, \mu\text{F} \) and \( C_2 = 2 \, \mu\text{F} \).
Final Answer: \[ \boxed{12 \, \mu\text{F}, 2 \, \mu\text{F}} \]
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