Question:
Three capacitors of capacitances $3\ \mu\text{F}$, $3\ \mu\text{F}$ and $3\ \mu\text{F}$ are connected in series. The equivalent capacitance is}
Three capacitors of capacitances $3\ \mu\text{F}$, $3\ \mu\text{F}$ and $3\ \mu\text{F}$ are connected in series. The equivalent capacitance is}
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Remember, for series connections, the equivalent capacitance is always less than any individual capacitor's capacitance.
Updated On: May 31, 2026
- $1\ \mu\text{F}$
- $9\ \mu\text{F}$
- $3\ \mu\text{F}$
- $1.5\ \mu\text{F}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
When capacitors are connected in series, the equivalent capacitance \(C_{eq}\) is given by the formula: \[\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}\] where \(C_1, C_2, \ldots, C_n\) are the capacitances of individual capacitors.
Step 2: Meaning
This formula means that the reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of each individual capacitor's capacitance. In a series connection, the total charge stored in all capacitors must be the same, but the voltage across each capacitor will differ according to its capacitance.
Step 3: Analysis
Given three capacitors with capacitances \(3\ \mu\text{F}\), \(3\ \mu\text{F}\), and \(3\ \mu\text{F}\) connected in series: 1. The reciprocal of each capacitor's capacitance is: \[\frac{1}{3} = 0.3333\ \mu\text{F}^{-1}\] 2. Summing these reciprocals gives the total reciprocal of the equivalent capacitance: \[\frac{1}{C_{eq}} = 0.3333 + 0.3333 + 0.3333 = 1\ \mu\text{F}^{-1}\] 3. Therefore, the equivalent capacitance \(C_{eq}\) is: \[C_{eq} = \frac{1}{1} = 1\ \mu\text{F}\]
Step 4: Conclusion
The equivalent capacitance of three capacitors each with a capacitance of \(3\ \mu\text{F}\) connected in series is \(1\ \mu\text{F}\). Final Answer: (A)
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