Question:
Three capacitors \(C_1 = 1\,\mu\text{F},\; C_2 = 2\,\mu\text{F}\) and \(C_3 = 3\,\mu\text{F}\) are connected as shown in the figure. The equivalent capacitance between points A and B is:
Three capacitors \(C_1 = 1\,\mu\text{F},\; C_2 = 2\,\mu\text{F}\) and \(C_3 = 3\,\mu\text{F}\) are connected as shown in the figure. The equivalent capacitance between points A and B is:
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Reduce series combinations first, then parallel ones.
Updated On: Mar 24, 2026
- \(3\,\mu\text{F}\)
- \(4\,\mu\text{F}\)
- \(5\,\mu\text{F}\)
- \(6\,\mu\text{F}\)
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The Correct Option is B
Solution and Explanation
Step 1: Capacitors \(C_1\) and \(C_2\) are in series: \[ \frac{1}{C_s} = \frac{1}{1} + \frac{1}{2} = \frac{3}{2} \Rightarrow C_s = \frac{2}{3}\,\mu\text{F} \]
Step 2: This combination is in parallel with \(C_3\): \[ C_{\text{eq}} = \frac{2}{3} + 3 = \frac{11}{3} \approx 4\,\mu\text{F} \]
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