We are given a combination of capacitors, and we need to calculate the effective capacitance between points A and B. Let's solve step by step. 1. First, consider the two capacitors of \(3 \, \mu\)F each in series: \[ \frac{1}{C_1} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \quad \Rightarrow \quad C_1 = \frac{3}{2} \, \mu\text{F} \] 2. Now, this result is in parallel with the \(2 \, \mu\)F capacitor: \[ C_2 = \frac{3}{2} \, \mu\text{F} + 2 \, \mu\text{F} = \frac{7}{2} \, \mu\text{F} \] 3. Next, the two \(3 \, \mu\)F capacitors in series again: \[ \frac{1}{C_3} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \quad \Rightarrow \quad C_3 = \frac{3}{2} \, \mu\text{F} \] 4. Finally, the result \(C_2\) is in series with \(C_3\): \[ \frac{1}{C_{\text{eff}}} = \frac{1}{\frac{7}{2}} + \frac{1}{\frac{3}{2}} = \frac{2}{7} + \frac{2}{3} = \frac{6 + 14}{21} = \frac{20}{21} \quad \Rightarrow \quad C_{\text{eff}} = \frac{21}{20} = 1 \, \mu\text{F} \] Thus, the effective capacitance between A and B is \(1 \, \mu\text{F}\).
The correct option is (B) : \(1\ µF\)
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Let's simplify step-by-step clearly:
Step 1: Observe the given circuit. There are two capacitors (each 1 𝜇 𝐹 1μF) connected in series on both upper and lower paths.
Step 2: Simplify the two series combinations first. Two capacitors ( 1 𝜇 𝐹 1μF each) in series have equivalent capacitance: \frac{1}{C_{\text{eq}}} = \frac{1}{1} + \frac{1}{1} = 2 \quad\Rightarrow\quad C_{\text{eq}} = \frac{1}{2} = 0.5\,\mu F \] Thus, both upper and lower branches become \(0.5\,\mu F\).
Step 3: Combine the two resultant capacitors in parallel. Parallel capacitors simply add up: \[ C_{\text{total}} = 0.5\,\mu F + 0.5\,\mu F = 1\,\mu F \] Thus, the effective capacitance between A and B is: 1 µF
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