Question

The equivalent capacitance of the system shown in the following circuit is:

• A. \(9\mu F\)
• B. \(2\mu F\)
• C. \(3\mu F\)
• D. \(6\mu F\)

Answer 1

The Correct Option is B

To find the equivalent capacitance of the given circuit, we need to analyze the configuration of the capacitors. The circuit consists of three capacitors, each with a capacitance of \(3 \mu F\).

Here's the circuit diagram provided:

1. We start by noting the configuration:
   • The two capacitors at the top and bottom are in parallel, as they are connected to the same two nodes. This means their capacitances add up:
   • \(C_{\text{parallel}} = 3 \mu F + 3 \mu F = 6 \mu F\)
2. Next, the equivalent capacitor from the first step (\(6 \mu F\)) is in series with the third capacitor (\(3 \mu F\)) on the left:
   • The formula for series capacitance is given by:
   • \(\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2}\)
   • Substituting the values:
   • \(\frac{1}{C_{\text{series}}} = \frac{1}{6 \mu F} + \frac{1}{3 \mu F}\)
   • Calculating further:
   • \(\frac{1}{C_{\text{series}}} = \frac{1 + 2}{6} = \frac{3}{6} = \frac{1}{2}\)
   • Therefore, \(C_{\text{series}} = 2 \mu F\)

Thus, the equivalent capacitance of the system is \(2 \mu F\).

The correct answer is \(2 \mu F\).