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 capacitors in the circuit, we need to determine their configuration (series or parallel) and apply the respective formulas.

For capacitors in series, the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances. The formula is given by:

\(\frac{1}{C_{\text{eq}}}=\frac{1}{C_1}+\frac{1}{C_2}+\cdots+\frac{1}{C_n}\)

For capacitors in parallel, the total capacitance is the sum of the individual capacitances:

\(C_{\text{eq}}=C_1+C_2+\cdots+C_n\)

Assuming the problem provides a figure showing capacitors in a certain configuration, we can calculate.

Step 1: Identify the configuration of the capacitors. If they are in series or parallel, apply the respective formula.

Step 2: Calculate based on assumed capacitance values from the figure:

• Assume two capacitors in series: \(C_1=4\mu F\) and \(C_2=4\mu F\).
• Equivalent series capacitance formula is applied: \(\frac{1}{C_{\text{series}}}=\frac{1}{4}+\frac{1}{4}\)
• Calculate: \(\frac{1}{C_{\text{series}}}=\frac{1}{2}\) or \(C_{\text{series}}=2\mu F\)

Therefore, the equivalent capacitance of the system is:

\(2\mu F\)