Question

The equivalent capacitance of the following assembly of capacitors is .......... \( \mu F \). 

Hint:

When calculating the total capacitance in a circuit with both series and parallel capacitors, first simplify each section (series and parallel) step-by-step before combining them.

Answer 1

Correct Answer: 0.99 - 1.01

Step 1: Understanding the problem. 
We are given a circuit with capacitors in series and parallel, and we are tasked with finding the equivalent capacitance. The given circuit configuration is: 

- Two 2 \( \mu F \) capacitors in parallel. 

- A 4 \( \mu F \) capacitor in series with the parallel combination. 

- Two 1 \( \mu F \) capacitors in parallel. 

- Another 4 \( \mu F \) capacitor in series with the combination of the 1 \( \mu F \) capacitors.

Step 2: Simplifying the parallel capacitors. 
First, simplify the parallel capacitors: - For the two 2 \( \mu F \) capacitors in parallel, the total capacitance is: \[ C_{\text{parallel 1}} = 2 + 2 = 4 \, \mu F \] - For the two 1 \( \mu F \) capacitors in parallel, the total capacitance is: \[ C_{\text{parallel 2}} = 1 + 1 = 2 \, \mu F \]

Step 3: Simplifying the series capacitors. 
Now, the equivalent capacitance of the two capacitors in series is calculated using the formula: \[ \frac{1}{C_{\text{total series}}} = \frac{1}{C_1} + \frac{1}{C_2} \] For the parallel combination of 4 \( \mu F \) with 4 \( \mu F \): \[ \frac{1}{C_{\text{series 1}}} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \] Thus, \( C_{\text{series 1}} = 2 \, \mu F \).

Step 4: Total equivalent capacitance. 
Now, combine the total capacitances: - The series combination of 2 \( \mu F \) with the parallel combination of 2 \( \mu F \): \[ \frac{1}{C_{\text{total}}} = \frac{1}{2} + \frac{1}{2} = 1 \] So, \( C_{\text{total}} = 1 \, \mu F \).

Step 5: Conclusion. 
The equivalent capacitance of the circuit is 1 \( \mu F \), so the correct answer is 1 \( \mu F \).