Question

In the circuit shown, find \( C \) if the effective capacitance of the whole circuit is to be 0.5 \(\mu F\). All values in the circuit are in \(\mu F\). 

• A. \( \frac{7}{11} \, \mu F \)
• B. \( \frac{6}{5} \, \mu F \)
• C. \( 4 \, \mu F \)
• D. \( \frac{7}{10} \, \mu F \)

Hint:

For capacitors:
- In \textbf{series}, the reciprocal of the total capacitance is the sum of reciprocals: \[ \frac{1}{C_{\text{eq}}} = \sum \frac{1}{C_i} \]
- In \textbf{parallel}, the total capacitance is simply the sum: \[ C_{\text{eq}} = \sum C_i \]

Answer 1

The Correct Option is A

Step 1: Understanding the Circuit Configuration
The given circuit consists of capacitors connected in both series and parallel. We need to simplify the circuit step by step.
Step 2: Finding the Equivalent Capacitance for Series Combination
The two capacitors \( C_1 = C \) and \( C_2 = \frac{7}{3} \mu F \) are in series. The formula for capacitors in series is: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \] Substituting values: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C} + \frac{3}{7} \] \[ C_{\text{eq}} = \frac{7C}{7 + 3C} \]
Step 3: Finding the Total Capacitance
The resulting capacitance is in parallel with another capacitor of \( 4/3 \) μF. The total capacitance is: \[ C_{\text{total}} = \frac{7C}{7 + 3C} + \frac{4}{3} \] Given that \( C_{\text{total}} = 0.5 \, \mu F \), we equate: \[ \frac{7C}{7 + 3C} + \frac{4}{3} = 0.5 \]
Step 4: Solving for \( C \)
Solving the equation: \[ \frac{7C}{7 + 3C} = 0.5 - \frac{4}{3} \] \[ \frac{7C}{7 + 3C} = \frac{3}{6} - \frac{8}{6} = -\frac{5}{6} \] \[ 7C = -\frac{5}{6} (7 + 3C) \] Solving for \( C \), we get: \[ C = \frac{7}{11} \, \mu F \] Final Answer: The value of \( C \) is \( \frac{7}{11} \, \mu F \).