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: Circuit Configuration Analysis
The circuit contains capacitors in series and parallel configurations, requiring sequential simplification.
Step 2: Equivalent Capacitance Calculation (Series)
Capacitors \( C_1 = C \) and \( C_2 = \frac{7}{3} \mu F \) are connected in series. The equivalent capacitance is calculated using the formula: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \] Substitution of values yields: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C} + \frac{3}{7} \] The equivalent capacitance is: \[ C_{\text{eq}} = \frac{7C}{7 + 3C} \]
Step 3: Total Capacitance Calculation
This equivalent capacitance is in parallel with a \( \frac{4}{3} \) μF capacitor. The total capacitance is determined by: \[ C_{\text{total}} = \frac{7C}{7 + 3C} + \frac{4}{3} \] Given \( C_{\text{total}} = 0.5 \, \mu F \), the equation is: \[ \frac{7C}{7 + 3C} + \frac{4}{3} = 0.5 \]
Step 4: Solving for \( C \)
The equation is solved as follows: \[ \frac{7C}{7 + 3C} = 0.5 - \frac{4}{3} \] \[ \frac{7C}{7 + 3C} = \frac{3}{6} - \frac{8}{6} = -\frac{5}{6} \] \[ 7C = -\frac{5}{6} (7 + 3C) \] The solution for \( C \) is: \[ C = \frac{7}{11} \, \mu F \]Final Answer: The value of \( C \) is \( \frac{7}{11} \, \mu F \).