Question

The effective capacitance of the network between points A and B shown in the figure is:

• A. $\frac{4}{3}$ $\mu$F
• B. 2 $\mu$F
• C. $\frac{3}{2}$ $\mu$F
• D. 6 $\mu$F

Hint:

When dealing with combinations of capacitors, always simplify the series and parallel combinations step by step to find the total capacitance.

Answer 1

The Correct Option is D

The given network of capacitors consists of series and parallel combinations. Let's break down the calculation
Step by
Step: 1. Capacitors \(6 \, \mu F\) and \(12 \, \mu F\) are in parallel between points C and D. The equivalent capacitance for parallel capacitors is given by: \[ C_{1} = 6 + 12 = 18 \, \mu F \] 2. The \(18 \, \mu F\) capacitor is in series with the \(1 \, \mu F\) capacitor. The total capacitance for series capacitors is given by: \[ \frac{1}{C_{2}} = \frac{1}{18} + \frac{1}{1} = \frac{1}{18} + 1 = \frac{19}{18} \] \[ C_{2} = \frac{18}{19} \, \mu F \] 3. Next, the \(3 \, \mu F\) capacitor is in parallel with \(C_{2}\). The total capacitance for parallel capacitors is: \[ C_{3} = 3 + \frac{18}{19} = \frac{57}{19} \, \mu F \] 4. Finally, the equivalent capacitor \(C_{3}\) is in parallel with the remaining \(6 \, \mu F\) capacitor. The total capacitance for these parallel capacitors is: \[ C_{eff} = 6 + \frac{57}{19} = \frac{114}{19} \approx 6 \, \mu F \] Thus, the effective capacitance of the network is approximately \(6 \, \mu F\).