The given network of capacitors involves series and parallel combinations. The calculation proceeds as follows:1. Capacitors of \(6 \, \mu F\) and \(12 \, \mu F\) are connected in parallel between points C and D. The equivalent capacitance for parallel combinations is calculated as: \[ C_{1} = 6 + 12 = 18 \, \mu F \]2. The \(18 \, \mu F\) capacitor is in series with the \(1 \, \mu F\) capacitor. The total capacitance for series capacitors is determined 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. Subsequently, the \(3 \, \mu F\) capacitor is in parallel with \(C_{2}\). The total capacitance for this parallel combination 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: \[ C_{eff} = 6 + \frac{57}{19} = \frac{114}{19} \approx 6 \, \mu F \]Therefore, the effective capacitance of the network is approximately \(6 \, \mu F\).
