In the given figure all the capacitors \(\mathrm{C}_1, \mathrm{C}_2, \mathrm{C}_3\) and \(\mathrm{C}_4\) have the same capacitance \(4 \mu \mathrm{F}\). The emf of the cell is 12 V. What is the effective capacitance ?

The circuit consists of three parallel branches connected across the 12 V source. Let the left node be A and the right node be B.

Branch 1 consists of capacitors $C_1$ and $C_2$ in series. Their equivalent capacitance $C_{12}$ is given by $\dfrac{1}{C_{12}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{2}{4} = \dfrac{1}{2}$. Thus, $C_{12} = 2 \mu F$.

Branch 2 consists of capacitor $C_3$. Thus, $C_{branch2} = C_3 = 4 \mu F$.

Branch 3 consists of capacitor $C_4$. Thus, $C_{branch3} = C_4 = 4 \mu F$.

Since these three branches are connected in parallel between nodes A and B, the effective capacitance $C_{eff}$ is the sum of the capacitances of the individual branches:

$C_{eff} = C_{12} + C_{branch2} + C_{branch3} = 2 \mu F + 4 \mu F + 4 \mu F = 10 \mu F$.

Answer: \(C_{\text {eff }}=10 \mu \mathrm{~F}\)