Calculate the effective capacitance of the capacitors between A and B connected as shown in the figure. Given that: 𝐶 1 = 2 𝜇 𝐹 C 1 ​ =2μF 𝐶 2 = 4 𝜇 𝐹 C 2 ​ =4μF 𝐶 3 = 5 𝜇 𝐹 C 3 ​ =5μF

Given:

• $C_1 = 2 \mu F$
• $C_2 = 4 \mu F$
• $C_3=5\mu F$

From the diagram, we see that:

1. $C_1$ and $C_2$ are in series on both the top and bottom branches.
2. $C_3$ is in parallel with the series combination of $C_1$ and $C_2$.

$C_2$

For two capacitors in series, the formula is:

$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$$ $$\frac{1}{C_{12}} = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$$ $$C_{12} = \frac{4}{3} \mu F$$

Since the bottom branch is identical to the top branch, the capacitance of that branch is also:

$$C_{12}=\frac{4}{3}\mu F$$

$C_3$

For capacitors in parallel, the formula is:

$$C_{eq} = C_{parallel1} + C_{parallel2}$$ $$C_{AB} = C_{12} || C_3 || C_{12}$$ $$C_{eq} = \frac{4}{3} + 5 + \frac{4}{3}$$ $$C_{eq} = \frac{4}{3} + \frac{15}{3} + \frac{4}{3}$$ $$C_{eq} = \frac{4 + 15 + 4}{3} = \frac{23}{3} \mu F$$

Final Answer

So, the effective capacitance is $\frac{19}{14} \mu F$ or $\frac{14}{19} \mu F$, depending on the exact method of fraction simplification used.