Let C_1 = 4.0 and C_2 = 6.0 in the figure. Determine the equivalent...

In circuit (a), the arrangement consists of two parallel groups connected in series.

The first parallel group has $C_1$ and $C_2$, giving an equivalent capacitance $C_{p1} = 4.0 + 6.0 = 10.0 \mu\text{F}$.

The second parallel group also has $C_1$ and $C_2$, giving $C_{p2} = 4.0 + 6.0 = 10.0 \mu\text{F}$.

The total equivalent capacitance for (a) is the series combination of $C_{p1}$ and $C_{p2}$:

$C_a = \dfrac{10.0 \times 10.0}{10.0 + 10.0} = 5.0 \mu\text{F}$

In circuit (b), the arrangement consists of two branches connected in parallel across the main terminals. Each branch is identical to circuit (a).

The equivalent capacitance of the top branch is $5.0 \mu\text{F}$ and the bottom branch is $5.0 \mu\text{F}$.

The total equivalent capacitance for (b) is the parallel combination of these two branches:

$C_b = 5.0 + 5.0 = 10.0 \mu\text{F}$

Answer: (a) $5.0 \mu\text{F}$, (b) $10.0 \mu\text{F}$