An equivalent capacitor of 10 that can be connected across a 200 V...

Let $n$ be the number of capacitors connected in series in each branch.

The voltage across each capacitor in the series combination is $\dfrac{V}{n}$.

Since the total voltage is $200 \text{ V}$ and each capacitor can withstand a maximum of $50 \text{ V}$, we have:

$\dfrac{200}{n} \le 50 \Rightarrow n \ge 4$

Taking the minimum number of capacitors in series, $n = 4$.

The equivalent capacitance of one such branch containing 4 capacitors in series is:

$C_s = \dfrac{C}{n} = \dfrac{10}{4} = 2.5 \mu\text{F}$

Let $m$ be the number of such branches connected in parallel to achieve the desired total capacitance.

The total equivalent capacitance is:

$C_{eq} = m \times C_s$

Given that the required equivalent capacitance is $10 \mu\text{F}$, we get:

$m \times 2.5 = 10 \Rightarrow m = 4$

Thus, 4 parallel branches, each containing 4 capacitors in series, are required.