Physics - Capacitor Question with Solution | TestHub

Let the capacity of each capacitor be $C$.

When three capacitors are connected in parallel, their equivalent capacity is $C_p=C+C+C=3C$.

When a fourth capacitor of capacity $C$ is connected in series with this combination, the resultant capacity $C_{eq}$ is given by:

$\frac{1}{C_{eq}}=\frac{1}{C_p}+\frac{1}{C}$

$\frac{1}{C_{eq}}=\frac{1}{3C}+\frac{1}{C}=\frac{4}{3C}$

$C_{eq}=\frac{3C}{4}$

Given that $C_{eq}=4.5\mu \mathrm{F}$, we have:

$\frac{3C}{4}=4.5$

$3C=18$

$C=6\mu \mathrm{F}$

Answer: $6\mu \mathrm{F}$