Question

four capacitor of equal capacitance when connected in series the net capacity cs and when connected in parallel have a net capacity CP what will be the value of CS in CP

Answer 1

The value of Cs is
C/4

The value of Cp is C+C+C+C =4C

C=Cp/4

so Cs =4/(Cp/4)
Cs=16/Cp

Answer 2

Answer:

The value of $C_{S}$ is $\dfrac{C_{P}}{16}$

Explanation:

Given that,

Four capacitors of equal capacitance.

When the capacitors are connected in parallel then the net capacity is

$C_{P}=C_{1}+C_{2}+C_{3}+C_{4}$

$C_{P}=C+C+C+C=4C$

When the capacitors are connected in series then the net capacity is

$\dfrac{1}{C_{S}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}}+\dfrac{1}{C_{4}}$

$\dfrac{1}{C_{S}}=\dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C}$

$\dfrac{1}{C_{S}}=\dfr{4}{C}$

$C_{S}=\dfrac{C}{4}$

The ratio of $C_{S}$ and $C_{P}$ is

$\dfrac{C_{P}}{C_{S}}=\dfrac{4C}{\dfrac{C}{4}}$

$\dfrac{C_{P}}{C_{S}}=16$

$C_{S}=\dfrac{C_{P}}{16}$

Hence, The value of $C_{S}$ is $\dfrac{C_{P}}{16}$