Question

Three capacitors of capacitance 2uf are connected in series .find resultant capacitance.

Plz give answer quickly plz

Answer 1

GiveN :

• Three Capacitors each of capacitance $\sf{2 \mu F}$ are connected in series.

To FinD :

• Equivalent capacitance

SolutioN :

We've formula :

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$\implies \boxed{\boxed{\sf{\dfrac{1}{C_{eq}}\ =\ \sum^n _{i\ =\ 1} \: \dfrac{1}{C_i}}}} \\ \\ \\ \\ \implies \sf{\dfrac{1}{C_{eq}}\ =\ \dfrac{1}{C_1}\ +\ \dfrac{1}{C_2}\ +\ \dfrac{1}{C_3}} \\ \\ \\ \\ \implies \sf{\dfrac{1}{C_{eq}}\ =\ \dfrac{1}{2}\ +\ \dfrac{1}{2}\ +\ \dfrac{1}{2}} \\ \\ \\ \\ \implies \sf{\dfrac{1}{C_{eq}}\ =\ \dfrac{1\ +\ 1\ +\ 1}{2}} \\ \\ \\ \\ \implies \sf{\dfrac{1}{C_{eq}}\ =\ \dfrac{3}{2}} \\ \\ \\ \\ \implies {\underline{\boxed{\sf{C_{eq}\ =\ \bigg( \dfrac{2}{3} \bigg) \: \mu F}}}}$