Question

Three capacitors of capacity 2uF, 3uF and 5uF are connected in series.
The equivalent capacity of the combination is....
a) 2uF
b) 3uF
c) 5 uF
d) 1 uF​

Answer 1

AnswEr :

• Three capacitors of capacitance 2μF, 3μF and 5μF are connected in series

Formula for Capacitors in series is :

$\dashrightarrow{\underline{ \boxed{\large\displaystyle \mathfrak{\dfrac{1}{C_{eq}}\ =\ \sum^n _{i\ \to\ 1} \dfrac{1}{C_i}}}}}$

Now, Putting values

$\longrightarrow \frak{\dfrac{1}{C_{eq}}\ =\ \dfrac{1}{C_1}\ +\ \dfrac{1}{C_2}\ +\ \dfrac{1}{C_3}} \\ \\ \\ \longrightarrow \frak{\dfrac{1}{C_{eq}}\ =\ \dfrac{1}{2}\ +\ \dfrac{1}{3}\ +\ \dfrac{1}{5}} \\ \\ \\ \longrightarrow \frak{\dfrac{1}{C_{eq}}\ =\ \dfrac{15\ +\ 10\ +\ 6}{30}} \\ \\ \\ \longrightarrow \frak{\dfrac{1}{C_{eq}}\ =\ \dfrac{31}{30}} \\ \\ \\ \longrightarrow \frak{C_{eq}\ =\ \dfrac{30}{31}}\ \mu \sf F$

Answer 2

GIVEN :-

• Capacitor , c₁ = 2µF.
• Capacitor , c₂ = 3µF.
• Capacitor , c₃ = 5µF.

TO FIND :-

• Equal capacitance of the combination.

SOLUTION :-

As we know that when the capacitors are connected in series combination the equivalent capacitance is given by,

$: \implies \displaystyle \sf \: \frac{1}{c_{eq}} = \frac{1}{c_1} + \frac{1}{c_2} + \frac{1}{c_3} + .... + \frac{1}{c_n}$

$: \implies \displaystyle \sf \: \frac{1}{c_{eq}} = \frac{1}{2 \mu F} + \frac{1}{3 \mu F } + \frac{1}{5 \mu F}$

$: \implies \displaystyle \sf \: \frac{1}{c_{eq}} = \frac{15}{30\mu F} + \frac{10}{30 \mu F } + \frac{6}{30 \mu F}$

$: \implies \displaystyle \sf \: \frac{1}{c_{eq}} = \frac{31}{30\mu F}$

$: \implies \displaystyle \sf \: c_{eq} = \frac{30\mu F}{31}$

$: \implies \displaystyle \sf \: c_{eq} =0.96\mu F$

$: \implies \underline{ \boxed{ \displaystyle \sf \: c_{eq} \approx 1\mu F}}$

$\therefore{ \underline{\sf Equal \: capacitance \: of \: the \: combination \: is \: d) \: 1 \mu F}}$