Question

three capacitors of capacitance 1 microfarad are connected in parallel to this combination another capacitor of 1 micro farad is connected in series then find the net capacitance ​

Answer 1

★ Given :-

• C 1 = 1 μ F
• C 2= 1 μ F
• C 3 = 1 μ F
• C 4 = 1 μ F

Let , C 1 = C 2 = C 3 = C 4 = C

★ To find :-

• Net capacitance

★ Solution :-

Three capacitors of equal capacitance are connected in parallel to each other

Equivalent capacitance of capacitors connected in parallel is given by the formula ,

$\implies\boxed{\mathtt{ C_p = C_1 + C_2 + C_3 +...C_n}}$

Hence ,

$\implies\mathtt{ C_p = C_1 + C_2+ C_3 }$

$\implies\mathtt{ C_p = C + C+ C }$

$\implies\mathtt{ C_p =3 C}$

$\implies\mathtt{ C_p =3 \times 1}$

$\implies\mathtt{ C_p =3 \mu \: F}$

Now C p and C 4 are connected in Series with each other ,

Equivalent capacitance of capacitors connected in series is given by the formula ,

$\implies\boxed{\mathtt{ \dfrac{1}{C_s}= \dfrac{1}{C_1 }+\dfrac{1}{ C_2 }+\dfrac{1}{ C_3} +...\dfrac{1}{C_n}}}$

hence ,

$\implies\mathtt{ \dfrac{1}{C_s}= \dfrac{1}{C_1 }+\dfrac{1}{ C_2 }}$

$\implies\mathtt{ \dfrac{1}{C_s}= \dfrac{1}{C}+\dfrac{1}{ C_p }}$

$\implies\mathtt{ \dfrac{1}{C_s}= \dfrac{1}{1}+\dfrac{1}{3}}$

$\implies\mathtt{ \dfrac{1}{C_s}= \dfrac{3 + 1}{3}}$

$\implies\mathtt{ \dfrac{1}{C_s}= \dfrac{4}{3}}$

$\implies\mathtt{ \boxed{ C_s= \dfrac{3}{4 } \mu \: F}}$

The net capacitance is 3/4 μ F