Question

two capacitance of 5 micro farade and 7micro Farande are connected in series find their equivalent capacitance ​

Answer 1

$\huge\underline{\underline{\bf \green{Question-}}}$

Two capacitance of 5 micro farade and 7micro Farande are connected in series find their equivalent capacitance.

$\huge\underline{\underline{\bf \green{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• ${\sf C_1}$ = 5μF
• ${\sf C_2}$ = 7 μF

$\large\underline{\underline{\sf To\:Find:}}$

• Equivalent Capacitance ${\sf (C_{eq})}$

✪$\large{\boxed{\bf \blue{\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}} }}$

$\implies{\sf \dfrac{1}{5}+\dfrac{1}{7} }$

$\implies{\sf \dfrac{7+5}{5×7} }$

$\implies{\sf \dfrac{12}{35}}$

$\implies{\sf C_{eq}=\dfrac{35}{12}}$

$\implies{\bf \red{C_{eq}=2.91\:\mu F}}$

$\huge\underline{\underline{\bf \green{Answer-}}}$

Equilateral Capacitance is ${\bf \red{2.91\:\mu F}}$

Answer 2

Given ,

The two capacitors i.e 5 micro f and 7 micro f are connected in series arrangement

We know that ,

The equivalent capacitance in series arrangement is given by

$\large \mathtt{\fbox{ \frac{1}{C} = \frac{1}{ C_{1} } + \frac{1}{ C_{2} } + ... + \frac{1}{ C_{n} }}}$

Thus ,

1/C = 1/5 + 1/7

1/C = (7 + 5)/35

1/C = 12/35

C = 35/12

C = 2.91 micro f

Hence , the equivalent capacitance is 2.91 micro f

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