Question

two capiciters are connected in series and parallel respectively then the resulted capacities 2 and 9 what is the capacities?​

Answer 1

Answer:

The capacitors are 6F and 3F.

Explanation:

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When we connect capacitors parallel

Ceq = C1 + C2

Here it is given as 9

That is C1 + C2 = 9

so take C2 = 9-C1

Then if we connect capacitors in series

1/Ceq = 1/C1 + 1/C2

It is given as 2

So 1/2 = 1/C1 + 1/(9-C1)

Solving this will give C1 = 6

Therefore C1 = 6 and C2 = 3

Answer 2

$\huge\sf\bold\red{\underline{\underline{ANSWER:-}}}$

Let -

$\: \: \: \: \:$$\sf{Capicity \: of \: first \: capacitor = C1}$

$\:$$\sf{Capicity \: of \: second \: capacitor = C2}$

When C1 and C2 are connected in series combination , then -

$\: \:$$\sf{Resulted \: Capicity = 2}$

$\: \: \: \:$$\sf{\dfrac{1}{C1} + \dfrac{1}{C2} =\dfrac{1}{2}}$

$\: \:$$\sf{\dfrac{(C1 + C2)}{C1 C2} =\dfrac{1}{2}}$

$\: \:$$\sf{2(C1 + C2) = C1 C2}$

$\: \: \:$$\sf{(C1 + C2) =\dfrac{1}{2}C1 C2 ---(1)}$

When C1 and C2 are connected in parallel combination , then -

$\: \:$$\sf{Resulted \: Capicity = 9}$

$\: \:$$\sf{(C1 + C2) = 9 ---(2)}$

$\: \: \:$FROM EQ. (1) & EQ. (2) -

$\: \: \: \: \:$$\sf{\dfrac{1}{2}C1 C2 = 9}$

$\: \: \: \: \: \: \: \:$$\sf{C1 C2 = 18}$

$\: \: \: \: \:$$\sf{4C1 C2 = 72}$

Now ,

$\: \: \:$$\sf{(C1 - C2) =\sqrt{(C1 + C2)^{2} - 4C1 C2}}$

$\: \: \:$$\sf{(C1 - C2) =\sqrt{(9)^{2} - 72}}$

$\: \: \:$$\sf{(C1 - C2) =\sqrt{81 - 72}=\sqrt{9}}$

$\: \: \:$$\sf{(C1 - C2) = 3 ----(3)}$

FROM Eq. (2) + Eq. (3) -

$\: \: \: \: \: \: \:$$\sf{2C1 = 12}$

$\: \: \: \: \: \: \:$$\sf{C1 = 6}$

then -

$\: \: \: \: \: \: \:$$\sf{C2 = 3}$

The Capicity of first capacitor is 6 and the capacity of second capacitor is 3 .