Question

capacitance of a capacitor increase 4 times then its resonance frequency is?​

Answer 1

Answer:

Let us learn about cross multiplication formula first:

We consider two linear equations

$\quad\quad \mathsf{a_{1}x+b_{1}y+c_{1}=0}$

$\quad\quad \mathsf{a_{2}x+b_{2}y+c_{2}=0}$

Using the formula for cross multiplication, we get

$\quad\mathsf{\dfrac{x}{b_{1}c_{2}-b_{2}c_{1}}=\dfrac{y}{c_{1}a_{2}-c_{2}a_{1}}=\dfrac{1}{a_{1}b_{2}-a_{2}b_{1}}}$

Now we proceed to solve the given problem:

The given equations are

$\quad\quad \mathsf{\dfrac{15}{x}+\dfrac{2}{y}=17}$

$\quad\quad \mathsf{\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{36}{5}}$

Then the given equations become

$\quad\quad \mathsf{15p+2q=17}$15p+2q=17

$\quad\quad \mathsf{p+q=\frac{36}{5}}$

$\quad\quad \mathsf{15p+2q-17=0}15p+2q−17=0$

$\quad\quad \mathsf{5p+5q-36=0}5p+5q−36=0$

Now comparing these equations with the general equations, we have

$\quad\quad \mathsf{a_{1}=15,\:b_{1}=2,\:c_{1}=-17}$

$\quad\quad \mathsf{a_{2}=5,\:b_{2}=5,\:c_{2}=-36}$

Using the formula for cross multiplication, we get

$\quad\mathsf{\dfrac{p}{b_{1}c_{2}-b_{2}c_{1}}=\dfrac{q}{c_{1}a_{2}-c_{2}a_{1}}=\dfrac{1}{a_{1}b_{2}-a_{2}b_{1}}}$

$\to \mathsf{\dfrac{p}{2\times (-36)-5\times (-17)}=\dfrac{q}{(-17)\times 5-(-36)\times 15}=\dfrac{1}{15\times 5-5\times 2}}$

$\to \mathsf{\dfrac{p}{-72+85}=\dfrac{q}{-85+540}=\dfrac{1}{75-10}}$

$\to \mathsf{\dfrac{p}{13}=\dfrac{q}{455}=\dfrac{1}{65}}$

$\quad\quad \mathsf{\dfrac{p}{13}=\dfrac{1}{65}}$

$\implies \mathsf{p=\dfrac{13}{65}}$⟹

$\implies \mathsf{p=\dfrac{1}{5}}$⟹p

$\implies \mathsf{\dfrac{1}{x}=\dfrac{1}{5}\quad [\because p=\dfrac{1}{x}]}$

$\implies \mathsf{\dfrac{1}{y}=7\quad [\because q=\dfrac{1}{y}]}$

Therefore the required solution is

$\quad\quad \underline{\boxed{\bold{x=5,\:y=\dfrac{1}{7}}}}$

Answer 2

Answer:

I hope this will help you

Explanation:

Be brainly