Question

find the total energy stored in the capacitors in the given network

Answer 1

Hope this help you........

Answer 2

Answer:

The total energy stored in the capacitors is $3.9\times10^{-5}\ J$.

Explanation:

According to figure,

Voltage V = 6 V

Capacitor $C_{1}= 1 \mu f$

Capacitor $C_{2}= 1 \mu f$

Capacitor $C_{3}= 2 \mu f$

Capacitor $C_{4}= 2 \mu f$

Capacitor $C_{5}= 2 \mu f$

We need to calculate the equivalent capacitance

Firstly, C₂ and C₃ are connected in series

The equivalent capacitance is

$\dfrac{1}{C'}=\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}}$

$\dfrac{1}{C'}=\dfrac{1}{1}+\dfrac{1}{2}$

$C'=\dfrac{2}{3}$

Now, C' and C₅ are connected in parallel

The equivalent capacitance is

$C''=C'+C_{5}$

$C''=\dfrac{2}{3}+2$

$C''=\dfrac{8}{3}$

Now, C'' and C₄ are connected in series

The equivalent capacitance is

$\dfrac{1}{C'''}=\dfrac{3}{8}+\dfrac{1}{2}$

$C'''=\dfrac{8}{7}$

Now, C''' and C₁ are connected in parallel

$C= \dfrac{8}{7}+1$

$C=\dfrac{15}{7}\ \mu f$

Formula of the energy is defined as:

$E = \dfrac{1}{2}CV^2$

$E = \dfrac{1}{2}\times\dfrac{15}{7}\times10^{-6}\times(6)^2$

$E = 3.9\times10^{-5}\ J$

Hence, The total energy stored in the capacitors is $3.9\times10^{-5}\ J$.