Question

Two identical capacitors C. and C, of equal
capacitance are connected as shown in the
circuit. Terminals a and b of the key k are
connected to charge capacitor C, using battery of
emf V volt. Now disconnecting a and b the
terminals b and c are connected. Due to this,
what will be the percentage loss of energy​

Answer 1

Answer:

Ans is 50%

Explanation:

The concept of sharing of charge and common potential is used....................

Answer 2

Given that,

Two identical capacitors C₁ and C₂ of equal  capacitance.

Emf = V

When terminal a and b are connected then,

The energy is

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

When terminal b and c are connected then C₁ and C₂ in serise

The net capacitor is

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

The energy is

$\Delta E=\dfrac{1}{2}\times\dfrac{C_{1}C_{2}}{C_{1}+C_{2}}\times V^2$

$C_{1}=C_{2}=C$

Put the value into the formula

$\Delta E=\dfrac{1}{2}\times\dfrac{C^2}{2C}\times V^2$

$\Delta E=\dfrac{1}{4}CV^2$

We need to calculate the percentage loss of energy​

Using formula of energy loss

$\text{ percentage loss of energy​}=\dfrac{\Delta E}{E}\times100$

Put the value into the formula

$\text{ percentage loss of energy​}=\dfrac{\dfrac{1}{4}CV^2}{\dfrac{1}{2}CV^2}\times100$

$\text{ percentage loss of energy​}=50\%$

Hence, The percentage loss of energy​ is 50%.