Two identical capacitors C₁ and C₂ of equal capacitance are connected as shown in the circuit. Terminals a and b of the key k are connected to charge capacitor C₁ 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 ?

CONCEPT:

The energy stored in the capacitor is written as;

\(E = \frac{1}{2} CV^2\)

Here we have C as the capacitance, and V is the potential.

CALCULATION:

When we connect the a and b poles we have energy,

\(E_1 = \frac{1}{2} CV^2\)

and when we connected the poles b and c we see that capacitors C₁,C2  are in series, therefore the capacitance is written as;

\(\frac {1}{C}= \frac{1}{C_1}+ \frac{1}{C_2}\)

Therefore, \(C = \frac{ C}{2C}\)

⇒C = 1/2

The energy in the b and c is written as;

\(E_2 = \frac{1}{4} CV^2\)

Now, percentage loss is written as;

\(\frac{E_2}{E_1} \)% = \(\frac{\frac{1}{4} CV^2}{\frac{1}{2} CV^2}\times 100\)

⇒ \(\frac{E_2}{E_1}\)% = 50 %

Hence option 1) is the correct answer.