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 ?

Question

Question

Answer 1

CONCEPT:

The energy stored in the capacitor is written as;

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

CALCULATION:

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

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

Therefore,

⇒C = 1/2

The energy in the b and c is written as;

Now, percentage loss is written as;

% =

⇒ % = 50 %

Hence option 1) is the correct answer.

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