Question

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 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.

In 1st case its across C1 only one capacitor.

In 2nd case its across C1 & C2 where sharing of charge takes place.

Answer 2

The energy stored in a capacitor is in the form of  Electrostatic potential energy (U) and is given by :

           $U = \frac{1}{2} CV^{2}$

Where , C = Capacitance of the given capacitor

              V = potential difference across the capacitor.

From the given diagram ,

  $U_{i} = \frac{1}{2} CV^{2}$         (initial)

Loss of energy = $\frac{1}{2} \frac{C.C}{(C+C)} [V-0]^{2}$

                         = $\frac{1}{4} CV^{2}$

Therefore, the percentage loss of energy = $\frac{\frac{1}{4}CV^{2} }{\frac{1}{2} CV^{2} }$ ×100

                                                                     = $\frac{1}{2}$ × 100

                                                                     = 50%

Therefore, the percentage loss of energy is 50%.

#SPJ3