A capacitor of capacitance 900 µ F is charged by a 100 V battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as x × 10 - 2 J . The value o

Step-by-step solution

When a charged capacitor is connected to an uncharged capacitor, then there is loss of energy in the form of heat as charges flow from higher potential to lower potential. So, heat ∆ H = 1 2 C 1 C 2 C 1 + C 2   V 1 - V 2 2 As the other capacitor is identical therefore charge is equally divided. = 1 2 C 2 2 C ( 100 - 0 ) 2 = 1 2 900 × 10 - 6 2 × 10 4 = 9 4 Joule = 2 . 25 Joule. Hence, the value of x = 225 .