Question

A capacitor of capacitance 2mF is charged to a potential of 10V. The capacitor is now connected to an identical capacitor charged to a voltage 20V such that the plates of same polarity are connected together. The loss of energy in redistribution of the charges is

1) 2.8J
2) 0.5J
3) 0.05J
4) 6.6J​

Answer 1

Answer:

0.05

Have a gud day!

Answer 2

Given:

C₁ = 2mF

V₁ = 10V

C₂ = 2mF

V₂ = 20V

To find:

During redistribution of the charge, the loss of energy.

Solution:

It is given, Capacitance of capacitor 1, C₁ = 2mF

                  Potential around capacitor 1, V₁ = 10V

                 Capacitance of capacitor 2, C₂ = 2mF

                 Potential around capacitor 2, V₂ = 20V

and,  as we know, By law of conservation of energy we get,

  $\frac{1}{2}C_1V_1^2 + \frac{1}{2}C_2V_2^2 = \frac{1}{2}C_1V^2 + \frac{1}{2}C_2V^2 + H$

here, H = loss of energy

now putting all the values,

  $\frac{1}{2}2(10)^2 + \frac{1}{2}2(20)^2 = \frac{1}{2}2(10)^2 + \frac{1}{2}2(10)^2 + H$

  (10)² + (20)² = (10)² + (10)² + H

  100 + 400 =  100 + 100 + H

     500 = 200 + H

    H = 500 - 200

    H = 300μJ

So, energy loss in redistribution of the charges is 300μJ.