Question

A 5mF capacitor is charged to 100 V supply and supply is disconnected. now the capacitor is connected to a 3mf and charged capacitor. how much energy of the first capacitor get lost in attaining steady state ?​

Answer 1

Answer:

The charged capacitor has a charge of 500mC. A charge X mC comes to new capacitor.

Then

$\frac{500 - x}{5} = 5x \\ = > 1500 - 3x = 5x \\ = > 8x = 1500 \\ = > x = \frac{1500}{8} mc$

U1 = stored energy in 5mF before the connection

$\frac{1}{2} \times 5 \times {10}^{4} mj = > 2.5 \times {10}^{4} mj$

U2 = stored energy in 5mF after the connection

$( { \frac{25}{8}) }^{2} ( \frac{ {10}^{4} }{2 \times 5} )mj = > 0.98 \times {10}^{4} mj$

Now Loss energy = U1 - U2

$= (2.5 - 0.98) \times {10}^{4} mj \\ = 1.52 \times {10}^{4} mj = 15.2j$