Question

please solved this.......... please don't spamm ... Do fast it's urgent​

Answer 1

Didi Bhut Mehnat Lagi Hai Thodi Glti Bhi Hogi Thoda Manage Kr Lena

Given,

capacitance of the capacitor-C= 5 µF potential applied across the capacitance-V=

100V

.. Initial energy stored in capacitor,

1 1 2 cv2

1 2 = 10-6x (100)? ×5×10

= 2.5 x 10-2J.

Charge on 5 uF capacitor-q=CV

9- = 5 x 10-6 × 100

= 5 x 10-4c.

When, supply is disconnected and, charged capacitor is connected to another uncharged capacitor of 3 uF then, the two capacitors attain

a common potential V.=

total charge total capacitor

5x 10-4

(5+3) x 10-6 V.

125

2

Energy stored in the capacitor, after combination

U, = C,v,² 1

1 1 = X (5+3) x 10-6 x

2

= 1.56 X 10-2J

Therefore,

Electrostatic energy lost in the process of attaining the steady state = U; - UF

= (2.5 – 1.56) x 10-2 = 0.94 x 10-2 J.

125

2

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