SATHEE: Electrostatics Question 109

Electrostatics Question 109

Question: A body of capacity $4 μ F$ is charged to $80 V$ and another body of capacity $6 μ F$ is charged to 30V. When they are connected the energy lost by $4 μ F$ capacitor is [EAMCET 2001]

Options:

A) 7.8 mJ

B) 4.6 mJ

C) 3.2 mJ

D) 2.5 mJ

Show Answer

Answer:

Correct Answer: A

Solution:

Initial energy of body of capacitance 4 mF is $U_{i} = \frac{1}{2} \times (4 \times 10^{- 6}) {(80)}^{2} = 0.0128 J$

Final potential on this body after connection is $V = \frac{4 \times 80 + 6 \times 30}{4 + 6} = 50 V .$

So final energy on it $U_{f} = \frac{1}{2} \times 4 \times 10^{- 6} {(50)}^{2} = 0.005 J$

Energy lost by this body $= U_{i} U_{f} = 7.8 m J$

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Electrostatics Question 109

Question: A body of capacity 4μF is charged to 80V and another body of capacity 6μF is charged to 30V. When they are connected the energy lost by 4μF capacitor is [EAMCET 2001]

Options:

Answer:

Solution:

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Question: A body of capacity $4 μ F$ is charged to $80 V$ and another body of capacity $6 μ F$ is charged to 30V. When they are connected the energy lost by $4 μ F$ capacitor is [EAMCET 2001]