A capacitor of capacitance 900 μF 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 the uncharged capacitor is connected to the positive plate and another plate of the uncharged capacitor is connected to the negative plate of the charged capacitor. The loss of energy in this process is measured as \(x \times 10^{-2}\) J. The value of \(x\) is:
Correct Answer: 225
Solution and Explanation
The initial energy stored in the charged capacitor is given by the formula:
\(E_i = \frac{1}{2} C V^2\)
Substitute the values:
\(E_i = \frac{1}{2} \times 900 \times 10^{-6} \times (100)^2 = 4.5 \, \text{J}\)
After connecting the uncharged capacitor, the total charge gets shared between the two capacitors. The final voltage across each capacitor becomes half of the initial voltage because the two capacitors are identical:
\(V_f = \frac{V_i}{2} = \frac{100}{2} = 50 \, \text{V}\)
The final energy stored in each capacitor is:
\(E_f = \frac{1}{2} C V_f^2 = \frac{1}{2} \times 900 \times 10^{-6} \times (50)^2 = 2.25 \, \text{J}\)
The total final energy in both capacitors is:
\(2 \times E_f = 2 \times 2.25 = 4.5 \, \text{J}\)
The energy loss during this process is the difference between the initial and final energy, which is:
\(\Delta E = E_i - 2 \times E_f = 4.5 - 4.5 = 0 \, \text{J}\)
The energy loss is measured as \(x \times 10^{-2}\ J\), where \(x = 225.\)
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