Question

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:

Answer 1

Correct Answer: 225

To solve this problem, we need to calculate the energy loss when a charged capacitor is connected to an uncharged identical capacitor. Initially, the 900 μF capacitor is charged to 100 V. The initial energy stored in the charged capacitor is given by \( U_i = \frac{1}{2} C V^2 \).

Substituting the given values:

\( U_i = \frac{1}{2} \times 900 \times 10^{-6} \, \text{F} \times (100 \, \text{V})^2 \)

\( U_i = 0.5 \times 900 \times 10^{-6} \times 10000 \, \text{J} \)

\( U_i = 4.5 \, \text{J} \)

After connection, the charge is redistributed equally between the two capacitors, as they are identical. The total capacitance in parallel is \( 2C = 1800 \, \mu\text{F} \) and the total charge remains the same, so \( Q = C \times V = 900 \times 10^{-6} \times 100 = 0.09 \, \text{C} \).

The final voltage \( V_f \) across both capacitors connected in parallel can be found using \( Q = 2C \times V_f \).

\( 0.09 \, \text{C} = 1800 \times 10^{-6} \times V_f \)

\( V_f = \frac{0.09}{1800 \times 10^{-6}} \, \text{V} \)

\( V_f = 50 \, \text{V} \)

The final energy stored in the system is \( U_f = \frac{1}{2} \times 1800 \times 10^{-6} \times (50)^2 \).

\( U_f = 0.5 \times 1800 \times 10^{-6} \times 2500 \, \text{J} \)

\( U_f = 2.25 \, \text{J} \)

The energy loss is given by the difference between initial and final energies:

\( \Delta U = U_i - U_f = 4.5 - 2.25 = 2.25 \, \text{J} \)

The problem states the energy loss as \( x \times 10^{-2} \, \text{J} \), so we equate:

\( x \times 10^{-2} = 2.25 \)

\( x = 225 \)

Thus, the value of \( x \) is 225, which fits within the expected range of 225,225.