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Answer :
The decrease in the energy of the system when a 20F charged capacitor is connected in parallel with an uncharged 30F capacitor is (D.) 150J. This is calculated using the energy storage formula for capacitors and the conservation of charge in a closed system.
The original energy stored in the 20F capacitor when charged to 5V is given by the formula:
E = 1/2 C V^2
Substituting the values, we get:
E = 1/2 * 20F * (5V)^2
= 250J
When the charged 20F capacitor is connected in parallel with the uncharged 30F capacitor, the total capacitance becomes the sum of both, which is 50F. The charge remains the same, as no external circuit is involved to add or remove charge. The total charge (Q) is:
Q = C * V
= 20F * 5V
= 100C
The new voltage (V') across the combined system can be found by:
V' = Q / (C1 + C2)
= 100C / 50F
= 2V
Now, the new energy stored in the system is:
E' = 1/2 * (C1 + C2) * V'^2
= 1/2 * 50F * (2V)^2
= 100J
Therefore, the decrease in energy (ΔE) is the difference between the original energy and the new energy:
ΔE = E - E'
= 250J - 100J
= 150J
Thus, the decrease in the energy of the system will be 150J, which corresponds to option D.
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