A capacitor of capacitance C has charge Q and stored energy is U if the charge is increased to 2Q then energy will be A)4U B)2U
C)U/2 D)U/4
1 answer:
Answer:
2u
Explanation:
2u
W = Vq
q = CV
W = V.CV
W = CV²
W/C = V²
V = √(W/C)
√(W1/C1) = √(W2/C2)
√(u/c) = √(x/2c)
x = 2u
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V = [4/3]π r^3 => [dV / dr ] = 4π r^2
[dV/dt] = [dV/dr] * [dr/dt]
[dV/dt] = [4π r^2] * [ dr/ dt]
r = 60 mm, [dr / dt] = 4 mm/s
[dV / dt ] = [4π(60mm)^2] * 4mm/s = 180,955.7 mm/s