Question

A variable capacitor is connected to a 200 V battery. If its capacitance is changed from 2 µF to X µF, the decrease in energy of the capacitor is 2.0 × 10⁻² J. Calculate the value of X.

पर्याय

• 1 µF

• 2 µF

• 3 µF

• 4 µF

Answer 1

1 µF

Explanation:

Energy = `1/2` CV²

Decrease in energy = 2.0 × 10⁻² J

`1/2("C"_1 - "C"_2)"V"^2 = 2.0 xx 10^-2`

(C₁ − C₂) = `(2.0 xx 10^-2 xx 2)/"V"^2`

(C₁ − C₂) = `(2.0 xx 10^-2 xx 2)/200^2`

(C₁ − C₂) = 1.0 × 10⁻⁶ F

(2.0 μF − X) = 1.0 µF

X = 1.0 µF