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A capacitor in an LC oscillator experiences a maximum potential difference of 88V and a maximum energy of 2002 uJ. At a certain instant the energy in the capacitor is 125 uJ. At that same instant the potential difference across the capacitor is closest to:

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Answer:

Step-by-step explanation:

maximum energy of capacitor

E = 1/2 C V ²

C is capacitance of capacitor and V is potential difference

given V = 88 V

E = 2002 x 10⁻⁶ J

Putting the values

2002 x 10⁻⁶ = 1/2 x C x 88²

C = .517 x 10⁻⁶ F .

In the second case

Energy E = 125 x 10⁻⁶ J .

C = .517 x 10⁻⁶

V = ?

E = 1/2 C V ²

125 x 10⁻⁶ = 1/2 x .517 x 10⁻⁶ x V²

V² = 483.55

V = 21.98 V .

User Joran Beasley
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