an l-c circuit consists of a 65.0 mh inductor and a 300 Î¼f capacitor. the initial charge on the capacitor is 6.50 Î¼c, and the initial current in the inductor is zero.
a. What is the maximum voltage across the capacitor?
b. What is the maximum current in the inductor?
c. What is the maximum energy stored in the inductor?
d. When the current in the inductor has half its maximum value, what is the charge on the capacitor?
e. When the current in the inductor has half its maximum value, what is the energy stored in the inductor?

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Smecloudbird17go Smecloudbird17go · Answer 1

The maximum voltage across the capacitor is 2.1 * 10⁻² V and the initial energy in the capacitor is 7.042 * 10⁻⁶ J.

In the L-C oscillation, energy is transferred between capacitor and inductor with a certain periodicity.

Given that, Inductance L = 65 * 10⁻³ h

Capacitance C = 300 * 10⁻⁴ F

Initial charge q = 6.5 * 10⁻⁴ C

We know that,

1/2 q²/C + 1/2 L² = 1/2 C* v₀²

From the above equation, we can write v₀ = q/C

Substituting the values in the above equation, we get

v₀ = q/C = (6.5 * 10⁻⁴) / (300 * 10⁻⁴) = 0.021 V = 2.1 * 10⁻² V

The formula to calculate initial energy in the capacitor is 1/2* q²/C

where q is charge and C is capacitance

E = 1/2 * q²/C = (6.5 * 10⁻⁴)²/ 2 * (300 * 10⁻⁴)

⇒ (42.25 * 10⁻⁸)/ (6 * 10⁻²)

⇒ 7.042 * 10⁻⁶ J

Thus, maximum voltage across the capacitor is 2.1 * 10⁻² V and the initial energy in the capacitor is 7.042 * 10⁻⁶ J.

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