Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
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.
To know more about LC circuit:
https://brainly.com/question/13200678
#SPJ4
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
