Energy conservation in LC circuits ensures smooth oscillations between capacitor and inductor.
Step 1: Use the energy conservation in LC circuits - Maximum energy in capacitor = Maximum energy in inductor: \[ \frac{1}{2} L I_{\text{max}}^2 = \frac{1}{2} \frac{q_{\text{max}}^2}{C}. \]
Step 2: Solve for \(I_{\text{max}}\) - \[ I_{\text{max}} = \frac{q_{\text{max}}}{\sqrt{L C}}. \] Substituting values: \[ I_{\text{max}} = \frac{\sqrt{2.7 \times 10^{-6}}}{\sqrt{75 \times 10^{-3} \cdot 1.2 \times 10^{-6}}}. \] Simplifying: \[ I_{\text{max}} = 9 \times 10^{-3} \, \text{A} = 9 \, \text{mA}. \]
Final Answer: The maximum current in the circuit is 9 mA.
Find output voltage in the given circuit.
A | B | Y |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 1 |
1 | 1 | 0 |
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: