Multiple Choice
In an oscillating LC circuit in which the capacitance C = 4μF and the maximum voltage across the capacitor V = 1.50V, the maximum current measured across the inductor is 50mA. What is the angular frequency of this LC circuit?
30. Induction and Inductance
LC Circuits
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An LC circuit with an inductor of 0.05 H and a capacitor of 35 μF begins with the current of -1A. The capacitor plates have a maximum charge of 2.65mC at any time during the oscillation. What is the phase angle of this oscillation?
A
45o
B
15o
C
30o
D
60o
Verified step by step guidance1
Identify the given values: Inductance \( L = 0.05 \text{ H} \), Capacitance \( C = 35 \mu\text{F} = 35 \times 10^{-6} \text{ F} \), Initial current \( I_0 = -1 \text{ A} \), and Maximum charge \( Q_{max} = 2.65 \text{ mC} = 2.65 \times 10^{-3} \text{ C} \).
Recall the formula for the angular frequency of an LC circuit: \( \omega = \frac{1}{\sqrt{LC}} \). Substitute the given values of \( L \) and \( C \) to find \( \omega \).
The general solution for the charge \( Q(t) \) on the capacitor in an LC circuit is \( Q(t) = Q_{max} \cos(\omega t + \phi) \), where \( \phi \) is the phase angle.
The current \( I(t) \) is the derivative of the charge with respect to time: \( I(t) = -Q_{max} \omega \sin(\omega t + \phi) \). Use the initial condition \( I(0) = -1 \text{ A} \) to set up the equation \( -Q_{max} \omega \sin(\phi) = -1 \).
Solve for the phase angle \( \phi \) using the equation \( \sin(\phi) = \frac{1}{Q_{max} \omega} \). Calculate \( \phi \) to find the phase angle of the oscillation.
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