Let's craft a similar LC circuit problem and solve it step by step.

Problem:

An LC circuit contains a 50.0-mH inductor and a 10.0-µF capacitor that initially carries a 100-µC charge. The switch is open for $t < 0$ and is closed at $t = 0$.

• (a) Find the frequency (in hertz) of the resulting oscillations.
• (b) At $t = 2.00 \, \text{ms}$, find the charge on the capacitor.
• (c) At $t = 2.00 \, \text{ms}$, find the current in the circuit.

Solution:

(a) Find the frequency of oscillations

The angular frequency $\omega_0$ of an LC circuit is given by the formula: $\omega_0 = \frac{1}{\sqrt{LC}}$ where:

• $L = 50.0 \, \text{mH} = 50.0 \times 10^{-3} \, \text{H}$
• $C = 10.0 \, \mu\text{F} = 10.0 \times 10^{-6} \, \text{F}$

Substitute these values into the formula: $\omega_0 = \frac{1}{\sqrt{(50.0 \times 10^{-3})(10.0 \times 10^{-6})}}$

First, calculate the denominator: $L \times C = (50.0 \times 10^{-3})(10.0 \times 10^{-6}) = 5.0 \times 10^{-7}$

Now, take the square root: $\sqrt{5.0 \times 10^{-7}} = 7.071 \times 10^{-4}$

Thus: $\omega_0 = \frac{1}{7.071 \times 10^{-4}} = 1413.7 \, \text{rad/s}$

The frequency $f$ in hertz is related to angular frequency by: $f = \frac{\omega_0}{2\pi}$

Substitute $\omega_0 = 1413.7 \, \text{rad/s}$: $f = \frac{1413.7}{2\pi} \approx \frac{1413.7}{6.283} \approx 225 \, \text{Hz}$

So, the frequency of oscillations is approximately 225 Hz.

(b) Find the charge on the capacitor at $t = 2.00 \, \text{ms}$

The charge on the capacitor at time $t$ is given by: $q(t) = q_0 \cos(\omega_0 t)$ where:

• $q_0 = 100 \, \mu\text{C} = 100 \times 10^{-6} \, \text{C}$
• $\omega_0 = 1413.7 \, \text{rad/s}$
• $t = 2.00 \, \text{ms} = 2.00 \times 10^{-3} \, \text{s}$

First, calculate $\omega_0 t$: $\omega_0 t = 1413.7 \times 2.00 \times 10^{-3} = 2.827$

Now, calculate the cosine: $\cos(2.827) \approx -0.949$

Substitute into the equation for $q(t)$: $q(t) = (100 \times 10^{-6}) \times (-0.949) = -94.9 \times 10^{-6} \, \text{C} = -94.9 \, \mu\text{C}$

Thus, the charge on the capacitor at $t = 2.00 \, \text{ms}$ is approximately -94.9 µC.

(c) Find the current in the circuit at $t = 2.00 \, \text{ms}$

The current in an LC circuit is the rate of change of the charge: $i(t) = -\frac{dq(t)}{dt} = \omega_0 q_0 \sin(\omega_0 t)$

Substitute the known values: $i(t) = (1413.7) \times (100 \times 10^{-6}) \times \sin(2.827)$

First, calculate $\sin(2.827)$: $\sin(2.827) \approx 0.316$

Now, substitute into the current equation: $i(t) = 1413.7 \times 100 \times 10^{-6} \times 0.316 \approx 44.7 \times 10^{-3} \, \text{A} = 44.7 \, \text{mA}$

Thus, the current at $t = 2.00 \, \text{ms}$ is approximately 44.7 mA.

Would you like any more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How does the inductor affect the frequency of oscillation in an LC circuit?
2. What would happen to the oscillation frequency if the capacitance were doubled?
3. How does energy transfer between the inductor and the capacitor during oscillations?
4. What is the maximum current in the circuit, and when does it occur?
5. How would the time period of oscillation change if the inductance was halved?

Tip: The oscillation frequency of an LC circuit depends only on the values of the inductor and the capacitor, not on the initial charge or current.