(a) In the LC circuit shown in the figure, the current is in the direction shown and the charges on the capacitor have the signs shown. At this time,
a) the current is increasing and the charge on the positive plate is increasing.
b) the current is increasing and the charge on the positive plate is decreasing.
c) the current is decreasing and the charge on the positive plate is increasing.
d) the current is decreasing and the charge on the positive plate is decreasing.
e) the current and the charge on the positive plate are constant in time.
f) none of the above
For the following questions, assume that the inductance is 0.05 H, the capacitance is 4 [mathjaxinline]\mu[/mathjaxinline]F, and the maximum magnitude of the current in the circuit is 90 milliAmps.
(b) How many times per second will the magnitude of the current have this maximum value?
(c) The energy stored in the capacitor oscillates up and down as a function of time. What is the magnitude of the difference (in Joules) between the largest and smallest amounts of energy ever stored in the capacitor?
(d) What is the time averaged power (in Watts) flowing into or out of the self inductor over 3 full cycles? Define power into the inductor as positive.
To answer these questions, we need to understand the behavior of an LC circuit.
An LC circuit consists of an inductor (L) and a capacitor (C) connected in series. When the circuit is closed, the current starts to flow and energy is exchanged between the inductor and the capacitor.
(a) Let's analyze the given circuit. Unfortunately, there is no figure provided, so we cannot directly determine the answer option based on the circuit diagram. However, based on the given information, we can deduce the answer. Since the current is increasing, it implies that the voltage across the capacitor is increasing. According to the charge conservation principle, the charge on the positive plate of the capacitor must be decreasing since the charge on the negative plate is increasing to balance it. Therefore, the correct answer is (b) the current is increasing, and the charge on the positive plate is decreasing.
Now, let's proceed to the remaining questions:
(b) The magnitude of the current in the LC circuit oscillates sinusoidally with time. The formula for the current in the circuit at any given time is
\[I(t) = I_{max} \cdot \cos(\omega t)\]
where I(t) is the current at time t, I_max is the maximum magnitude of the current, and ω is the angular frequency.
The angular frequency ω is given by \(\omega = \frac{1}{\sqrt{LC}}\), where L is the inductance (0.05 H) and C is the capacitance (4 μF).
The maximum magnitude of the current is given as 90 milliAmps (0.09 A).
To find the number of times per second the magnitude of the current reaches its maximum value, we need to find the time period (T), which is the time taken for one complete oscillation of the current.
The time period can be calculated using the formula T = \( \frac{2 \pi}{\omega} \).
Substituting the given values, we have T = \( \frac{2 \pi}{\frac{1}{\sqrt{LC}}} \).
Once we have the time period, we can find the frequency (f) using the formula f = \( \frac{1}{T} \).
Substituting the value of T into the formula, we can calculate the frequency.
(c) The energy (E) stored in a capacitor is given by the formula E = \( \frac{1}{2} C V^2 \), where C is the capacitance and V is the voltage across the capacitor.
In the LC circuit, the voltage across the capacitor varies sinusoidally with time. The maximum voltage across the capacitor (V_max) can be calculated using the formula V_max = \( I_{max} \left(\frac{1}{\omega C}\right) \).
To find the difference in energy between the largest and smallest amounts stored in the capacitor, we need to calculate the energy at the maximum and minimum voltages across the capacitor.
Substituting the respective values into the energy formula, we can find the difference in energy.
(d) The time-averaged power (P) flowing into or out of the inductor can be calculated using the formula P = \( \frac{1}{T} \int_{0}^{T} V(t) I(t) dt \), where V(t) is the voltage across the inductor at time t.
Since the inductor and capacitor exchange energy, the average power over a complete cycle is zero. However, to find the average power over 3 full cycles, we need to calculate the power over a period of time equal to 3 cycles.
Using the given formulas, we can perform the integration over 3 cycles and calculate the time-averaged power.
Now we have the step-by-step explanations to solve each question within the LC circuit context.