Problem 21 - Fundamentals of Physics

Question

In an oscillating \(L C\) circuit with \(C=64.0 \mu \mathrm{F}\), the current is given by \(i=(1.60) \sin (2500 t+0.680),\) where \(t\) is in seconds, \(i\) in amperes, and the phase constant in radians. (a) How soon after \(t=0\) will the current reach its maximum value? What are (b) the inductance \(L\) and (c) the total energy?

Step-by-Step Solution

Answer

(a) Solve \(2500t + 0.680 = \frac{\pi}{2}\), (b) use \(L = \frac{1}{(2500)^2 \times 64 \times 10^{-6}}\), (c) calculate \(U = \frac{1}{2} C V^2\) or \(U = \frac{1}{2} L I^2\).

Step 1: Identify the Maximum Current Condition

The current function is given by: \(i = (1.60) \sin(2500t + 0.680)\).\ To reach the maximum current, the value inside the sine function must be \(\frac{\pi}{2}\), where \(\sin\left(\frac{\pi}{2}\right) = 1\).

Step 2: Solve for Time to Maximum Current

Set \(2500t + 0.680 = \frac{\pi}{2}\).\ Solving for \(t\), we get:\[2500t + 0.680 = \frac{\pi}{2}\]\[2500t = \frac{\pi}{2} - 0.680\]\[t = \frac{\frac{\pi}{2} - 0.680}{2500}\]

Step 3: Calculate Inductance from Angular Frequency

The angular frequency \(\omega\) in the current equation is \(2500\), where \(\omega = \frac{1}{\sqrt{LC}}\).\ Rearrange to find \(L\):\[L = \frac{1}{\omega^2 C}\]Substitute \(\omega = 2500\) and \(C = 64.0 \times 10^{-6} \ \mathrm{F}\) to find \(L\).

Step 4: Calculate Total Energy in the Circuit

The total energy \(U\) stored in the circuit can be calculated using either the electric or magnetic formula:\[U = \frac{1}{2} C V^2\]or\[U = \frac{1}{2} L I^2\]First determine \(V\) or \(I\) maximum using \(I = 1.60\ A\) (current amplitude) and substitute into the formula to find \(U\).

Key Concepts

Inductance

An oscillating LC circuit consists of an inductor (L) and a capacitor (C) working together to create electrical oscillations. One of the critical elements that determine the behavior of an LC circuit is the inductance, which is represented by \( L \). Inductance is a property of an inductor that measures its ability to store energy in a magnetic field. In mathematical terms, it is defined by the relationship between the current change in the circuit and the electromotive force (EMF) induced.
To find the inductance in an LC circuit, the angular frequency (\( \omega \)) is a key parameter. This is because \( \omega \) is related to both inductance and capacitance by the formula:
\[\omega = \frac{1}{\sqrt{LC}}\]Rearranging the formula to solve for \( L \):
\[L = \frac{1}{\omega^2 C}\]Given \( \omega = 2500 \) rad/s and \( C = 64.0 \times 10^{-6} \) F, one can substitute these values into the formula to compute the inductance.

Current Amplitude

The current amplitude in an LC circuit refers to the maximum value of the oscillating current described by the current function. It gives us the peak current flowing through the circuit. In the given problem, the current is defined as:
\(i = (1.60) \sin(2500t + 0.680)\)
Here, the amplitude is the coefficient of the sine function, which is 1.60 A. This value indicates the maximum strength of current that flows through the circuit during one complete cycle of oscillation.

• The amplitude tells us how strong the source is energizing the circuit.
• It's crucial for calculating other important parameters like energy.

Understanding the role of current amplitude is essential for analyzing how circuits perform under different conditions.

Angular Frequency

Angular frequency is a measure of how rapidly the oscillations in an LC circuit occur. It is denoted by \( \omega \) and is related to the frequency \( f \) by the equation:
\[\omega = 2\pi f\]For an LC circuit, \( \omega \) is more conveniently expressed in relation to the inductance \( L \) and capacitance \( C \):
\[\omega = \frac{1}{\sqrt{LC}}\]In the exercise given, the angular frequency is \( 2500 \) rad/s, dictating how quickly the oscillations happen in terms of radians per second.

• High \( \omega \) implies faster oscillations.
• It determines the rate at which energy is transferred between the inductor and capacitor.

Angular frequency is crucial in timing and energy calculations within the circuit.

Total Energy in Circuit

The total energy in an LC circuit can be calculated from the energy stored either in the electric field of the capacitor or the magnetic field of the inductor. This energy is retained in the form of alternating current and voltage oscillations within the circuit.
The formulas to calculate this energy are:
\[U = \frac{1}{2} C V^2\] (using capacitance)
\[U = \frac{1}{2} L I^2\] (using inductance)Using either formula requires knowledge of the maximum voltage \( V \) across the capacitor or the current amplitude \( I \) in the circuit. For this circuit, you can compute energy using the known current amplitude \( I = 1.60 \) A.

• The energy remains constant and oscillates between the inductor and capacitor.
• Understanding total energy helps in analyzing the circuit's efficiency.

Energy calculations reveal how well the circuit sustains and utilizes energy through each oscillation cycle.