Chapter 31: Problem 21
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 \((\mathrm{c})\) the total energy?
Short Answer
Step by step solution
Identify when current i reaches its maximum value
Solve for time t
Determine the Inductance L
Calculate the Inductance Value L
Determine Total Energy in the Circuit
Calculate the Total Energy
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Oscillating Circuits
When the capacitor releases its stored charge, it creates a current through the inductor, storing energy in its magnetic field. This current decreases, causing the capacitor to charge again with reversed polarity. Consequently, the circuit oscillates at a specific resonant frequency, dictated by its inductance and capacitance.
Key characteristics of oscillating circuits include:
- The resonant frequency, which determines the rate of oscillation, is given by \( f = \frac{1}{2\pi\sqrt{LC}} \).
- No external energy source is needed after initial charge, as energy is conserved and oscillates between the components.
- Damping can occur due to resistance, causing oscillations to decrease over time.
Current in LC Circuits
Key points about current in LC circuits include:
- The amplitude of the current is determined by the initial charge and resistance (though this example assumes ideal conditions).
- The phase of the current is influenced by the initial conditions of the circuit.
- The maximum value of the current occurs when the sine function reaches its peak value of 1. At this point, the argument of the sine function fulfills \( 2500t + 0.680 = \frac{\pi}{2} + 2\pi n \) for integer \( n \).
- Solving for \( t \) gives the precise moment when current first reaches maximum after \( t=0 \).
Inductance Calculation
For the exercise, the angular frequency \( \omega \) is given as 2500 rad/s. The relationship between inductance and angular frequency can be expressed as \( \omega = \frac{1}{\sqrt{LC}} \).
Steps to calculate the inductance include:
- Rearranging the formula to solve for \( L \), we get \( L = \frac{1}{(\omega)^2 C} \).
- Substitute \( C = 64.0 \times 10^{-6} F \) and \( \omega = 2500 \) rad/s into the rearranged formula.
- Solving gives \( L = 2.5 \times 10^{-3} H \) or 2.5 mH (millihenrys).
Energy in LC Circuits
\[ E = \frac{1}{2} L I^2 \]
where \( I \) is the peak current and \( L \) is the inductance.
Steps for energy calculation in the provided exercise include:
- Calculate the maximum current \( I = 1.60 A \), obtained from the current formula.
- Substitute the calculated inductance \( L = 2.5 \times 10^{-3} H \) and \( I = 1.60 A \) into the energy formula.
- Perform the calculation to find \( E = 3.2 \times 10^{-3} \) joules or 3.2 millijoules.