Chapter 30: Problem 37
An \(L\)-\(C\) circuit containing an 80.0-mH inductor and a 1.25-nF capacitor oscillates with a maximum current of 0.750 A. Calculate: (a) the maximum charge on the capacitor and (b) the oscillation frequency of the circuit. (c) Assuming the capacitor had its maximum charge at time \(t = 0\), calculate the energy stored in the inductor after 2.50 ms of oscillation.
Short Answer
Step by step solution
Identify the Formula for Maximum Charge
Calculate Maximum Charge (Part a)
Determine the Formula for Oscillation Frequency
Calculate Oscillation Frequency (Part b)
Find Energy Formula in the Inductor
Calculate Angular Frequency (\( \omega \))
Determine Current at Given Time
Compute Energy in Inductor (Part c)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Maximum Charge
An easy way to remember the formula for maximum charge is by using this relationship:
- \( Q_{max} = I_{max} \times \sqrt{L/C} \)
Given:
- \( L = 0.0800 \) H (inductance of the inductor)
- \( C = 1.25 \times 10^{-9} \) F (capacitance of the capacitor)
Oscillation Frequency
The formula to find the oscillation frequency \( f \) is:
- \( f = \frac{1}{2\pi} \sqrt{\frac{1}{LC}} \)
- \( L = 0.0800 \) H
- \( C = 1.25 \times 10^{-9} \) F
Energy Stored in Inductor
- \( E_L = \frac{1}{2} L I^2(t) \)
With \( \omega = 2\pi f \), and from our previous section, \( f \approx 503.29 \text{ kHz} \). Thus, \( \omega \approx 3162389 \text{ rad/s} \).
Let's calculate the current at \( t = 2.5 \) ms.
- First, find \( \omega t \): \( 3162389 \times 0.0025 \approx 7905.97 \)
- Next, use the sine function: \( I(t) = 0.750 \times \sin(7905.97) \approx 0.6495 \text{ A} \)