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Chapter 30: Problem 25

An LC circuit consists of a 1.00 -mH inductor and a fully charged capacitor. After \(2.10 \mathrm{~ms}\), the energy stored in the capacitor is half of its original value. What is the capacitance?

Short Answer

Expert verified
Answer: The capacitance of the capacitor is approximately \(1.59 \mu F\).

Step by step solution

01

Energy stored in a charged capacitor

The energy stored in a charged capacitor is given by the formula: \(E(t) = \frac{1}{2}CV^2(t)\) Where \(E(t)\) is the energy at time \(t\), \(C\) is the capacitance, and \(V(t)\) is the voltage across the capacitor at time \(t\).
02

Energy as half of original value

Given that after 2.10 ms, the energy stored in the capacitor is half of its original value, we can write the equation: \(\frac{1}{2}CV^2(2.10\mathrm{ms}) = \frac{1}{2} \cdot \frac{1}{2}CV^2(0)\) We can simplify this equation as: \(V^2(2.10\mathrm{ms}) = \frac{1}{2}V^2(0)\)
03

Period of oscillation in an LC circuit

In an LC circuit, the period of oscillation \(T\) is given by: \(T = 2\pi\sqrt{LC}\) Since we are given the time 2.10 ms as half of the oscillation, it means that \(T = 4.20\mathrm{ms}\). We can substitute and solve for \(C\): \(4.20\mathrm{ms} = 2\pi\sqrt{1.00\mathrm{mH} \cdot C}\)
04

Solve for capacitance

Now, we can solve for the capacitance \(C\): \(C = \frac{T^2}{4\pi^2 L} = \frac{(4.20\mathrm{ms})^2}{4\pi^2 \cdot 1.00\mathrm{mH}}\) Converting values into SI units: \(C = \frac{(4.20 \times 10^{-3}\mathrm{s})^2}{4\pi^2 \cdot 1.00 \times 10^{-3}\mathrm{H}}\) Calculating the value gives: \(C \approx 1.59 \times 10^{-6}\mathrm{F}\) So, the capacitance of this capacitor is approximately \(1.59\mu F\).

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Most popular questions from this chapter

The phase constant, \(\phi\), between the voltage and the current in an AC circuit depends on the ______. a) inductive reactance b) capacitive reactance c) resistance d) all of the above

Which statement about the phase relation between the electric and magnetic fields in an LC circuit is correct? a) When one field is at its maximum, the other is also, and the same for the minimum values. b) When one field is at maximum strength, the other is at minimum (zero) strength. c) The phase relation, in general, depends on the values of \(L\) and \(C\).

a) A loop of wire \(5.00 \mathrm{~cm}\) in diameter is carrying a current of \(2.00 \mathrm{~A}\). What is the energy density of the magnetic field at its center? b) What current has to flow in a straight wire to produce the same energy density at a point \(4.00 \mathrm{~cm}\) from the wire?

The figure shows a simple FM antenna circuit in which \(L=8.22 \mu \mathrm{H}\) and \(C\) is variable (the capacitor can be tuned to receive a specific station). The radio signal from your favorite FM station produces a sinusoidal time-varying emf with an amplitude of \(12.9 \mu \mathrm{V}\) and a frequency of \(88.7 \mathrm{MHz}\) in the antenna. a) To what value, \(C_{0}\), should you tune the capacitor in order to best receive this station? b) Another radio station's signal produces a sinusoidal time-varying emf with the same amplitude, \(12.9 \mu \mathrm{V}\), but with a frequency of \(88.5 \mathrm{MHz}\) in the antenna. With the circuit tuned to optimize reception at \(88.7 \mathrm{MHz}\), what should the value, \(R_{0}\), of the resistance be in order to reduce by a factor of 2 (compared to the current if the circuit were optimized for \(88.5 \mathrm{MHz}\) ) the current produced by the signal from this station?

In an RL circuit with alternating current, the current lags behind the voltage. What does this mean, and how can it be explained qualitatively, based on the phenomenon of electromagnetic induction?
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