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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

A variable capacitor used in an RLC circuit produces a resonant frequency of \(5.0 \mathrm{MHz}\) when its capacitance is set to \(15 \mathrm{pF}\). What will the resonant frequency be when the capacitance is increased to \(380 \mathrm{pF} ?\)

Why is RMS power specified for an AC circuit, not average power?

In Solved Problem 30.1 , the voltage supplied by the source of time-varying emf is \(33.0 \mathrm{~V}\), the voltage across the resistor is \(V_{R}=I R=13.1 \mathrm{~V}\), and the voltage across the inductor is \(V_{L}=I X_{L}=30.3 \mathrm{~V}\). Does this circuit obey Kirchhoff's rules?

A series RLC circuit has resistance \(R\), inductance \(L\), and capacitance \(C\). At what time does the energy in the circuit reach half of its initial value?

A 10.0 - \(\mu\) F capacitor is fully charged by a 12.0 - \(\mathrm{V}\) battery and is then disconnected from the battery and allowed to discharge through a 0.200 - \(\mathrm{H}\) inductor. Find the first three times when the charge on the capacitor is \(80.0-\mu \mathrm{C},\) taking \(t=0\) as the instant when the capacitor is connected to the inductor.
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