Question

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

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

Step-by-step solution

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 \(t\).

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: \(V^2(2.10\mathrm{ms}) = \frac{1}{2}V^2(0)\)

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 \(C\): \(4.20\mathrm{ms} = 2\pi\sqrt{1.00\mathrm{mH} \cdot C}\)

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 \(1.59\mu F\).