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 LC circuit is approximately \(1.52 \times 10^{-6} \mathrm{~F}\).

Step-by-step solution

Write down the energy decay formula

The energy decay process in an LC circuit can be written as: \(E(t) = E_0 e^{-\frac{t}{2RC}}\) where: - \(E(t)\) is the energy at time \(t\), - \(E_0\) is the initial energy, - \(R\) is the resistance, - \(C\) is the capacitance, - \(t\) is the time taken for the energy to decay. Since the energy stored in the capacitor is half of its original value, we have: \(\frac{1}{2}E_0 = E_0 e^{-\frac{t}{2RC}}\)

Find the resistance of the circuit

For an ideal LC circuit, the resistance is negligible. Therefore, the equation from the previous step simplifies to: \(\frac{1}{2}E_0 = E_0 e^{-\frac{t}{2LC}}\)

Solve for the capacitance

We want to find the capacitance \(C\). Dividing both sides of the equation by \(E_0\): \(\frac{1}{2} = e^{-\frac{t}{2LC}}\) Now, let's take the natural logarithm on both sides: \(\ln\left(\frac{1}{2}\right) = -\frac{t}{2LC}\) Next, we can rearrange the equation to solve for \(C\): \(C = -\frac{t}{2L \ln\left(\frac{1}{2}\right)}\) Finally, plug in the given values for \(t\) and \(L\): \(C = -\frac{2.10 \times 10^{-3} \mathrm{~s}}{2(1.00 \times 10^{-3} \mathrm{~H}) \ln\left(\frac{1}{2}\right)}\)

Calculate the capacitance

Now, we can calculate the capacitance value: \(C \approx -\frac{2.10 \times 10^{-3}}{2(1.00 \times 10^{-3}) (-0.693)}\) \(C \approx 1.52 \times 10^{-6} \mathrm{~F}\) Therefore, the capacitance of the LC circuit is approximately \(1.52 \times 10^{-6} \mathrm{~F}\).