Question

The time-varying current in an LC circuit where \(C=10.0 \mu \mathrm{F}\) is given by \(i(t)=(1.00 \mathrm{~A}) \sin (1200 . t),\) where \(t\) is in seconds. a) At what time after \(t=0\) does the current reach its maximum value? b) What is the total energy of the circuit? c) What is the inductance, \(L\) ?

Short answer

Answer: The first time the current reaches its maximum value is \(t = \frac{\pi}{2400} \mathrm{s}\). The total energy in the circuit is \(0.05 \mathrm{J}\). The inductance L of the LC circuit is approximately \(6.944 \times 10^{-5} \mathrm{H}\).

Step-by-step solution

a) Finding the time when the current reaches its maximum value

The current in the circuit is given by the sinusoidal function: \(i(t)=(1.00 \mathrm{~A}) \sin (1200 . t)\). Since the sine function achieves its maximum value (1) when its argument is equal to an odd multiple of \(\frac{\pi}{2}\), we can set up the equation for the time: \(1200t = (2n+1)\frac{\pi}{2}\) Solving for t, we have: \(t = \frac{(2n + 1)\pi}{2400}\) The first time it reaches its maximum value is when n=0, so: \(t = \frac{(2(0) + 1)\pi}{2400}\) \(t = \frac{\pi}{2400} \mathrm{s}\)

b) Finding the total energy of the circuit

The total energy of an LC circuit is the sum of the energy stored in the magnetic field of the inductor (\(W_L\)) and the energy stored in the electric field of the capacitor (\(W_C\)). We know the total energy is constant, and we can find the energy in the capacitor at a given moment when there is maximum current (since at maximum current, there's no energy in inductor hence energy in capacitor is equal to total energy). The energy in the capacitor is given by: \(W_C = \frac{1}{2}Cv^2\) We know that \(v = \frac{q}{C}\) and the current \(i(t) = \frac{dq}{dt} = (1.00 \mathrm{~A}) \sin (1200 . t)\). When the current is at maximum, the derivative of the charge with respect to time is maximum, so we have \(i_{max} = 1.00 \mathrm{~A}\). To find q(t), we integrate both sides: \(q(t) = \int i(t) dt = -\frac{1}{1200}\cos(1200t) + C\) We know that at \(t = 0\), \(q(0) = 0\), which implies that C = 0. Therefore, \(q(t) = -\frac{1}{1200}\cos(1200t)\). Now we can find the voltage at the maximum current (when t = \(\frac{\pi}{2400}\) s): \(v_{max} = \frac{q(\frac{\pi}{2400})}{C} = \frac{-\frac{1}{1200}\cos(1200 \cdot \frac{\pi}{2400})}{10\times10^{-6}} = 100 \mathrm{V}\) Now we can find the total energy of the circuit: \(W_{total} = W_C = \frac{1}{2}Cv_{max}^2 = \frac{1}{2} \cdot (10\times10^{-6}) \cdot (100)^2 = 0.05 \mathrm{J}\)

c) Finding the inductance L

We know that the angular frequency \(\omega\) of an LC circuit is given by: \(\omega = \sqrt{\frac{1}{LC}}\) We also know that the frequency of the current sine function is \(1200\) rad/s. So \(\omega = 1200\) rad/s and the capacitance is given as \(C = 10.0 \times 10^{-6} \mathrm{F}\). We can use these values to solve for the inductance L: \(L = \frac{1}{C\omega^2} = \frac{1}{(10.0 \times 10^{-6}) \cdot (1200)^2} = 6.944 \times 10^{-5} \mathrm{H}\) The inductance L of the LC circuit is approximately \(6.944 \times 10^{-5} \mathrm{H}\).

Key concepts

Time-Varying Current in an LC Circuit

Understanding how current flows in an LC circuit is key to grasp the intricate dance between electric and magnetic fields. An LC circuit, comprised of an inductor (L) and a capacitor (C), exhibits a time-varying current that oscillates due to the continuous energy transfer between these two components.

When a capacitor discharges, it creates a current that then