Electromagnetic induction > AC electrical circuits > Problem 11.4.5

Statement

$11.4.5.$ In the circuit shown in the figure, the diode $D$ and the inductor $L$ are connected to an alternating voltage source $V =V_0 cosωt$ using a key $K$. At time $t = 0$, the key $K$ is closed. Determine the current in the coil as a function of time.Plot the graph of this function. The diode and coil are considered ideal. Ignore the internal resistance of the source.

Solution

Let's write down Kirchhoff's rule:

$$L\frac{dI}{dt} = V_0 cos(\omega t)$$

Integrate and get:

$$I = \frac{V_0}{\omega L} sin(\omega t)$$

This equation obeys the current strength until the current goes in the opposite direction and the diode closes, that is, until the time $\frac{\pi}{\omega}$. Then, until the voltage at the source changes sign to positive again, this is the moment $\frac{3\pi}{2\omega}$. Then the current starts to rise again, but now let's look at the equation.

$$I = \int^{t}_{\frac{3\pi}{2\omega}} \frac{V_0}{\omega L} cos(\omega t) = \frac{V_0}{\omega L} (sin(\omega t)+1)$$

The period is the same, but the amplitude is twice as high, and this type of dependence persists.

Answer