Capacitors and Inductors in AC Circuits

The current through a capacitor is given by: \(i_c = C\frac{dV}{dt}\)

The voltage across a 1F capacitor is the triangular wave shown below. Find the current running through the capacitor.

This sawtooth wave has peaks at 1 and -1 with a period of \(2\pi\). The current through the capacitor is given by: \(I = C\frac{dV}{dt}\) The sawtooth has a slope of \(\pm \frac{2}{\pi}\) so the derivative is a square wave with peaks at \(\pm \frac{2}{\pi}\) and frequency \(2\pi\).

When we apply a sine wave across a capacitor we get a scaled version of that same sine wave but shifted left by \(90^\circ\) (or "leading" by \(90^\circ\)).

A capacitive circuit has the current leading the voltage.

The voltage across an inductor is given by: \(V_L = L\frac{dI_L}{dt}\) Which means the current is given by: \(I_L = \frac{1}{L}\int V_Ldt\)

The voltage across a 1H inductor is the pulse waveform shown below. Find the current through it.

The voltage pulse has peaks at \(\pm 1\) and a period of \(2\pi\). The current is given by our formula: \(I_L = \frac{1}{L}\int V_Ldt\) So our current is the triangular waveform with peaks at \(\pm \pi\) and a period of \(2\pi\).

When we apply a sine wave across an inductor we get a scaled version of that same sine wave but shifted right by \(90^\circ\) (or "lagging" by \(90^\circ\)).

An inductive circuit has the current lagging the voltage.