The Inductor and the Capacitor

The inductor

A pure inductor is a coil of inductance \(L\) and no resistance. The chapter Eddy Currents, Generators and Inductance showed that a changing current \(i\) in it produces an emf \(-L\,di/dt\) that opposes the change. Connect the coil across the source \(v = v_0\sin\omega t\). Going round the circuit, the source voltage must equal the drop across the coil. So at every instant \[ L\dfrac{di}{dt} = v_0\sin\omega t. \]

The current in an inductor reaches its peak a quarter cycle after the voltage does. We say the current lags the voltage by \(90°\), or \(\pi/2\). The reason is the back emf. The source must push hardest when the current is changing fastest. The current changes fastest as it passes through zero, so the voltage peaks when the current is zero.

The peak current is \(i_0 = v_0/\omega L\). So \(\omega L\) plays the part of a resistance. It is called the inductive reactance, \(X_L = \omega L\), and it is measured in ohms. Dividing both peaks by \(\sqrt2\) gives \(V_{rms} = I_{rms}X_L\).

Play the three chips of the figure above. With the resistor, the voltage and current phasors lie along one line. With the inductor, the current phasor stays a quarter turn behind the voltage phasor all the way round. So the blue current wave peaks a quarter period after the orange voltage wave. With the capacitor, the current phasor runs a quarter turn ahead.

The capacitor

Now connect a capacitor \(C\) across the same source. The chapter Capacitance and Dielectrics gave its charge, \(q = Cv\). So the charge on its plates follows the voltage, \(q = Cv_0\sin\omega t\). The current in the leads is the rate at which that charge changes: \[ \begin{gathered} i = \dfrac{dq}{dt} = \omega Cv_0\cos\omega t \\ i = \dfrac{v_0}{1/\omega C}\sin\left(\omega t + \dfrac{\pi}{2}\right). \end{gathered} \] No charge crosses the gap between the plates. Yet charge flows to and fro in the wires all the time, so an alternating current flows through the circuit.

This current reaches its peak a quarter cycle before the voltage. The current leads the voltage by \(90°\). The reason is again a rate. The current is largest when the charge is changing fastest, which is as the voltage passes through zero. By the time the voltage peaks, the plates are full and the current is zero.

The peak current is \(v_0/(1/\omega C)\). So the capacitive reactance is \(X_C = 1/\omega C\), again in ohms, and \(V_{rms} = I_{rms}X_C\).

Reactance and frequency

A resistor opposes a current equally at every frequency. The two reactances do not. \(X_L = \omega L\) grows in proportion to the frequency. A faster change of current makes a larger back emf, so the inductor opposes it more. \(X_C = 1/\omega C\) falls as the frequency rises. At a higher frequency the plates need less charge per half cycle to carry the same current, so the voltage across them is smaller.

At the far left of Fig 3.1 the frequency tends to zero, which is direct current. Then \(X_L \to 0\): an inductor passes a steady current as a plain wire does. And \(X_C \to \infty\): a capacitor blocks a steady current completely. At \(100\,\text{Hz}\) the inductor has \(62.8\,\Omega\) and the capacitor only \(15.9\,\Omega\). The two curves cross near \(50\,\text{Hz}\), where each is about \(31\,\Omega\). Lesson 5 shows that such a crossing is where resonance happens.

No power on average

In the inductor \(v = v_0\sin\omega t\) and \(i = -i_0\cos\omega t\). So the power delivered to it is \[ \begin{gathered} p = vi = -v_0 i_0\sin\omega t\cos\omega t \\ p = -\tfrac12 v_0 i_0\sin 2\omega t. \end{gathered} \] This is a sine of twice the frequency, positive for one quarter cycle and equally negative for the next. Its average over a cycle is zero. While \(p > 0\) the current grows and the source builds up the energy \(\tfrac12 Li^2\) in the coil's magnetic field. While \(p < 0\) the current falls and the field returns that energy to the source.

The capacitor behaves the same way. Its energy \(\tfrac12 Cv^2\) is stored in the electric field as the voltage rises and returned as it falls. So a pure inductor or a capacitor takes no power on average, although a current flows through it. Lesson 6 returns to this.