7.1 AC Voltage Applied to a Resistor, an Inductor and a Capacitor

A label on a hair dryer reads "$230$ V ~ $50$ Hz, $1200$ W". The wavy line means alternating: the voltage at the socket is not $230$ V at any steady moment. It rises to about $325$ V, falls through zero, reaches $-325$ V and comes back, fifty times a second. So what does "$230$ V" mean, and how can a single number describe the power of something whose current keeps reversing? A second puzzle: a capacitor is a gap in the circuit, so no charge ever crosses it, yet an AC ammeter in series with a capacitor shows a steady reading. And a coil of thick copper wire, which barely resists a steady current, can hold an alternating current down to a fraction of an ampere.

This section answers all three. We describe an alternating voltage by its peak value and frequency, define the rms value that meters and labels use, and introduce phasors — rotating arrows that turn sine waves into simple geometry. Then we connect a sinusoidal source to a resistor, an inductor and a capacitor in turn and find the current in each, how large it is (through the reactances $X_L=\omega L$ and $X_C=1/\omega C$) and whether it is in step with the voltage, behind it or ahead of it.

Alternating Voltage, the Resistor and RMS Values

An AC generator (Chapter 6) with a coil turning at a steady rate produces an emf that varies sinusoidally with time. Mains supplies, the signals in audio equipment and the output of a laboratory signal generator are all close to sinusoidal, and any repeating signal can be built from sinusoids, so the sine wave is the case to understand.

For the Indian mains, $\nu=50$ Hz, so $\omega=2\pi(50)=100\pi\approx314$ rad/s and $T=20$ ms.

A resistor. Connect a resistor $R$ to the source. At every instant Kirchhoff's loop rule (Section 3.4) gives $v_m\sin\omega t-iR=0$, so $$i=\frac{v_m}{R}\sin\omega t=i_m\sin\omega t,\qquad i_m=\frac{v_m}{R}.$$ The current reaches its peak, passes through zero and reverses at exactly the same moments as the voltage.

Over one complete cycle the current is positive for half the time and negative for the other half, so its average value is zero. Its heating effect is certainly not zero: the instantaneous power is $$p=i^2R=i_m^2R\sin^2\omega t=\tfrac12 i_m^2R\,(1-\cos2\omega t).$$ The term $\cos2\omega t$ is positive as often as it is negative, so its average over a cycle is zero, and the average power is $\bar p=\tfrac12 i_m^2R$. A steady current $I$ would give the same heating if $I^2R=\tfrac12 i_m^2R$, that is $I=i_m/\sqrt2$. This "equivalent steady value" is the square root of the mean of $i^2$.

So "$230$ V" is the rms value of the mains; its peak is $230\sqrt2\approx325$ V. The insulation of every cable and switch must withstand the peak, not just the rms value.

Phasors

Sine waves are awkward to add and compare on a graph. There is a neater picture, familiar from SHM (Class 11 Physics, Chapter 13): a point moving round a circle at a steady angular speed $\omega$ has a vertical coordinate that varies sinusoidally. So we represent each sinusoidal quantity by a rotating arrow.

A phasor ahead of another in the anticlockwise sense leads it: its quantity reaches its peak earlier. For the resistor the voltage and current phasors point the same way at every instant (phase difference zero). Phasors are not vectors in space — they are a bookkeeping device for amplitude and phase — but they add like vectors, which is what makes them so useful in Section 7.2.

AC Voltage Applied to an Inductor

Connect the source to a coil of inductance $L$ whose resistance is negligible (a "pure" inductor). A changing current through the coil induces a back emf $-L\,\dfrac{di}{dt}$ (Chapter 6). The loop rule at each instant is $v_m\sin\omega t-L\dfrac{di}{dt}=0$, so $$\frac{di}{dt}=\frac{v_m}{L}\sin\omega t .$$ (Uses integration.) The current is a function whose derivative is $\dfrac{v_m}{L}\sin\omega t$: $$i=-\frac{v_m}{\omega L}\cos\omega t+\text{constant}.$$ The constant is a steady current that the source does not drive, so it is zero. Since $-\cos\omega t=\sin\!\left(\omega t-\dfrac\pi2\right)$, $$i=i_m\sin\!\left(\omega t-\frac{\pi}{2}\right),\qquad i_m=\frac{v_m}{\omega L}.$$

Why does no power go in? The instantaneous power is $p=vi=v_m\sin\omega t\cdot\left(-i_m\cos\omega t\right)$ $=-\tfrac12v_mi_m\sin2\omega t$, which is positive for one quarter-cycle and negative for the next. While $|i|$ grows the coil takes energy from the source and stores it as $\tfrac12Li^2$ in its magnetic field; while $|i|$ falls it hands all of it back. Physically, the lag comes from the back emf: the current can grow only while the applied voltage is positive, so it peaks when the voltage has just fallen to zero.

AC Voltage Applied to a Capacitor

Now connect the source across a capacitor $C$. No charge crosses the gap, but charge flows onto one plate and off the other as the voltage changes, so a current flows in the connecting wires. At every instant the charge is $q=Cv=Cv_m\sin\omega t$, and the current is $$i=\frac{dq}{dt}=\omega Cv_m\cos\omega t,$$ that is, $$i=i_m\sin\!\left(\omega t+\frac\pi2\right),\qquad i_m=\frac{v_m}{1/\omega C}.$$

The current is largest when the voltage is changing fastest — as $v$ passes through zero — and zero when $v$ is at its peak and momentarily steady. So the current reaches its peak a quarter-cycle before the voltage does. Again $p=vi=\tfrac12v_mi_m\sin2\omega t$ averages to zero: the energy $\tfrac12Cv^2$ stored in the electric field while the capacitor charges is returned while it discharges.

Exercises