sinusoidal current reactance and impedance

In a sinusoidal-current circuit the voltage and current vary harmonically: $u(t)=U_0\sin(\omega t)$, $i(t)=I_0\sin(\omega t-\varphi)$, where $\omega=2\pi f$ is the angular frequency.

A resistor provides the active resistance $R$: across it the current and voltage are in phase, and the amplitudes obey Ohm's law $U_R=I_0R$. An inductor has the inductive reactance $X_L=\omega L$, and its current lags the voltage by $\pi/2$; a capacitor has the capacitive reactance $X_C=\dfrac{1}{\omega C}$, and its current leads the voltage by $\pi/2$.

In a series connection the resistances cannot simply be added: the active and reactive parts form a right triangle, so the impedance is $Z=\sqrt{R^2+(X_L-X_C)^2},$ and the current amplitude is $I_0=\frac{U_0}{Z}.$

The phase shift is given by $\tan\varphi=\frac{X_L-X_C}{R}.$ If $X_L>X_C$, the circuit is inductive: $\varphi>0$ and the current lags the voltage. If $X_L<X_C$, the circuit is capacitive: $\varphi<0$ and the current leads the voltage.

When $X_L=X_C$, the reactances cancel each other: resonance occurs, $Z=R$ is minimal, and the current is maximal. The resonant frequency $\omega_0=\frac{1}{\sqrt{LC}}$ coincides with the Thomson formula for free oscillations in an LC circuit — the same relation between $L$ and $C$.

An inductor $L=50$ mH and a capacitor $C=20$ µF are connected to an AC circuit with frequency $f=50$ Hz. Find the inductive reactance $X_L$ and the capacitive reactance $X_C$.

A series RLC circuit has $R=30\ \Omega$, $X_L=70\ \Omega$ and $X_C=30\ \Omega$. The source voltage amplitude is $U_0=100$ V. Find the impedance $Z$, the current amplitude $I_0$ and the phase shift $\varphi$. Does the current lag or lead the voltage?

In a series RLC circuit the inductor has $L=0.2$ H and the resistor $R=25\ \Omega$. Choose the capacitance $C$ at which the circuit resonates at $f_0=50$ Hz, and state the impedance and the phase shift in this regime.

Safe self-check in a simulator. Build a series RLC circuit in an online AC simulator (for example, the Falstad Circuit Simulator or PhET "AC Electrical Circuits"). Use $R=30\ \Omega$, $L=50$ mH, $C=20$ µF and a source with amplitude $5$ V.

1. Compute $X_L$, $X_C$, $Z$, $I_0$ and $\varphi$ for $f=50$ Hz and compare with the simulator readings.
2. Sweep the frequency from $20$ to $300$ Hz, recording the current amplitude. Plot $I_0(f)$.
3. Find the frequency at which the current is maximal; compare it with $f_0=\dfrac{1}{2\pi\sqrt{LC}}$.
4. Repeat step 3 for $L=0.2$ H and check the value $C\approx 51$ µF from Example 3.

Do not build a real circuit connected to the $220$ V mains: it is dangerous. Carry out all observations only in a simulator or with a low-voltage source (up to $12$ V).

Show the geometric relation between the active and reactive resistances and explain why the impedance is minimal at resonance.

Two parts in one figure.

Left — the impedance phasor triangle: the horizontal leg $R$, the vertical leg $X_L-X_C$ (upward when $X_L>X_C$, downward when $X_L<X_C$), and the hypotenuse $Z=\sqrt{R^2+(X_L-X_C)^2}$; the angle between $R$ and $Z$ equals the phase shift $\varphi$.

Right — a graph of the quantities versus $\omega$: the decreasing curve $X_C=\dfrac{1}{\omega C}$, the increasing straight line $X_L=\omega L$, and the curve $Z(\omega)$ with its minimum $Z=R$ at the intersection $X_L=X_C$; the resonant frequency $\omega_0=\dfrac{1}{\sqrt{LC}}$ is marked on the axis.

A refrigerator compressor motor is plugged into a $230\ \text{V}$, $50\ \text{Hz}$ socket. Its winding has resistance $R = 40\ \Omega$ and inductance $L = 0.5\ \text{H}$, so at $50\ \text{Hz}$ ($\omega = 2\pi f = 314\ \text{rad/s}$) the inductive reactance is $X_L = \omega L = 157\ \Omega$ and the impedance is $Z = \sqrt{R^2 + X_L^2} = \sqrt{40^2 + 157^2} \approx 162\ \Omega$. The running current is $I = U/Z = 230/162 \approx 1.4\ \text{A}$, and it lags the voltage by $\varphi = \arctan(157/40) \approx 76^\circ$.

Both parts of the theory work here: inductive reactance $X_L = \omega L$ and impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$ with $X_C = 0$. It is the inductance, not the resistance, that gives $162\ \Omega$ instead of $40\ \Omega$, and it also fixes the phase shift through $\tan\varphi = X_L/R$.

In a phone the audio output reaches the $32\ \Omega$ headphones through a series coupling capacitor $C = 100\ \mu\text{F}$. On a bass note of $100\ \text{Hz}$ its reactance is $X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{2\pi \cdot 100 \cdot 100\times10^{-6}} \approx 16\ \Omega$, while on a $5\ \text{kHz}$ cymbal it is only $X_C \approx 0.3\ \Omega$. The impedance of the chain is $Z = \sqrt{32^2 + 16^2} \approx 36\ \Omega$ at $100\ \text{Hz}$ and $Z \approx 32\ \Omega$ at $5\ \text{kHz}$, so the bass loses a noticeable part of the voltage across the capacitor while the treble loses almost none.

This is capacitive reactance $X_C = 1/(\omega C)$, which falls as frequency rises, together with the impedance of a series $R$–$C$ circuit $Z = \sqrt{R^2 + X_C^2}$. The voltage division between $X_C$ and $R$ is what weakens the bass; the phase shift here is negative, $\tan\varphi = -X_C/R$, so the current leads the voltage.

An older AC tram motor is supplied at $380\ \text{V}$, $50\ \text{Hz}$; its field winding has $R = 6\ \Omega$ and $L = 0.15\ \text{H}$. Then $X_L = 2\pi \cdot 50 \cdot 0.15 = 47.1\ \Omega$, $Z = \sqrt{6^2 + 47.1^2} = 47.5\ \Omega$, the current is $I = 380/47.5 \approx 8.0\ \text{A}$, and the phase lag is $\varphi = \arctan(47.1/6) \approx 83^\circ$.

Resistance alone would give $380/6 \approx 63\ \text{A}$ and burn the winding; the current is held at $8\ \text{A}$ by $X_L = \omega L$ inside $Z = \sqrt{R^2 + X_L^2}$ and by $I = U/Z$. Because $X_L \gg R$, the current lags the voltage by almost $90^\circ$, the signature of a nearly purely inductive load.