Resonance

Pushing a swing

Push a child on a swing at the right rhythm, and small pushes build up a big swing. Push at the wrong rhythm, and the swing hardly moves. A series circuit does the same with its current. For one special frequency the current is far larger than at any other. This is resonance.

The condition for resonance

Keep the source voltage fixed and change only the frequency. The current is $I = V/Z$. It is largest when $Z$ is smallest. In $Z = \sqrt{R^2 + (X_L - X_C)^2}$ the smallest value is $R$, and it happens when the two reactances are equal:

Solve for the angular frequency. We want $\omega_0$, so $\omega_0 = \dfrac{1}{\sqrt{LC}}$. The frequency in hertz is

At resonance three things happen together:

• The impedance is the smallest it can be, $Z = R$.
• The current is the largest it can be, $I_{max} = V/R$.
• The phase angle is $\phi = 0$. The current is in step with the voltage, and the circuit acts like a plain resistor.

Take $L = 0.10$ H and $C = 10\ \mu\text{F}$. Then $LC = 1.0 \times 10^{-6}\ \text{s}^2$, so $\omega_0 = 1000$ rad/s and $f_0 = 159$ Hz.

The voltages at resonance

Look at the voltages for the same circuit with $R = 20\ \Omega$ and a source of $10$ V (rms).

1. The current. $I = \dfrac{V}{R} = \dfrac{10\ \text{V}}{20\ \Omega} = 0.50$ A.
2. The coil. $X_L = \omega_0L = 1000\ \text{rad/s} \times 0.10\ \text{H} = 100\ \Omega$. So $V_L = IX_L = 0.50\ \text{A} \times 100\ \Omega = 50$ V.
3. The capacitor. $X_C = 100\ \Omega$ too, so $V_C = 50$ V.

The coil and the capacitor each hold $50$ V, which is five times the source voltage! There is no mistake. At every instant $v_L$ and $v_C$ are exactly opposite, so they cancel. The source only has to supply the $10$ V across the resistor. The two large voltages go back and forth between the coil and the capacitor. A real circuit with a large $Q$ and a strong source can give a dangerous voltage across its capacitor.

The quality factor

The ratio of $V_L$ to the source voltage is called the quality factor $Q$:

It is a pure number. In our circuit $Q = \dfrac{100\ \Omega}{20\ \Omega} = 5$. At resonance, the voltage across the coil, and across the capacitor, is $Q$ times the source voltage.

How sharp is the peak?

Is the resonance a spike or a broad hill? Look at the frequencies where the current falls to $\dfrac{1}{\sqrt{2}}$ of its largest value. The power in the resistor is $I^2R$, so there the power is half of its largest value. We call them the half-power frequencies.

The current falls to $I_{max}/\sqrt{2}$ when $Z = \sqrt{2}\,R$. That means $|X_L - X_C| = R$. Close to $\omega_0$, a small step $\delta$ in the angular frequency changes the net reactance by $2L\delta$. Setting $2L\delta = R$ gives $\delta = R/2L$. So the two half-power frequencies lie close to $\omega_0 \pm \dfrac{R}{2L}$. The gap between them is the bandwidth, and an exact calculation gives the same value:

In hertz, $\Delta f = \dfrac{f_0}{Q}$. A large $Q$ means a narrow peak. A circuit with a large $Q$ is selective: it picks out one frequency and ignores the others close to it. A small resistance gives a large $Q$.

For our circuit, $\Delta\omega = \dfrac{R}{L} = \dfrac{20\ \Omega}{0.10\ \text{H}} = 200$ rad/s. That is $\Delta f = 31.8$ Hz, which is $159\ \text{Hz}/5$. The current is above $71\ \%$ of its largest value only between about $144$ Hz and $176$ Hz.

Set $R = 5\ \Omega$ and then $R = 40\ \Omega$. Watch $Q$ and the width of the solid curve. Which gives the more selective circuit? Now change $L$ and $C$ and watch the position of the peak, but not its width, when $R$ is kept the same.

Tuning a radio

A radio is a resonance machine. Many stations send out waves at once, and each makes a tiny voltage of its own frequency in the aerial. The tuning circuit is a coil and a variable capacitor. Turning the knob changes $C$ until $f_0$ matches the station. At resonance the current of that station is large, and the others are weak.

A worked example. A coil of $250\ \mu\text{H}$ is used to tune a station at $1000$ kHz. Find the capacitance. The resistance of the circuit is $10\ \Omega$. Find $Q$ and the bandwidth.

1. The angular frequency. $\omega_0 = 2\pi f_0 = 2\pi \times 1.0 \times 10^6\ \text{Hz} = 6.28 \times 10^6\ \text{rad/s}$.
2. The capacitance. At resonance $\omega_0^2LC = 1$. We want $C$, so $C = \dfrac{1}{\omega_0^2L}$. The bottom is $(6.28 \times 10^6\ \text{s}^{-1})^2 \times 250 \times 10^{-6}\ \text{H} = 9.87 \times 10^{9}$ in SI units. So $C = 1.01 \times 10^{-10}\ \text{F} = 101\ \text{pF}$.
3. The quality factor. The coil has $X_L = \omega_0L = 6.28 \times 10^6\ \text{rad/s} \times 250 \times 10^{-6}\ \text{H} = 1571\ \Omega$. So $Q = \dfrac{X_L}{R} = \dfrac{1571\ \Omega}{10\ \Omega} = 157$.
4. The bandwidth. $\Delta f = \dfrac{f_0}{Q} = \dfrac{1000\ \text{kHz}}{157} = 6.4$ kHz.

The circuit lets through a band about $6.4$ kHz wide. A station $20$ kHz away is far outside it, and its current is small.

Practise with new numbers

Try it in Python

The program sweeps the frequency of the source from $60$ Hz to $280$ Hz and prints the current as a row of bars. It then compares the largest current with the formulas.

Common mistakes

• Saying the impedance is zero at resonance. It is $R$. The reactances cancel, but the resistance stays.
• Thinking the voltage across the coil is zero at resonance. The coil and the capacitor each have $Q$ times the source voltage. Only their difference is zero.
• Saying a large $R$ gives a sharp peak. It is the other way round. A small $R$ gives a large $Q$ and a sharp peak.
• Giving $\omega_0$ in hertz. The formula $\omega_0 = 1/\sqrt{LC}$ gives rad/s. Divide by $2\pi$ for hertz.
• Forgetting the prefixes. mH is $10^{-3}$ H, $\mu\text{F}$ is $10^{-6}$ F, pF is $10^{-12}$ F.
• Thinking the peak current depends on $L$ and $C$. At resonance $I_{max} = V/R$. Only the position of the peak depends on $L$ and $C$.

Remember these

• Resonance: $X_L = X_C$, so $\omega_0 = \dfrac{1}{\sqrt{LC}}$ and $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$.
• At resonance $Z = R$, $I_{max} = V/R$ and $\phi = 0$.
• $Q = \dfrac{\omega_0L}{R} = \dfrac{1}{\omega_0CR} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}$. At resonance $V_L = V_C = QV$.
• Bandwidth: $\Delta\omega = \dfrac{R}{L}$ and $\Delta f = \dfrac{f_0}{Q}$.
• A small $R$ gives a large $Q$, a narrow peak and a selective circuit.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A coil of $5.0$ mH and a capacitor of $20\ \mu\text{F}$ are in series. What is the resonant angular frequency?

2. At resonance the source voltage is $6.0$ V and the voltage across the capacitor is $90$ V. What is $Q$?

3. The resistance of a tuned circuit is doubled, with $L$ and $C$ unchanged. What happens to $f_0$, to $Q$ and to the bandwidth?

4. Why can the voltage across a capacitor in a series circuit be larger than the source voltage at resonance?