Every medium-wave station of All India Radio within range sends its signal into the aerial of a small transistor radio at the same moment. The aerial carries all of them together. Yet a turn of the tuning knob brings in one station clearly and silences the rest. Turn a little further, and a different station comes in alone.
When the reactances cancel
In the last lesson \(X_L\) and \(X_C\) pointed opposite ways and partly cancelled. \(X_L = \omega L\) rises with frequency and \(X_C = 1/\omega C\) falls. So at one frequency the two are equal and cancel completely. There \(Z = \sqrt{R^2 + 0} = R\), the smallest impedance the circuit can have. The current \(V/R\) is then the largest it can be, and \(\tan\phi = 0\), so the current is in phase with the supply. This is resonance.
\[ X_L = X_C \]
Why. resonance is the frequency at which the two reactances cancel.
\[ \omega L = \dfrac{1}{\omega C} \]
Why. write out the two reactances.
?
Your turn. multiply both sides by \(\omega\) and divide by \(L\). Which line is it?
\[ \begin{gathered} \omega_0 = \dfrac{1}{\sqrt{LC}} \\ f_0 = \dfrac{1}{2\pi\sqrt{LC}} \end{gathered} \]
Read it. take the positive root; divide by \(2\pi\) for the frequency in hertz.
The resonant frequency depends only on \(L\) and \(C\). The resistance does not appear in it. The resistance decides only how large the current becomes at resonance. This is the electrical twin of the forced oscillator in the chapter Superposition, Damped and Forced Oscillations. There the steady swing peaked when the push matched the natural frequency \(\sqrt{k/m}\), and the damping set the height of the peak.
Slide the frequency in the figure above with \(R = 10\,\Omega\). At \(800\,\text{rad/s}\) the circuit is capacitive, \(Z = 46\,\Omega\) and \(I = 0.43\,\text{A}\), leading by \(78°\). At \(1000\,\text{rad/s}\) the current peaks at \(2\,\text{A}\) with \(\phi = 0\). At \(1200\,\text{rad/s}\) it has fallen to \(0.53\,\text{A}\), now lagging by \(75°\). Below resonance the capacitor's reactance wins; above it the inductor's wins.
A series LCR circuit is driven at a frequency below its resonant frequency. How does the current compare with the supply voltage?
A \(0.1\,\text{H}\) inductor, a \(10\,\mu\text{F}\) capacitor and a \(10\,\Omega\) resistor are in series across \(20\,\text{V}\) rms of variable frequency. Find the resonant angular frequency and frequency, and the impedance, current and phase angle at resonance.
\(\omega_0\) \(= 1/\sqrt{0.1 \times 10^{-5}}\) \(= 1/\sqrt{10^{-6}}\) \(= 1000\,\text{rad/s}\), and \(f_0\) \(= 1000/2\pi\) \(= 159\,\text{Hz}\).
At resonance \(Z = R = 10\,\Omega\), so \(I = 20/10 = 2\,\text{A}\), the largest possible, and \(\phi = 0\).
Values used: the numbers given, with \(\pi = 3.14\).
A \(0.5\,\text{H}\) inductor and a \(20\,\mu\text{F}\) capacitor are in series with a resistor. Find the resonant frequency, in hertz.
A series LCR circuit is driven at its resonant frequency. Which statements are true?
Voltages at resonance
At resonance the current in the circuit of the example is \(2\,\text{A}\). The inductor's voltage is \(V_L\) \(= IX_L\) \(= 2 \times 1000 \times 0.1\) \(= 200\,\text{V}\). The capacitor's is \(V_C = IX_C = 200\,\text{V}\) too. So each is ten times the \(20\,\text{V}\) supply. They are equal and exactly opposite, so they cancel. The supply only has to drive the current through \(R\), and \(V_R = 20\,\text{V}\) equals the supply.
A large voltage across the inductor or the capacitor at resonance is a real hazard. A capacitor rated for \(50\,\text{V}\) in this circuit would fail, although the supply is only \(20\,\text{V}\). Energy sloshes between the magnetic field of the coil and the electric field of the capacitor. The supply need only make up what the resistor turns into heat.
A \(0.5\,\text{H}\) inductor, a \(20\,\mu\text{F}\) capacitor and a \(50\,\Omega\) resistor are in series across \(100\,\text{V}\) rms at the resonant frequency. Find the rms voltage across the inductor, in volts.
How sharp is the peak?
Switch the figure above to \(R = 40\,\Omega\). The peak stays at \(1000\,\text{rad/s}\), since \(\omega_0\) contains no \(R\). But its height falls to \(20/40 = 0.5\,\text{A}\), and it spreads out. Away from resonance the two curves nearly meet, because there the reactance, not \(R\), limits the current. So a small resistance gives a tall, narrow peak, and a large one a low, broad peak.
The width of the peak is measured between the two frequencies where the current has fallen to \(1/\sqrt2\) of its peak. At those points the power \(I^2R\) is half its peak value, so they are called the half-power points. The gap between them is the bandwidth, \(\Delta\omega = R/L\). The sharpness is measured by the quality factor, \[ Q = \dfrac{\omega_0 L}{R} = \dfrac{\omega_0}{\Delta\omega}. \] A high \(Q\) means a narrow peak. It also means large voltages at resonance, since \(V_L/V\) \(= I\omega_0L/(IR)\) \(= Q\).
For the circuit of the example, \(Q\) \(= 1000 \times 0.1/10\) \(= 10\) and \(\Delta\omega\) \(= 1000/10\) \(= 100\,\text{rad/s}\). The half-power points are at about \(951\) and \(1051\,\text{rad/s}\), where the current is \(1.41\,\text{A}\). With \(40\,\Omega\), \(Q = 2.5\) and \(\Delta\omega = 400\,\text{rad/s}\). The curve is four times as broad and its peak four times lower.
Match the circuit to a block on a spring. The inductance \(L\) is the mass: it resists a change of current as mass resists a change of speed. The capacitance plays the part of \(1/k\): a small capacitor is a stiff spring. The resistance is the damping. So \(\omega_0 = 1/\sqrt{LC}\) is the twin of \(\sqrt{k/m}\), and a small \(R\) gives a sharp peak just as light damping does.
A series circuit has \(L = 0.1\,\text{H}\), \(C = 10\,\mu\text{F}\) and \(R = 20\,\Omega\). Find its quality factor.
The resistance in a series LCR circuit is increased, with \(L\) and \(C\) unchanged. What happens to the resonance curve?
Tuning a radio
The aerial feeds the tuned circuit of the radio with the signals of every station at once, each at its own frequency. The circuit is a coil and a variable capacitor. Turning the knob changes \(C\), and so moves \(\omega_0\). The station whose frequency equals \(\omega_0\) drives a large current in the circuit. Every other station meets a large impedance and drives almost nothing.
A radio's tuned circuit has a \(1\,\text{mH}\) coil and a resistance of \(20\,\Omega\). Find the capacitance that tunes it to a station at \(1\,\text{MHz}\). Find also its \(Q\) and its bandwidth in kilohertz.
(a) \(C\) \(= \dfrac{1}{(2\pi f_0)^2 L}\) \(= \dfrac{1}{(6.28 \times 10^6)^2 \times 10^{-3}}\) \(= 2.5 \times 10^{-11}\,\text{F}\) \(= 25\,\text{pF}\).
(b) \(Q\) \(= \omega_0L/R\) \(= 6.28 \times 10^6 \times 10^{-3}/20\) \(= 314\). The bandwidth is \(f_0/Q\) \(= 10^6/314\) \(= 3.2\,\text{kHz}\).
Medium-wave stations broadcast between about \(530\) and \(1600\,\text{kHz}\). With the \(1\,\text{mH}\) coil, the knob must sweep the capacitor from about \(90\,\text{pF}\) down to \(10\,\text{pF}\). Neighbouring stations are only a few kilohertz apart. A bandwidth of \(3\,\text{kHz}\) lets the wanted station in and keeps its neighbours out. A low-\(Q\) circuit would let two stations through at once.
A radio is tuned to a station at \(600\,\text{kHz}\). To tune it to a station at \(1200\,\text{kHz}\) with the same coil, the capacitance must be
Back to the radio on the table. The aerial carries every station, but the tuned circuit resonates with only one. Only that one drives a large current, which the radio amplifies. Turning the knob changes \(C\), moves the resonant frequency to another station, and brings that station in instead.
A coil and a capacitor can also be connected in parallel across the supply. Then the currents in the two branches are opposite in phase and cancel at \(\omega_0 = 1/\sqrt{LC}\). The current drawn from the supply falls to a minimum, and the impedance is at its largest. Such a parallel circuit is called a rejector circuit, and it is used to block one frequency. The series circuit of this lesson, whose current is largest at \(\omega_0\), is called an acceptor circuit.
Early wireless stations sent out a spark that spread over a wide band of frequencies. Two stations close together jammed each other, and any receiver heard them both. On 26 April 1900 Guglielmo Marconi filed British patent number 7777, later called the four sevens. It tuned the sending and the receiving circuits, each a coil with a capacitor, to one resonant frequency. A station could then speak to its own receiver while others worked nearby. Every radio dial since has used the same resonance.
- Saying that \(X_L\) and \(X_C\) are zero at resonance.They are equal, not zero. Only their difference is zero, so \(Z = R\).
- Putting \(R\) into the resonant frequency.\(\omega_0 = 1/\sqrt{LC}\) depends on \(L\) and \(C\) only. \(R\) sets the height and width of the peak.
- Thinking no voltage can exceed the supply.At resonance \(V_L = V_C = QV\), which can be many times the supply.
- Giving \(f_0\) as \(1/\sqrt{LC}\).\(1/\sqrt{LC}\) is \(\omega_0\) in rad/s. Divide by \(2\pi\) for the frequency in hertz.