How a Radio Picks One Station Out of Many

Tuning a radio, stepping voltage down for a phone charger and generating power at a dam all rely on AC circuits combining resistance, inductance and capacitance. Their combined opposition — impedance — and the special case of resonance explain how.

This part covers impedance and phase in a series LCR circuit, resonance, power and power factor, and AC generators and transformers.

**In a series LCR circuit the same current flows through all three elements but their voltages are out of phase, so adding the phasors gives the impedance $Z=\sqrt{R^2+(X_L-X_C{)}^2}$ and the phase angle $\tan \varphi =\frac{X_L-X_C}{R}$.

Phasor addition:**

- $V_R$ is in phase with the current
- $V_L$ leads the current by $90^{\circ }$ and $V_C$ lags it by $90^{\circ }$
- $V_L$ and $V_C$ point in opposite directions, so their difference combines at right angles with $V_R$

$$Z=\sqrt{R^2+(X_L-X_C{)}^2},\tan \varphi =\frac{X_L-X_C}{R}$$

If $X_L>X_C$ the circuit is inductive and the current lags; if $X_L<X_C$ it is capacitive and the current leads.

Worked example. $R=30\Omega$, $X_L=80\Omega$ and $X_C=40\Omega$ on a $100$ V RMS supply:

$$Z=\sqrt{30^2+40^2}=50\Omega ,I=\frac{100}{50}=2\text{ A},\tan \varphi =\frac{40}{30}\Rightarrow \varphi \approx 53^{\circ }$$

An everyday example. **Adding the readings of voltmeters placed across $R$, $L$ and $C$ can give more than the supply voltage — because the voltages are out of phase and must be added as phasors.

The substance. The voltage across $L$ or $C$ alone can exceed the supply voltage**, even though the two partly cancel.

**Resonance occurs when the inductive and capacitive reactances are equal and cancel, so the impedance falls to its minimum value $R$ and the current is largest; this happens at ${\omega }_0=\frac{1}{\sqrt{LC}}$, or $f_0=\frac{1}{2\pi \sqrt{LC}}$.

Condition:**

$$X_L=X_C\Rightarrow {\omega }_{0}L=\frac{1}{{\omega }_{0}C}\Rightarrow {\omega }_0=\frac{1}{\sqrt{LC}}$$

At resonance, $Z=R$, $\varphi =0$, and the current is $\frac{V}{R}$.

Sharpness. A smaller $R$ gives a taller, narrower current peak; the quality factor is $Q=\frac{{\omega }_{0}L}{R}$.

Worked example. For $L=0.1$ H and $C=10\mu$F:

$${\omega }_0=\frac{1}{\sqrt{0.1\times 10^{-5}}}=1000{\text{ rad s}}^{-1},f_0=\frac{1000}{2\pi }\approx 159\text{ Hz}$$

With $R=10\Omega$ and a $20$ V supply, the current at resonance is $2$ A, and $Q=\frac{1000\times 0.1}{10}=10$.

An everyday example. Turning a radio's tuning knob changes the capacitance until the circuit resonates at one station's frequency, picking it out from all the others.

The substance. **Resonance needs both $L$ and $C$** — an RL or RC circuit has no resonant frequency.

**The average power in an AC circuit is $\overline{P}=VI\cos \varphi$, with RMS values $V$ and $I$ and power factor $\cos \varphi$; the part of the current $90^{\circ }$ out of phase with the voltage delivers no average power and is called wattless current.

Derivation.** With $v=v_m\sin \omega t$ and $i=i_m\sin (\omega t+\varphi )$, the instantaneous power averages over a cycle to

$$\overline{P}=\frac{v_m{i}_m}{2}\cos \varphi =VI\cos \varphi$$

Power factor, $\cos \varphi =\frac{R}{Z}$:

- Pure resistor — $\cos \varphi =1$, the maximum power
- Pure inductor or capacitor — $\cos \varphi =0$, no average power
- At resonance — $\cos \varphi =1$

Wattless current. Resolving the current along and perpendicular to the voltage, the component $I\sin \varphi$ carries no average power.

Worked example. For the circuit above, with $Z=50\Omega$, $R=30\Omega$, $100$ V and $2$ A:

$$\cos \varphi =\frac{30}{50}=0.6,\overline{P}=100\times 2\times 0.6=120\text{ W}$$

Check: $I^{2}R=4\times 30=120$ W.

An everyday example. Factories with many large motors connect capacitors to their supply to raise the power factor and cut the extra current that wastes energy in the wires.

The substance. All the average power is used in the resistance — inductors and capacitors only store energy and give it back.

**An AC generator rotates a coil in a magnetic field so the changing flux induces $\epsilon =NBA\omega \sin \omega t$, while a transformer uses mutual induction between two coils on a shared iron core, giving $\frac{V_s}{V_p}=\frac{N_s}{N_p}$ — more secondary turns step the voltage up, fewer step it down.

AC generator:**

- A coil of $N$ turns and area $A$ rotates at angular speed $\omega$ in a field $B$
- The flux linkage $NBA\cos \omega t$ changes, inducing $\epsilon =NBA\omega \sin \omega t$ with peak ${\epsilon }_0=NBA\omega$
- Slip rings and brushes lead the current out

Worked example — generator. $N=50$, $A=0.1$ m${}^2$ and $B=0.5$ T, rotating at $50$ revolutions per second:

$${\epsilon }_0=50\times 0.5\times 0.1\times 100\pi \approx 785\text{ V}$$

Transformer:

- Primary and secondary coils are wound on a laminated soft iron core
- For an ideal transformer, $\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}$
- Step-up: $N_s>N_p$; step-down: $N_s<N_p$

Worked example — transformer. A step-down transformer with $2300$ primary turns converts $230$ V to $5$ V, so its secondary has $2300\times \frac{5}{230}=50$ turns.

Losses. Eddy currents (reduced by laminations), hysteresis, heating of the windings and flux leakage.

An everyday example. **Electricity travels across the country at very high voltage and is stepped down near homes to $230$ V**, keeping the current, and the $I^{2}R$ loss, small on long lines.

The substance. A transformer cannot work on steady DC, because a constant current produces no changing flux.