Worked examples: RL on mains, RC high-pass, and an RLC tuning circuit

Example 1 — Series RL: 30 Ω + 127 mH on 50 Hz mains

A 30 Ω resistor sits in series with a 127 mH inductor across a 230 V, 50 Hz supply. Find $X_L$, the impedance magnitude and angle, the RMS current, and the voltages across each element.

1. Inductive reactance.
   $$X_L = 2\pi f L = 2\pi (50)(0.127) \approx 40\,\Omega$$
2. Impedance.
   $$Z = R + jX_L = 30 + j40 \;\Rightarrow\; |Z| = \sqrt{30^2 + 40^2} = 50\,\Omega, \;\varphi = \arctan(40/30) \approx 53.13°$$
   A classic 3-4-5 triangle.
3. RMS current.
   $$I = V_s / |Z| = 230 / 50 = 4.6\,\text{A}$$
   The current lags the source voltage by 53.13° (because $\varphi$ is positive).
4. Element voltages.
   $$V_R = RI = 30 \cdot 4.6 = 138\,\text{V}$$
   $$V_L = X_L I = 40 \cdot 4.6 = 184\,\text{V}$$
   Note $V_R + V_L = 322$ V — not 230 V. They sum as phasors, not as magnitudes: $\sqrt{138^2 + 184^2} \approx 230$ V, which matches the source. Phasor addition, not arithmetic addition.

Example 2 — Series RC: a 1 kHz high-pass filter

A 1 kΩ resistor is in series with a 100 nF capacitor. A 1 V signal at 1 kHz is applied. The capacitor is the “input” side, the resistor is the output — i.e. we take $V_\text{out} = V_R$. Find the magnitude and phase of the output.

1. Capacitive reactance.
   $$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi(1000)(10^{-7})} \approx 1591\,\Omega$$
2. Impedance.
   $$Z = R - jX_C = 1000 - j1591 \;\Rightarrow\; |Z| = \sqrt{1000^2 + 1591^2} \approx 1880\,\Omega$$
3. Voltage divider.
   $$V_\text{out} = V_\text{in} \cdot \frac{R}{Z} = 1 \cdot \frac{1000}{1000 - j1591}$$
   Magnitude: $1000 / 1880 \approx 0.532$. So $|V_\text{out}| \approx 0.53$ V. The signal gets through at ~53% of its input level — partly attenuated because we're not yet above the cutoff frequency.
4. Phase. $\angle Z = \arctan(-1591/1000) \approx -57.9°$. Output phase $= \angle(1000) - \angle Z = 0 - (-57.9°) = +57.9°$. Output leads the input — which is the fingerprint of an RC high-pass filter below its corner frequency. (Topic 9 will explore this in depth.)

Example 3 — Series RLC tuning: 50 Ω, 1 mH, 10 nF, swept

A series RLC filter with $R = 50$ Ω, $L = 1$ mH, and $C = 10$ nF is driven by a 1 V signal. What is the resonant frequency, and what current flows at resonance versus at 100 kHz and 1 MHz?

1. Resonant frequency.
   $$f_0 = \frac{1}{2\pi \sqrt{LC}} = \frac{1}{2\pi \sqrt{10^{-3} \cdot 10^{-8}}} \approx 50.3\,\text{kHz}$$
2. At f = 50.3 kHz (resonance). $X_L = X_C$, so $X = 0$, $|Z| = R = 50$ Ω, $I = 1/50 = 20$ mA. Peak current; the reactive opposition has vanished.
3. At f = 100 kHz (above resonance).
   $$X_L = 2\pi (10^5)(10^{-3}) \approx 628\,\Omega$$
   $$X_C = \frac{1}{2\pi(10^5)(10^{-8})} \approx 159\,\Omega$$
   $$Z = 50 + j(628 - 159) = 50 + j469 \;\Rightarrow\; |Z| \approx 471\,\Omega$$
   Current falls to $1/471 \approx 2.1$ mA — one tenth of the resonant peak. Net inductive (φ ≈ 84°).
4. At f = 1 MHz (well above). $X_L \approx 6283$ Ω, $X_C \approx 15.9$ Ω. The cap is effectively a short; the inductor dominates. $|Z| \approx 6283$ Ω, $I \approx 0.16$ mA. Almost nothing gets through.

That sharp peak at $f_0$ is exactly how AM-radio tuning works: the series RLC passes 820 kHz (or whatever frequency it's tuned to) and chokes off every other station.