7.4 RC/RL Time Constants

7.4 RC/RL Time Constants & Impedance Networks

Quick Answer: τ = RC or τ = L/R. In 1τ, charge/voltage (RC) or current (RL) moves about 63% of the way to the final value (37% remains when decaying). About 5τ to settle. Impedance |Z| = √[R² + (XL − XC)²] in series RLC. Resonance f0 = 1/(2π√(LC)); series |Z| → R only when XL = XC.

Topic 3-B finishes by connecting time-domain transients with frequency-domain impedance—exactly how radio circuits behave when you key a transmitter, charge a coupling network, or tune an LC stage.

RC time constant

For a resistor R and capacitor C:

$$\tau = R C$$

Charging (step toward final voltage $V_f$ on the capacitor, starting from 0):

$$V_C(t) = V_f \left(1 - e^{-t/\tau}\right)$$

Discharging toward zero from initial $V_0$:

$$V_C(t) = V_0 \, e^{-t/\tau}$$

Worked example — pool RC value

R = 10 kΩ, C = 100 µF. Time constant?

$$\tau = (10 \times 10^3)(100 \times 10^{-6}) = 1.0\,\mathrm{s}$$

After 1 s of charging from 0 toward a 12 V step, $V_C \approx 0.632 \times 12 \approx 7.6\,\mathrm{V}$. After 5 s, the capacitor is essentially at 12 V for exam purposes.

Worked example — shorter τ

R = 1 kΩ, C = 0.01 µF = 10 nF:

$$\tau = 10^3 \times 10^{-8} = 10^{-5}\,\mathrm{s} = 10\,\mu\mathrm{s}$$

Five time constants ≈ 50 µs—typical thinking for coupling and decoupling network recovery speeds.

RL time constant

For inductance L and resistance R:

$$\tau = \frac{L}{R}$$

Inductor current cannot jump; it rises or falls exponentially with the same 63%/37% and ~5τ rules, but the exponential applies to current (and to the inductor’s opposing voltage transient).

Worked example — RL

L = 5 H, R = 100 Ω (relay coil order-of-magnitude teaching example):

$$\tau = 5/100 = 0.05\,\mathrm{s} = 50\,\mathrm{ms}$$

Current is ~63% of final after 50 ms when a DC voltage is applied; ~5τ ≈ 250 ms to settle.

Why time constants matter in radio work

1. Keying and AGC networks set rise/fall with RC.
2. Power-supply filters charge and discharge with large RC products.
3. Relay and inductor kickback timing tracks L/R.
4. Coupling capacitors pass AC while blocking DC—but after a step, the DC path settles on a τ = RC schedule.

Impedance: R, XL, XC together

Resistance R is frequency-independent (ideal). Inductive reactance: $X_L = 2\pi f L$ (increases with f). Capacitive reactance: $X_C = 1/(2\pi f C)$ (decreases with f).

Impedance Z is the AC ratio of voltage to current, combining resistance and reactance. In rectangular form:

$$Z = R + j(X_L - X_C)$$

Magnitude for a series combination:

$$|Z| = \sqrt{R^2 + (X_L - X_C)^2}$$

Phase angle:

$$\theta = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$$

Worked example — series impedance magnitude

At a test frequency: R = 30 Ω, XL = 40 Ω, XC = 0 (series RL):

$$|Z| = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\,\Omega$$

If V_RMS = 100 V, then $I = V/|Z| = 2\,\mathrm{A}$. True power = $I^2 R = 120\,\mathrm{W}$; apparent power = $VI = 200\,\mathrm{VA}$.

Worked example — canceling reactances

R = 50 Ω, XL = 80 Ω, XC = 80 Ω:

$$|Z| = \sqrt{50^2 + (80-80)^2} = 50\,\Omega$$

Net reactance is zero; impedance equals R only—series resonance condition.

Series vs parallel RLC at resonance

Resonant frequency (ideal series or parallel LC):

$$f_0 = \frac{1}{2\pi \sqrt{LC}}$$

Exam hook: impedance of a series RLC circuit at resonance is equal to the resistance only (not zero unless R = 0; not equal to XL or XC alone).

Worked example — resonant frequency

L = 10 µH, C = 100 pF:

$$\sqrt{LC} = \sqrt{(10 \times 10^{-6})(100 \times 10^{-12})} = \sqrt{10^{-15}} = 3.162 \times 10^{-8}$$ $$f_0 = \frac{1}{2\pi \times 3.162 \times 10^{-8}} \approx 5.03\,\mathrm{MHz}$$

That 5.03 MHz result is a classic Element 3-style numeric. Distractors often misplace powers of ten (1.59 MHz, 15.9 MHz, 50.3 MHz).

Parallel impedance intuition (exam level)

Two pure resistances in parallel still use the resistor formula. Two pure reactances of the same type combine like parallel resistors but with j-sign bookkeeping. A parallel LC near resonance can present a very large impedance—the tank “rings” and draws little net line current at f0 (finite Q limited by losses).

For many Element 3 stems, you only need:

1. Series resonance → Z = R, current max.
2. Parallel resonance → Z high, line current min.
3. $f_0 = 1/(2\pi\sqrt{LC})$ for both ideal cases.

Combining networks step-by-step

1. Compute XL and XC at the operating frequency.
2. Series elements: add resistances; add reactances with sign (XL positive, XC negative in the jX sense).
3. Parallel elements: use reciprocal sums of complex admittances—or, for two pure equal/unlike cases, apply product-over-sum carefully.
4. Convert final Z to magnitude when the stem asks for ohms of impedance as a single number.
5. Apply Ohm’s Law for AC: (I = V / |Z|) for magnitude problems.

Bridging to later Element 3 topics

• Filters and IF transformers are resonant impedance networks with Q and bandwidth (Chapter on RLC resonant circuits).
• Oscillators use parallel tanks and time-constant bias networks.
• Power supplies use large RC filters (τ large → more smoothing, slower recovery).
• Antenna tuners cancel reactance so the transmitter sees a resistive match near 50 Ω.

Exam-day checklist for §7.4

1. τ = RC or L/R — units must produce seconds.
2. 1τ → ~63% / ~37%; ~5τ → settled.
3. Series |Z| = √[R² + (XL − XC)²].
4. f0 = 1/(2π√(LC)); series resonance Z → R.
5. XL rises with f; XC falls with f—opposites that can cancel.
6. Parallel tank at resonance → high impedance, not zero.

With Ohm’s Law networks, frequency/wavelength/waveform math, power and dB, and time-constant/impedance tools in place, you have completed Element 3 Topic 3-B electrical mathematics and are ready for passive and active component topics that apply these formulas on real parts.