E5B

Time constants and phase angle

Work out RC time constants, the phase angle of series RLC circuits, and convert impedance to admittance.

12 pool questions1 on the exam

Syllabus: Time constants and phase relationships: RL and RC time constants; phase angle in reactive circuits and components; admittance and susceptance

One time constant gets you 63.2% of the way

A capacitor charges through a resistor by closing a fixed fraction of the remaining gap every time constant. One τ is 63.2% charged, or 36.8% left when discharging.

Time constantsRC: τ = R × C (seconds). RL: τ = L ÷ R.

Each step of 1τ closes 63.2% of the remaining gap.

Resistance (R)100 kΩCapacitance (C)10 µF

τ = R × C = 100 kΩ × 10 µF1 s

Time	Charging	Discharging (left)
1τ	63.2%	36.8%
2τ	86.5%	13.5%
3τ	95.0%	5.0%
5τ	99.3%	0.7%

Combine parts first. Parallel capacitors add: 2 × 220 µF = 440 µF. Parallel equal resistors divide: 1 MΩ ÷ 2 = 0.5 MΩ. τ = 0.5 MΩ × 440 µF = 220 s.

Rememberτ = R × C. After one τ: 63.2% charged, or 36.8% left.

In an inductor voltage leads. In a capacitor current leads.

Memory hookELI the ICE man: voltage (E) leads current (I) in an inductor (L). Current (I) leads voltage (E) in a capacitor (C).

Each is 90° out of phase. In a series RLC circuit, subtract the reactances, then compare to R. The bigger reactance wins.

Phase angleθ = arctan((XL − XC) ÷ R)

Whichever wave peaks first is leading.

Resistance (R)100 ΩInductive reactance (XL)75 ΩCapacitive reactance (XC)25 Ω

θ = arctan((XL − XC) ÷ R) = arctan(50 ÷ 100)26.6° voltage leads

R, XL, XC	XL − XC	θ	Voltage
1000, 250, 500	−250	arctan(−0.25) = −14.0°	lags
100, 100, 300	−200	arctan(−2) = −63.4°	lags
100, 75, 25	+50	arctan(0.5) = +26.6°	leads

RememberNet inductive (positive): voltage leads. Net capacitive (negative): voltage lags.

Admittance is impedance turned upside down

Admittance (Y) is the inverse of impedance, in siemens. Its parts mirror impedance: conductance G (real, from R) and susceptance B (imaginary, from reactance).

AdmittanceY = 1 ÷ Z = G + jB

Admittance is the flip of impedance: invert the size, mirror the angle. Arrows are not to scale.

Impedance magnitude |Z|50 ΩImpedance angle30°

Y = G + jB17.3 − j10 mS

Impedance	→ Admittance
Polar Z = 50 Ω ∠30°	Y = 1 ÷ 50 = 0.02 S ∠−30°
Pure reactance X	susceptance replaced by its reciprocal, 1 ÷ X in size (sign flips)

RememberAdmittance: reciprocal of the magnitude, sign of the angle changed. Susceptance is its imaginary part, letter B.

Loading lesson…

Go deeper

Compendium articles on the ideas in this lesson, beyond what the exam asks.

Check what stuck

Practise the 12 pool questions for this group. Answers are shuffled, and every one comes with a one-line "why".

Practise E5B →