Egypt Secondary 3 Physics, Week 15: Review: AC circuits, reactance, impedance and resonance

Term 1 review: circuits, magnetism, induction and AC

Week 15 · Term 1 review (Chapters 1–4) and Darslab’s own review tasks · 3 lessons

Review: AC circuits, reactance, impedance and resonance

Physics Sec 3 Term 1, Chapters 1–4: Review

GoalCombine reactance, impedance, phase angle, power and resonance in multi-step tasks on series AC circuits.

New words

net reactance · X = XL − XC: positive for an inductive circuit, negative for a capacitive onephase angle · tan θ = X ÷ R; in this course θ is the angle by which the voltage is ahead of the currentresonant frequency · f₀ = 1 ÷ (2π√(LC)), where XL = XC

Explanation

Chapter 4 adds the AC toolbox. Start by finding the reactances at the given frequency: XL = 2πfL and XC = 1 ÷ (2πfC), with π = 3.14, mH and µF changed to H and F. Then the net reactance X = XL − XC, the impedance Z = √(R² + X²), the current I = V ÷ Z (effective values), the phase angle tan θ = X ÷ R and the power P = I²R. Never add R, XL and XC as plain numbers, and never add VR, VL and VC as plain numbers: R is at right angles to the reactances, and XL and XC point opposite ways. If X is positive the circuit is inductive and the voltage leads the current; if X is negative it is capacitive and the voltage lags; if X = 0 the circuit is at resonance, Z = R, the current is largest, and f₀ = 1 ÷ (2π√(LC)). Radio tuning changes C so that f₀ matches the wanted station. Take the frequency of the Egyptian supply as 50 Hz.

Topic	Tool
frequency	T = 1 ÷ f; 50 Hz → T = 0.02 s
reactances (π = 3.14)	XL = 2πfL; XC = 1 ÷ (2πfC)
series circuits	Z = √(R² + (XL − XC)²); tan θ = (XL − XC) ÷ R; I = V ÷ Z
phase	coil: voltage leads by 90°; capacitor: voltage lags by 90°; resistor: in phase
power	P = I²R
resonance	XL = XC; f₀ = 1 ÷ (2π√(LC)); Z = R

Worked examples

A series circuit has R = 30 Ω, XL = 70 Ω and XC = 30 Ω on a 150 V supply. X = 70 − 30 = 40 Ω. Z = √(30² + 40²) = 50 Ω. I = 150 ÷ 50 = 3 A. P = I²R = 9 × 30 = 270 W. tan θ = 40 ÷ 30 = 4/3, θ = 53°. X is positive, so the circuit is inductive and the voltage leads the current.

R = 30 Ω, L = 0.2 H and C = 50 µF on a 120 V, 50 Hz supply. XL = 2 × 3.14 × 50 × 0.2 = 62.8 Ω. XC = 1 ÷ (2 × 3.14 × 50 × 50 × 10⁻⁶) = 63.7 Ω. X = 62.8 − 63.7 = −0.9 Ω, almost zero. Z = √(30² + 0.9²) = 30.0 Ω and I = 120 ÷ 30.0 = 4.0 A. The resonant frequency is f₀ = 1 ÷ (2 × 3.14 × √(0.2 × 50 × 10⁻⁶)) = 50.4 Hz, so the supply is almost exactly at resonance and the circuit behaves like the resistor alone.

A series RC circuit has R = 40 Ω and C = 40 µF on a 100 V, 50 Hz supply. XC = 1 ÷ (2 × 3.14 × 50 × 40 × 10⁻⁶) = 79.6 Ω. Z = √(40² + 79.6²) = 89.1 Ω. I = 100 ÷ 89.1 = 1.12 A. VR = 1.12 × 40 = 44.9 V and VC = 1.12 × 79.6 = 89.4 V (their plain sum 134 V is more than the supply, but √(44.9² + 89.4²) = 100 V). tan θ = −79.6 ÷ 40 = −1.99, so the voltage lags the current by about 63°.

Class activity

Paper only. Make a one-page summary of Chapter 4 with the table of tools. Then, for R = 20 Ω, fill a table with three rows, (XL, XC) = (45, 20), (20, 45) and (30, 30) Ω, giving X, Z, whether the circuit is inductive, capacitive or at resonance, and whether the voltage leads, lags or is in phase.

Practice

R = 24 Ω, L = 0.15 H and C = 80 µF are in series on a 100 V, 50 Hz supply (π = 3.14). Find XL, XC, Z and I.

A coil of L = 0.5 H is to be in resonance with a capacitor on the 50 Hz supply (π = 3.14). Find the capacitance.

Spot the error. A student says: “At resonance the impedance of the series LCR circuit is the largest, so the current is the largest.” Correct the statement.

R = 40 Ω and XL = 30 Ω are in series on a 100 V supply. Find Z, I, VR, VL and the power, and check that VR and VL give 100 V.