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Alternating Current

AC circuits, impedance, LCR series circuit, resonance, transformers, and power factor.

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Reading time~8 min
Revision time~3 min
Last updated2026-07-19
1Read the chapter~8 min

🎯 Key Points

  • V_rms=V₀/√2; for R: V,I in phase; for L: I lags V by 90° (X_L=ωL); for C: I leads V by 90° (X_C=1/ωC)
  • Impedance Z=√(R²+(X_L−X_C)²); resonance when X_L=X_C → ω₀=1/√(LC), Z=R (minimum), current is maximum
  • Average power P=V_rms·I_rms·cosφ; power factor cosφ=R/Z (=1 for pure R, =0 for pure L or C — no real power dissipated)
  • Transformer: V₂/V₁=N₂/N₁=I₁/I₂ — step-up increases voltage but decreases current proportionally (power conserved ideally)
  • Quality factor Q=ω₀L/R — higher Q means sharper, more selective resonance (used in radio tuning)
V and I Phase Relationships in AC CircuitsPure Inductor: I lags V by 90°VIPure Capacitor: I leads V by 90°VIMemory aid "ELI the ICE man": in an inductor (L) E(V) leads I; in a capacitor (C) I leads E(V)For a pure resistor, V and I stay perfectly in phase (no lag or lead at all)

In a pure inductor the current lags the voltage by 90° (energy is briefly stored in the magnetic field before current responds); in a pure capacitor the current leads the voltage by 90° (current must flow to build up charge before voltage rises) — the mnemonic "ELI the ICE man" keeps these straight.

AC Basics

  • v = V₀·sin(ωt); i = I₀·sin(ωt + φ)
  • RMS values: V_rms = V₀/√2; I_rms = I₀/√2
  • Angular frequency: ω = 2πf

Pure Elements in AC

  • Resistor R: V and I in phase; V_R = IR
  • Inductor L: I lags V by 90°; X_L = ωL = 2πfL (inductive reactance)
  • Capacitor C: I leads V by 90°; X_C = 1/ωC = 1/2πfC (capacitive reactance)

Series LCR Circuit

  • Impedance: Z = √(R² + (X_L - X_C)²)
  • Phase angle: tan(φ) = (X_L - X_C)/R
  • Current: I = V/Z
  • Resonance: X_L = X_C → ω₀ = 1/√(LC); Z = R (minimum), I is maximum
  • Quality factor: Q = ω₀L/R = 1/ω₀CR = (1/R)√(L/C)
Series LCR circuit diagram with an AC voltage source V driving current I through a resistor R, an inductor L and a capacitor C connected one after another in a single loop

Series LCR circuit: Z = √(R² + (X₃ − X₊)²), and at resonance X₃ = X₊ so Z = R and the current is maximum. Image: V4711, CC BY-SA 3.0, via Wikimedia Commons.

Power in AC

  • Instantaneous power: p = vi
  • Average power: P_avg = V_rms·I_rms·cos(φ)
  • Power factor: cos(φ) = R/Z (= 1 for resistor, 0 for pure L or C)
  • Wattless current: component of I that contributes no power

Transformers

  • V₂/V₁ = N₂/N₁ = I₁/I₂ (ideal transformer)
  • Step-up: N₂ > N₁ (voltage increases, current decreases)
  • Step-down: N₂ < N₁ (voltage decreases, current increases)
  • Energy losses: copper loss (I²R), iron loss (eddy currents, hysteresis)
Transformer diagram showing a laminated core, a primary winding of N_P turns carrying primary current I_P at primary voltage V_P, magnetic flux circulating around the core, and a secondary winding of N_S turns delivering secondary current I_S at secondary voltage V_S

Transformer: flux linked through a common core gives Vₛ/Vₚ = Nₛ/Nₚ, and for an ideal transformer VₚIₚ = VₛIₛ. Image: BillC, CC BY-SA 3.0, via Wikimedia Commons.

RMS and Average Values

  • RMS (root mean square) value is the value of steady current that produces the same heating effect as the AC over a full cycle; for sinusoidal AC, I_rms = I₀/√2 ≈ 0.707 I₀
  • Average value of AC over a full cycle is zero; average over half cycle: I_avg = 2I₀/π ≈ 0.637 I₀
  • Form factor = RMS value / average value (over half cycle) = π/(2√2) ≈ 1.11 for sinusoidal AC

LC Oscillations

  • A charged capacitor connected to an inductor (no resistance) produces electrical oscillations analogous to SHM
  • Angular frequency of free oscillation: ω = 1/√(LC), same as the series resonance condition
  • Energy oscillates between the electric field of the capacitor and the magnetic field of the inductor, with total energy conserved (in the ideal, resistance-free case)

Half Power Frequencies and Sharpness of Resonance

  • At resonance, current is maximum (I₀ = V/R) and the LCR circuit behaves purely resistively
  • Sharpness of resonance is measured by quality factor Q; higher Q means a narrower resonance curve and better frequency selectivity (important in radio tuning circuits)
  • Bandwidth of resonance: Δω = ω₀/Q (the smaller the bandwidth, the sharper the resonance)

Frequency Dependence of Reactance

  • Inductive reactance X_L = ωL = 2πfL increases linearly with frequency — an inductor blocks high frequencies but passes DC freely (at f = 0, X_L = 0, so it behaves like a plain wire)
  • Capacitive reactance X_C = 1/ωC = 1/(2πfC) decreases with frequency — a capacitor passes high frequencies but blocks DC (at f = 0, X_C is infinite, so no steady current flows through it)
  • This opposite behaviour is why an inductor is used as a choke to filter high-frequency noise, while a capacitor is used to block DC and couple AC signals between stages

Phasor Representation of AC Quantities

  • A phasor is a rotating vector whose length represents the peak value of an AC quantity and whose angle represents its instantaneous phase; its projection on the vertical axis gives the instantaneous value
  • Because V and I are generally out of phase, they are drawn as phasors separated by the phase angle φ, and the voltages across R, L, and C are combined as vectors rather than as plain numbers
  • In a series LCR circuit the V_R phasor lies along the current, V_L leads it by 90°, and V_C lags it by 90°; combining them geometrically gives the resultant voltage and the impedance triangle (R, X_L − X_C, Z)

Power Factor and AC Power Transmission

  • The power factor cosφ = R/Z is the fraction of the apparent power (V_rms·I_rms) that is converted into useful work; the remainder merely oscillates back and forth between source and reactive elements
  • A low power factor forces a larger current to deliver the same real power, raising I²R losses in the lines — so industries improve it by adding capacitors to cancel inductive lag
  • AC is preferred for long-distance transmission because transformers can step the voltage up (reducing current and hence I²R line loss) and step it back down for safe domestic use

🚀 JEE Advanced Edge

Choke coil: A pure inductor (ideally zero resistance) used to limit AC current without dissipating power, since cosφ=0 for a pure inductor — used in fluorescent tube ballasts instead of a resistor, which would waste power as heat.

Phasor diagram method: Representing V_R, V_L, V_C as vectors (phasors) at 0°, 90°, and −90° respectively lets you add them vectorially to find total voltage/impedance — V_L and V_C phasors are anti-parallel and partially cancel, which is why impedance uses (X_L−X_C), not (X_L+X_C).

Worked problem: A series LCR circuit has R=30Ω, X_L=50Ω, X_C=10Ω, connected to a 200V (rms) AC source. Find the impedance, current, and power factor. Approach: Z=√(R²+(X_L−X_C)²)=√(900+1600)=√2500=50Ω. I=V/Z=200/50=4A. cosφ=R/Z=30/50=0.6 (lagging, since X_L>X_C).

2Revise~3 min before the exam

📐 Formula Sheet

  • RMS values: Irms = I₀/√2 ≈ 0.707I₀  |  Vrms = V₀/√2
  • Reactance: inductive XL = ωL  |  capacitive XC = 1/ωC
  • Impedance (series LCR): Z = √(R² + (XL − XC)²)
  • Phase angle: tanφ = (XL − XC)/R
  • Phase: in a pure inductor V leads I by 90°; in a pure capacitor V lags I by 90°; in R they are in phase
  • Resonance: XL = XC ⇒ ω₀ = 1/√(LC), f₀ = 1/2π√(LC); Z is minimum (= R) and I is maximum
  • Power: P = VrmsIrms·cosφ, where cosφ = R/Z is the power factor
  • Transformer: Vs/Vp = Ns/Np = Ip/Is (ideal)
  • Quality factor: Q = ω₀L/R = (1/R)√(L/C)
3Practiceapply it

✍️ Worked Examples

Example 1 — Resonant frequency
Q: A series LCR circuit has L = 2 H and C = 8 μF. Find its resonant frequency.
Step 1 — Use ω₀ = 1/√(LC): LC = 2 × 8 × 10⁻⁶ = 1.6 × 10⁻⁵.
Step 2 — √(LC) = √(1.6 × 10⁻⁵) ≈ 4 × 10⁻³, so ω₀ ≈ 250 rad/s.
Step 3 — Convert: f₀ = ω₀/2π = 250/6.28 ≈ 39.8 Hz.
Answer: ≈ 39.8 Hz. Note: at resonance the circuit behaves as if purely resistive — the power factor is 1.

Example 2 — Impedance of a series LCR circuit
Q: A circuit has R = 30 Ω, XL = 80 Ω and XC = 40 Ω. Find the impedance and power factor.
Step 1 — Net reactance: XL − XC = 80 − 40 = 40 Ω.
Step 2 — Impedance: Z = √(30² + 40²) = √(900 + 1600) = √2500 = 50 Ω.
Step 3 — Power factor: cosφ = R/Z = 30/50 = 0.6.
Answer: Z = 50 Ω, power factor 0.6 (inductive, since XL > XC). Trap: reactances subtract before squaring — never add them directly to R.

Example 3 — Step-down transformer
Q: An ideal transformer converts 240 V to 12 V. If the primary has 1000 turns, find the secondary turns and the primary current when the secondary draws 2 A.
Step 1 — Turns ratio: Ns/Np = Vs/Vp = 12/240 = 1/20 ⇒ Ns = 50 turns.
Step 2 — Ideal transformers conserve power: VpIp = VsIs.
Step 3 — Solve: 240 × Ip = 12 × 2 = 24 ⇒ Ip = 0.1 A.
Answer: 50 turns, 0.1 A. Note: stepping voltage down steps current up — a transformer trades one for the other.

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Frequently Asked Questions — Alternating Current

What are the key concepts in Alternating Current?
AC circuits, impedance, LCR series circuit, resonance, transformers, and power factor.
Is Alternating Current important for NEET & JEE?
Yes. Alternating Current is part of the Physics Class 12 NCERT syllabus and is directly tested in NEET and JEE examinations. StudyHub provides structured notes, diagrams, and practice questions covering all exam-level subtopics.
How can I practice Alternating Current questions on StudyHub?
Open StudyHub and select Physics → Alternating Current. Choose Easy, Medium, or Hard difficulty. Hard-tier questions are at NEET & JEE level with full step-by-step explanations.

References

  1. NCERT Class 12 Physics Textbook — Chapter: Alternating Current
  2. CBSE Curriculum — Physics (Class 12)
  3. NTA NEET UG Official Syllabus — subject-wise topic list
  4. NTA JEE Main Official Syllabus — subject-wise topic list