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State Board · Class 12 · Physics

Lesson 13: AC Circuits

7 concepts · 56 practice questions with answers · Free to read

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Peak, average and rms values of alternating emf and current

From daily life

The 220 V at home is not the value at every instant; it is the rms value.

Simple definition

Alternating current changes in direction and size : i = I₀ sin ωt.

AC : peak, rms and averageI₀I_rms = I₀ ÷ √2I_avg (half cycle) = 2I₀ ÷ π

In the figure : I₀, I_rms and the average.

Rule 1: Peak and rms values — I_rms = I₀ ÷ √2 = 0.707 I₀, V_rms = V₀ ÷ √2. The rms value is the steady DC that gives the same heating.

  • Easy : I₀ = 5 A ⇒ I_rms = 3.54 A.
  • Medium : V₀ = 311 V ⇒ V_rms = 220 V (the 220 V mains).
  • Exam (CET) : V_rms = 220 ⇒ V₀ = ? (1) 311 V (2) 220 V (3) 155 V (4) 440 V → answer 1) Trick : 220 × 1.414.

Rule 2: Average value — Over a full cycle the average is 0; over a half cycle I_avg = 2 I₀ ÷ π = 0.637 I₀.

  • Easy : I₀ = 5 A ⇒ I_avg = 3.18 A.
  • Medium : Full-cycle mean 0 ⇒ a DC ammeter reads zero.
  • Exam (CET) : Over a full cycle I_avg? (1) 0 (2) 0.637I₀ (3) I₀ (4) 0.707I₀ → answer 1)

Rule 3: AC meters and period — AC meters read rms. T = 1/f, ω = 2πf. In India f = 50 Hz.

  • Easy : f = 50 Hz ⇒ T = 0.02 s; ω = 314 rad/s.
  • Medium : i = 10 sin 100πt ⇒ I₀ = 10 A, f = 50 Hz, I_rms = 7.07 A.
  • Exam (CET) : In i = 8 sin 200t, ω = ? (1) 200 (2) 8 (3) 100 (4) 400 → answer 1)

Derivation : I_rms = I₀ ÷ √2

  • Step 1 : i = I₀ sin ωt; i² = I₀² sin² ωt.
  • Step 2 : The mean of sin² over a cycle = ½.
  • Step 3 : ⟨i²⟩ = I₀² ÷ 2.
  • Step 4 : I_rms = √⟨i²⟩ = I₀ ÷ √2.
Remember

I_rms = I₀/√2; I_avg(½) = 2I₀/π; ω = 2πf.

Common mistake / CET trap

Meters show rms, not peak; the full-cycle average is 0.

Shortcut

Memory clue : "rms = 0.707 of peak".

Spot the difference

QuantityFormulaValue
rmsI₀/√20.707 I₀
Average (half)2I₀/π0.637 I₀
ω2πfrad/s

Practice questions

Easy
  1. 1What is rms?
  2. 2Mains frequency in India?
  3. 3Full-cycle mean?
Medium
  1. 4I₀ = 8 A : I_rms?
  2. 5V_rms = 110 V : V₀?
  3. 6f = 60 Hz : T?
Challenging
  1. 7i = 6 sin 314t : I_rms and f?
  2. 820 Ω with I_rms = 2 A : heating power?
Show answers (try them yourself first)
  1. 1The DC that gives the same heating
  2. 250 Hz
  3. 3Zero
  4. 4I_rms = 5.66 A
  5. 5V₀ = 156 V
  6. 6T = 0.0167 s
  7. 7I_rms = 4.24 A; f = 50 Hz
  8. 8P = 80 W

AC through R, L and C : phasors and reactance

From daily life

A fan capacitor changes the speed — a capacitor offers its own opposition to AC.

Simple definition

Reactance is the opposition of an inductor or capacitor to AC; no energy is lost.

Phasor diagrams : R, L, CR : v and i in phaseL : v leads i by 90°C : v lags i by 90°X_L = ωL ; X_C = 1 ÷ ωC

In the figure : Phasors : R along the axis, L up, C down.

Live figure

AC through R, L and C : phasors and reactance

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Rule 1: Pure resistor — v and i are in phase; I = V ÷ R; phase difference 0.

  • Easy : V_rms = 220, R = 55 Ω ⇒ I_rms = 4 A.
  • Medium : R does not depend on f.
  • Exam (CET) : Phase difference in R? (1) 0 (2) 90° (3) 180° (4) 45° → answer 1)

Rule 2: Inductor — X_L = ω L = 2π f L; v leads i by 90°. For DC X_L = 0.

  • Easy : L = 0.2 H, f = 50 Hz ⇒ X_L = 62.8 Ω.
  • Medium : f doubled ⇒ X_L doubles.
  • Exam (CET) : X_L ∝ ? (1) f (2) 1/f (3) f² (4) constant → answer 1)

Rule 3: Capacitor — X_C = 1 ÷ ωC = 1 ÷ 2π f C; v lags i by 90°. For DC X_C = ∞.

  • Easy : C = 50 µF, f = 50 Hz ⇒ X_C = 63.7 Ω.
  • Medium : f doubled ⇒ X_C halves.
  • Exam (CET) : A capacitor in DC? (1) open circuit (2) short (3) half (4) like R → answer 1) Trick : CIVIL.

Derivation : X_L = ω L

  • Step 1 : i = I₀ sin ωt; e_L = −L di/dt.
  • Step 2 : v = L di/dt = L I₀ ω cos ωt.
  • Step 3 : = (ωL) I₀ sin(ωt + 90°).
  • Step 4 : V₀ = I₀ (ωL) ⇒ X_L = ω L; v leads 90°.
Remember

X_L = ωL; X_C = 1/ωC; CIVIL.

Common mistake / CET trap

X_L and X_C are reactances, not resistances; no energy is dissipated.

Shortcut

Memory clue : "CIVIL : in C, I leads; in L, V leads".

Spot the difference

ElementReactancePhase of v w.r.t. i
RR0°
LωL+90° (leads)
C1/ωC−90° (lags)

Practice questions

Easy
  1. 1Formula for X_L?
  2. 2Unit of X_C?
  3. 3What does CIVIL mean?
Medium
  1. 4L = 0.5 H, f = 50 : X_L?
  2. 5C = 100 µF, f = 50 : X_C?
  3. 6X_L = 100 Ω, V = 200 V : I?
Challenging
  1. 7f 50 → 200 Hz : ratios of X_L and X_C?
  2. 8Phase difference between L and C voltages?
Show answers (try them yourself first)
  1. 12πfL
  2. 2Ohm
  3. 3C : I leads V; L : V leads I
  4. 4X_L = 157.1 Ω
  5. 5X_C = 31.8 Ω
  6. 6I = 2 A
  7. 7X_L × 4; X_C ÷ 4
  8. 8180°

Series LCR circuit and impedance

From daily life

Tuning a radio station uses an LCR circuit.

Simple definition

Impedance Z is the combined opposition of R, X_L and X_C.

Series LCR : impedance phasorRX_L − X_CZφZ = √(R² + (X_L − X_C)²)

In the figure : The right-angled triangle of R, X_L − X_C and Z.

Rule 1: Impedance — Z = √(R² + (X_L − X_C)²); I = V ÷ Z; angle tan φ = (X_L − X_C) ÷ R.

  • Easy : R = 30, X_L = 60, X_C = 20 Ω ⇒ Z = 50 Ω; φ = 53.1°.
  • Medium : V = 200 V ⇒ I = 4 A.
  • Exam (CET) : If X_L > X_C? (1) V leads i (2) lags (3) in phase (4) 180° → answer 1)

Rule 2: Nature of the circuit — X_L > X_C : inductive; X_L < X_C : capacitive; X_L = X_C : resistive (resonance).

  • Easy : X_L = 40, X_C = 70, R = 40 ⇒ capacitive; Z = 50 Ω.
  • Medium : X_L = X_C ⇒ Z = R, the minimum.
  • Exam (CET) : Z is minimum when? (1) X_L = X_C (2) X_L > X_C (3) X_C = 0 (4) R = 0 → answer 1)

Rule 3: Voltage division — V_R = I R, V_L = I X_L, V_C = I X_C; V = √(V_R² + (V_L − V_C)²); V_L or V_C can exceed the supply V.

  • Easy : Above : V_R = 120, V_L = 240, V_C = 80 V ⇒ V = 200 V.
  • Medium : Add as phasors, not arithmetically.
  • Exam (CET) : V = V_R + V_L + V_C? (1) wrong (2) right (3) sometimes (4) in DC → answer 1)

Derivation : Z = √(R² + (X_L − X_C)²)

  • Step 1 : Same i in series; V_R along i, V_L 90° ahead, V_C 90° behind.
  • Step 2 : V_L and V_C oppose ⇒ net (V_L − V_C).
  • Step 3 : V² = V_R² + (V_L − V_C)².
  • Step 4 : Dividing by I, Z² = R² + (X_L − X_C)².
Remember

Z = √(R²+(X_L−X_C)²); tanφ = (X_L−X_C)/R.

Common mistake / CET trap

Arithmetic V_R + V_L + V_C is wrong — add as phasors.

Shortcut

Memory clue : "L ahead, C behind, R along".

Spot the difference

ConditionφNature
X_L > X_C+inductive
X_L < X_C−capacitive
X_L = X_C0resistive

Practice questions

Easy
  1. 1Formula for Z?
  2. 2When is Z minimum?
  3. 3Unit of Z?
Medium
  1. 4R = 8, X_L = 10, X_C = 4 : Z?
  2. 5R = 12, X_L − X_C = 16 : φ?
  3. 6Z = 50, V = 100 : I?
Challenging
  1. 7R = 6 Ω, I = 2 A, V_L = 20 V, V_C = 12 V : V?
  2. 8How can V_C exceed the supply?
Show answers (try them yourself first)
  1. 1√(R²+(X_L−X_C)²)
  2. 2X_L = X_C
  3. 3Ohm
  4. 4Z = 10 Ω
  5. 5φ = 53.1°
  6. 6I = 2 A
  7. 7V = 14.4 V (V_R = 12 V)
  8. 8Near resonance V_L and V_C become large

Power and power factor in AC circuits

From daily life

Factories are penalised on their bill if the power factor is low.

Simple definition

The power factor cos φ is the ratio of real to apparent power.

Power factor : cos φ = R ÷ ZP = V I cosφ ; cosφ = R ÷ Zv (blue), i (red) : φ

In the figure : Angle φ between v and i; P = VI cosφ.

Rule 1: Average power — P_av = V_rms I_rms cos φ; cos φ = power factor = R ÷ Z. Unit W.

  • Easy : V = 220 V, I = 2 A, φ = 60° ⇒ P = 220 W.
  • Medium : R = 30, Z = 50 ⇒ cos φ = 0.6.
  • Exam (CET) : P = ? (1) VI cosφ (2) VI (3) VI sinφ (4) V/I → answer 1)

Rule 2: Special cases — Pure R : cos φ = 1; pure L or C : cos φ = 0 ⇒ P = 0 (wattless current); at resonance cos φ = 1.

  • Easy : Pure L with V = 100, I = 5 ⇒ P = 0.
  • Medium : φ = 90° ⇒ cos φ = 0.
  • Exam (CET) : P of a pure C? (1) 0 (2) VI (3) V²/R (4) maximum → answer 1)

Rule 3: Why power factor matters — A low cos φ needs more I for the same P ⇒ more I²R loss. Industries therefore add capacitors.

  • Easy : P = 2.2 kW, V = 220 : cos φ = 1 ⇒ I = 10 A; cos φ = 0.5 ⇒ I = 20 A.
  • Medium : Loss is 4 times (I²).
  • Exam (CET) : cos φ halved ⇒ I (P fixed)? (1) doubles (2) halves (3) four times (4) same → answer 1)

Derivation : P_av = V I cos φ

  • Step 1 : v = V₀ sin ωt, i = I₀ sin(ωt − φ).
  • Step 2 : p = v i = V₀I₀ sin ωt sin(ωt − φ).
  • Step 3 : Cycle average : ½ V₀I₀ cos φ.
  • Step 4 : P = V_rms I_rms cos φ.
Remember

P = VI cosφ; cosφ = R/Z; pure L, C : P = 0.

Common mistake / CET trap

Do not write P = VI; never forget cos φ.

Shortcut

Memory clue : "angle 90° ⇒ zero power".

Spot the difference

QuantityFormulaUnit
Average powerVI cosφW
Apparent powerVIV A
cos φR/Z—

Practice questions

Easy
  1. 1What is power factor?
  2. 2Power in a pure L?
  3. 3Unit of apparent power?
Medium
  1. 4V = 100, I = 4, φ = 30° : P?
  2. 5R = 8, Z = 10 : cos φ?
  3. 6P = 500 W, V = 250, cos φ = 0.8 : I?
Challenging
  1. 7cos φ 1 → 0.5, P fixed : ratio of loss?
  2. 8What is wattless current?
Show answers (try them yourself first)
  1. 1cos φ = R/Z
  2. 2Zero
  3. 3V A
  4. 4P = 346.4 W
  5. 5cos φ = 0.8
  6. 6I = 2.5 A
  7. 74 : 1
  8. 8Current at φ = 90°; P = 0

LC oscillations

From daily life

A swing is still at the top and fastest at the bottom; in LC, charge and current do the same.

Simple definition

LC oscillations are the swapping of electrical energy between a capacitor and an inductor.

LC oscillations : energy swappingCLω = 1 ÷ √(LC)½CV² ⇄ ½LI²

In the figure : Energy swaps between C and L.

Rule 1: LC oscillations — A charged C connected to L : energy swaps C ⇄ L; ω = 1 ÷ √(LC), f = 1 ÷ 2π√(LC).

  • Easy : L = 0.1 H, C = 10 µF ⇒ ω = 1000 rad/s; f = 159.2 Hz.
  • Medium : C four times ⇒ f halves.
  • Exam (CET) : f ∝ ? (1) 1/√(LC) (2) √(LC) (3) LC (4) 1/LC → answer 1)

Rule 2: Energy and current — Total energy ½ Q₀²/C is constant; I_max = Q₀ ω. E.g. Q₀ = 100 µC.

  • Easy : Above : I_max = 100 × 10⁻⁶ × 1000 = 0.1 A.
  • Medium : Energy = ½ Q²/C = 5 × 10⁻⁴ J.
  • Exam (CET) : When all energy is in L, q = ? (1) 0 (2) Q₀ (3) Q₀/2 (4) 2Q₀ → answer 1)

Rule 3: Mechanical analogy — LC ⇔ spring–mass : L ⇔ m, 1/C ⇔ k, q ⇔ x, I ⇔ v. R ⇔ friction.

  • Easy : ω = √(k/m) ⇔ 1/√(LC).
  • Medium : An ideal LC oscillates forever; R damps it.
  • Exam (CET) : L ⇔ ? (1) m (2) k (3) x (4) v → answer 1)

Derivation : ω = 1 ÷ √(LC)

  • Step 1 : KVL : L dI/dt + q/C = 0, I = dq/dt.
  • Step 2 : L d²q/dt² + q/C = 0.
  • Step 3 : d²q/dt² = −(1/LC) q — SHM.
  • Step 4 : ω = 1 ÷ √(LC).
Remember

ω = 1/√(LC); I_max = Q₀ω; L ⇔ m, 1/C ⇔ k.

Common mistake / CET trap

Total energy is constant (ideal); when C's energy is 0 L's is maximum.

Shortcut

Memory clue : "LC = spring and mass".

Spot the difference

ElectricalMechanical
Lm
1/Ck
qx
Iv

Practice questions

Easy
  1. 1ω of an LC circuit?
  2. 2L ⇔ ?
  3. 3How does energy change?
Medium
  1. 4L = 2 mH, C = 8 µF : f?
  2. 5Q₀ = 50 µC, ω = 1000 : I_max?
  3. 6ω = 500, L = 0.4 H : C?
Challenging
  1. 7L doubled, C doubled : f?
  2. 8What does R do to LC oscillations?
Show answers (try them yourself first)
  1. 11/√(LC)
  2. 2m
  3. 3Alternately C ⇄ L
  4. 4f = 1258 Hz
  5. 5I_max = 0.05 A
  6. 6C = 1 × 10⁻⁵ F
  7. 7Halves
  8. 8Damped oscillations

Series and parallel resonance

From daily life

Pushing a swing in its own rhythm makes it swing high — resonance.

Simple definition

Resonance is maximum current at the frequency f₀.

Resonance curve : I versus ff₀ = 1 ÷ 2π√(LC)small Rlarge R

In the figure : A peak at f₀, lowered by a larger R.

Live figure

Series and parallel resonance

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Rule 1: Series resonance — X_L = X_C ⇒ f₀ = 1 ÷ 2π√(LC); Z = R minimum; I = V ÷ R maximum; cos φ = 1.

  • Easy : L = 0.5 H, C = 20 µF ⇒ f₀ = 50.3 Hz.
  • Medium : V = 100, R = 10 Ω ⇒ I = 10 A.
  • Exam (CET) : At series resonance Z = ? (1) R (2) maximum (3) X_L (4) ∞ → answer 1)

Rule 2: Parallel resonance — L and C in parallel : at resonance Z is maximum (≈ L/RC) and I minimum — acceptor/rejector circuits.

  • Easy : Series : I maximum; parallel : I minimum.
  • Medium : Parallel resonance rejects an unwanted frequency.
  • Exam (CET) : At parallel resonance Z is? (1) maximum (2) minimum (3) R (4) 0 → answer 1)

Rule 3: Radio tuning — Varying C sets f₀ to the desired station; that station's current is then maximum.

  • Easy : L = 200 µH, f₀ = 1 MHz ⇒ C = 126.7 pF.
  • Medium : To raise f₀ reduce C.
  • Exam (CET) : To change station? (1) vary C (2) vary R (3) vary V (4) nothing → answer 1)

Derivation : The resonance condition

  • Step 1 : Z = √(R² + (ωL − 1/ωC)²).
  • Step 2 : Z is minimum ⇒ ωL = 1/ωC.
  • Step 3 : ω₀² = 1/LC.
  • Step 4 : f₀ = 1 ÷ 2π√(LC); Z = R.
Remember

f₀ = 1/2π√(LC); series Z = R; parallel Z maximum.

Common mistake / CET trap

Series resonance has minimum Z, parallel has maximum — do not swap.

Shortcut

Memory clue : "series accepts, parallel rejects".

Spot the difference

TypeZI
SeriesR (minimum)maximum
Parallelmaximumminimum

Practice questions

Easy
  1. 1Resonance condition?
  2. 2cos φ at series resonance?
  3. 3I at parallel resonance?
Medium
  1. 4L = 1 H, C = 25 µF : f₀?
  2. 5V = 50, R = 25 : I at resonance?
  3. 6f₀ = 500 Hz, L = 0.1 H : C?
Challenging
  1. 7What happens to the curve if R increases?
  2. 8C to double f₀?
Show answers (try them yourself first)
  1. 1X_L = X_C
  2. 21
  3. 3Minimum
  4. 4f₀ = 31.8 Hz
  5. 5I = 2 A
  6. 6C = 1.01 × 10⁻⁶ F
  7. 7The peak lowers and broadens
  8. 8C becomes one quarter

Q factor, bandwidth and the choke coil

From daily life

A good radio does not mix neighbouring stations — it has a large Q.

Simple definition

The Q factor measures the sharpness of resonance; bandwidth is the gap between the half-power frequencies.

Q factor and bandwidthI₀/√2Δf = f₂ − f₁ ; Q = f₀ ÷ Δf

In the figure : A sharp peak ⇒ large Q, small Δf.

Rule 1: Quality factor — Q = ω₀ L ÷ R = 1 ÷ ω₀ C R = (1/R)√(L/C). A large Q means sharp resonance.

  • Easy : L = 0.2 H, C = 10 µF, R = 5 Ω ⇒ ω₀ = 707; Q = 28.3.
  • Medium : R halved ⇒ Q doubles.
  • Exam (CET) : Q depends on? (1) L, C, R (2) V (3) I (4) f only → answer 1)

Rule 2: Bandwidth — The spread between half-power frequencies : Δf = f₀ ÷ Q = R ÷ 2πL; where I = I₀/√2.

  • Easy : f₀ = 112.5 Hz, Q = 28.3 ⇒ Δf = 3.98 Hz.
  • Medium : Large Q ⇒ small Δf ⇒ selective.
  • Exam (CET) : Δf = ? (1) f₀/Q (2) f₀Q (3) Q/f₀ (4) f₀ → answer 1)

Rule 3: Choke coil — A coil of large L and small R cuts AC current without wasting energy (P ≈ 0) — better than a resistor.

  • Easy : Tubelight choke : reduces I with negligible heating.
  • Medium : A choke is useless for DC (X_L = 0).
  • Exam (CET) : A choke works for? (1) AC (2) DC (3) both (4) neither → answer 1)

Derivation : Δf = f₀ ÷ Q

  • Step 1 : Half power : Z = √2 R ⇒ ωL − 1/ωC = ± R.
  • Step 2 : The gap between ω₁ and ω₂ is Δω = R ÷ L.
  • Step 3 : Δf = R ÷ 2πL.
  • Step 4 : Q = ω₀ L ÷ R ⇒ Δf = f₀ ÷ Q.
Remember

Q = ω₀L/R; Δf = f₀/Q; choke : large L, small R.

Common mistake / CET trap

A larger Q means a smaller Δf; they do not rise together.

Shortcut

Memory clue : "large Q = selective".

Spot the difference

QuantityFormulaUnit
Qω₀L/R—
Δff₀/Q = R/2πLHz
f₀1/2π√LCHz

Practice questions

Easy
  1. 1Unit of Q?
  2. 2What does a large Q suggest?
  3. 3Why a choke?
Medium
  1. 4L = 0.1 H, R = 2 Ω, ω₀ = 200 : Q?
  2. 5f₀ = 1000 Hz, Q = 50 : Δf?
  3. 6L = 0.4 H, R = 8 : Δf?
Challenging
  1. 7R doubled : Q and Δf?
  2. 8L = 1 H, C = 4 µF, R = 10 : Q?
Show answers (try them yourself first)
  1. 1No unit
  2. 2Sharp resonance, small Δf
  3. 3To cut I without wasting energy
  4. 4Q = 10
  5. 5Δf = 20 Hz
  6. 6Δf = 3.18 Hz
  7. 7Q halves; Δf doubles
  8. 8Q = 50

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
rmsI = I₀/√2; V = V₀/√2A, V
ReactanceX_L = ωL; X_C = 1/ωCΩ
ImpedanceZ = √(R² + (X_L − X_C)²)Ω
PowerP = VI cosφ; cosφ = R/ZW
LCω = 1/√(LC)rad/s
Resonancef₀ = 1/2π√(LC); series Z = RHz
Qω₀L/R; Δf = f₀/Q—; Hz
Chokelarge L, small R—

🎯 Most important points for the exam

  • rms = peak/√2; meters show rms.
  • CIVIL : in C, I leads; in L, V leads.
  • Z = √(R² + (X_L − X_C)²); phasor addition.
  • P = VI cosφ; pure L, C ⇒ 0.
  • LC : ω = 1/√(LC); L ⇔ m, 1/C ⇔ k.
  • Series resonance — maximum I; parallel — minimum I.
  • Q = ω₀L/R; Δf = f₀/Q.
Study mantra

Find X_L, X_C → Z → I = V/Z → φ → P = VI cosφ.