NCERT Plus Two (Class 12) Physics • Chapter 7

Alternating Current

Q: Define RMS value of alternating current and derive the relation I_rms = iₘ / √2.

Definition: RMS current is that steady DC current which produces the same Joule heat in a resistor in one full cycle as the AC produces. I² R T = ∫₀ᵀ iₘ² sin²(ωt) R dt = ½ iₘ² R T ⇒ I = iₘ / √2 ≈ 0.707 iₘ .

Q: Show that the average power dissipated over one complete cycle in a pure inductor (or pure capacitor) connected to an AC source is zero.

In a pure inductor, v = vₘ sin(ωt) and i = iₘ sin(ωt − π/2) = −iₘ cos(ωt) . Average power P = V I cosφ = V I cos 90° = 0 (energy absorbed in one quarter-cycle is returned in the next).

Q: Using a phasor diagram, derive expressions for impedance Z and phase angle φ of a series LCR circuit driven by v = vₘ sin(ωt).

Take current I along reference axis: V_R = IR (0°), V_L = IX_L (+90°), V_C = IX_C (−90°). Net reactive voltage V_L − V_C is perpendicular to V_R : V = √[V_R² + (V_L − V_C)²] = I √[R² + (X_L − X_C)²] ⇒ Z = √[R² + (X_L − X_C)²] and tanφ = (X_L − X_C)/R .

Q: Obtain the expression for resonant frequency of a series LCR circuit. Define Quality factor (Q) and write its formula.

Resonance occurs when X_L = X_C ⇒ ω₀L = 1/(ω₀C) ⇒ ω₀ = 1/√(LC) ⇒ ν₀ = 1/(2π√(LC)) ( Z_min = R , I_max = V/R ). Q-Factor: Measure of sharpness of resonance, Q = ω₀ / (2Δω) = (1/R)√(L/C) .

Q: A 200 Ω resistor and 15.0 μF capacitor are in series on a 220 V, 50 Hz source. Why can the algebraic sum V_R + V_C exceed 220 V?

X_C = 1/(2π × 50 × 15×10⁻⁶) ≈ 212.3 Ω ⇒ Z = √(200² + 212.3²) ≈ 291.7 Ω ⇒ I = 0.755 A . V_R = 151 V , V_C = 160.3 V (sum 311.3 V > 220 V ) because V_R and V_C are 90° out of phase , and √(151² + 160.3²) = 220 V !

Q: What is power factor and wattless current in an AC circuit? How can power factor be improved?

Power factor: cosφ = R/Z in P = V I cosφ . Wattless current: Component I sinφ perpendicular to V , consuming zero average power. Improvement: Connect an appropriate capacitor in parallel with inductive loads to cancel lagging wattless current.

Q: State the working principle of a transformer. Mention any three energy losses in a transformer and how each is minimized.

Principle: Mutual induction ( V_s/V_p = N_s/N_p ). 1. Eddy current loss: Laminated soft-iron core sheets. 2. Hysteresis loss: Soft-iron core (narrow B–H loop). 3. Copper (I²R) loss: Thick copper wire. 4. Flux leakage: Coaxial winding on closed core.

Q: An ideal step-down transformer steps 2200 V down to 220 V. If the secondary delivers 10 A to a load, find (i) turns ratio N_s/N_p and (ii) primary current I_p.

(i) N_s / N_p = V_s / V_p = 220 / 2200 = 1 : 10 . (ii) Ideal power V_p I_p = V_s I_s ⇒ I_p = (220 × 10) / 2200 = 1.0 A .

Peak vs RMS AC values & rotating phasors, R/L/C reactances, series LCR impedance, resonance & Q-factor, power factor & transformers.

📄 Official NCERT Textbook PDF

Topic 7.2 – 7.3: Peak vs RMS AC Current/Voltage & Rotating Phasors

Your home wall socket in Kerala says 230 V AC (or 220 V), yet the real voltage whips up to +311 V and down to −311 V fifty times a second! If you average a sine wave over one complete cycle, positive and negative halves cancel to ZERO — yet an electric iron still gets blazing hot because heating goes as i²! That's why we use RMS (Root-Mean-Square)!

RMS VALUESI_rms = i_m√2 ≈ 0.707 i_m • V_rms = v_m√2 ≈ 0.707 v_m

CYCLE AVERAGESFull cycle avg = 0 • Half cycle avg = 2 i_mπ ≈ 0.637 i_m!

PURE RESISTOR & PHASORv and i stay IN PHASE (φ = 0°); P = V_rms I_rms = ½ v_m i_m = I² R!

🧠 Memory Trick: 220 V AC on a socket label is ALWAYS RMS! Peak voltage is 220 × √2 ≈ 311 V — that's why 220 V AC gives a sharper shock than 220 V DC!

💡 Simple Kid-Check: Think of RMS as the "DC Heating Equivalent": 2 A RMS AC boils the exact same cup of tea in the exact same time as steady 2 A DC, even though the AC peak actually surges up to 2 × 1.414 = 2.83 A!

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Why i_m / √2? Instantaneous power in a resistor R is p = i² R = i_m² R sin²(ωt). Over one full cycle, the average of sin²(ωt) is ½, so average power is ½ i_m² R = I² R ⇒ I = i_m√2 ≈ 0.707 i_m!
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What Is a Phasor? A rotating arrow (length = peak amplitude v_m or i_m) spinning anticlockwise at angular speed ω = 2πν. Its vertical y-projection equals instantaneous value v = v_m sin(ωt)!
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Purely Resistive AC Circuit: Applying v = v_m sin(ωt) across resistor R gives i = (v_m/R) sin(ωt) — both arrows rotate locked together (phase difference φ = 0°)!

🧒 Super-Easy Example: An electric bulb draws 100 W from a 220 V, 50 Hz supply.
• Peak voltage v_m = 220√2 ≈ 311 V.
• RMS current I = P/V = 100/220 ≈ 0.455 A • Peak current i_m = 0.455√2 ≈ 0.643 A!

Watch three classic exam traps: (1) AC voltmeters & ammeters ALWAYS read RMS values, NEVER peak values or simple averages! (2) Don't confuse Half-Cycle Average (2 i_m / π ≈ 0.637 i_m) with RMS (i_m / √2 ≈ 0.707 i_m)! (3) Average power over a cycle in a resistor uses V_rms I_rms (or ½ v_m i_m — NEVER v_m i_m without the ½!).

Topic 7.4 – 7.5: Pure Inductive (X_L) & Capacitive (X_C) AC Circuits & Phase

Feed steady DC to an inductor coil and it passes effortlessly; feed fast-wiggling AC and the coil snaps back hard (X_L = ωL)! Now feed DC to a capacitor and it blocks current completely like an open bridge; feed fast AC and charges slosh back and forth effortlessly (X_C = 1/ωC)!

PURE INDUCTOR (L)X_L = ωL = 2πνL (Ω) • Current LAGS Voltage by 90° (π/2) • P_avg = 0!

PURE CAPACITOR (C)X_C = 1ωC = 12πνC (Ω) • Current LEADS Voltage by 90° (π/2) • P_avg = 0!

DC (ν = 0) LIMITInductor: X_L = 0 (passes DC!). Capacitor: X_C = ∞ (blocks DC!).

🧠 Memory Trick: Remember the word CIVIL! In a Capacitor, I comes BEFORE V (I leads V by 90°); in an Inductor (L), V comes BEFORE I (I lags V by 90°)!

💡 Simple Kid-Check: Why does a capacitor let AC through even with empty insulation between its plates? Because AC keeps reversing — as fast as one side fills with electrons, the other side empties and pushes electrons out, so alternating current keeps flowing in the wires!

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Pure Inductor: Applying v = v_m sin(ωt) across L gives L(di/dt) = v_m sin(ωt) ⇒ integrating gives i = i_m sin(ωt − π/2) where i_m = v_m / X_L and Inductive Reactance X_L = ωL = 2πνL. Current lags voltage by 90°!
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Pure Capacitor: Charge q = C v_m sin(ωt) ⇒ differentiating gives i = dq/dt = i_m sin(ωt + π/2) where i_m = v_m / X_C and Capacitive Reactance X_C = 1/(ωC) = 1/(2πνC). Current leads voltage by 90°!
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Why Average Power = Zero in Both! During one quarter-cycle an inductor or capacitor borrows energy from the source (storing ½LI² or ½CV²), and during the very next quarter-cycle it returns 100% back — zero net heat!

🧒 NCERT Example 7.5 Concept: An open inductor coil and bulb sit in series on an AC mains supply. Slide an iron rod inside the coil → core permeability jumps → L rises → X_L = 2πνL rises → circuit current drops → bulb glow DECREASES!

Watch three classic traps: (1) Graphs vs frequency ν: X_L = 2πνL is a straight line through origin (∝ ν), whereas X_C = 1/(2πνC) is a rectangular hyperbola (∝ 1/ν)! (2) Both X_L and X_C are measured in Ohms (Ω), NOT Henry or Farad! (3) Pure L and pure C consume ZERO average power over a full cycle!

Topic 7.6.1: Series LCR Circuit, Impedance Z & Phasor Solution

Put a Resistor, Inductor, and Capacitor in single file (series) on an AC source. Same current I marches through all three, yet their voltmeter readings point in THREE different directions — right, straight up, and straight down! That's why reading 150 V across R and 160 V across C can STILL come from a single 220 V plug — you must add them like perpendicular arrows using Pythagoras!

IMPEDANCE ZZ = √[ R² + (X_L − X_C)² ] (Ω) • Current: I = V / Z!

PHASOR VOLTAGEV = √[ V_R² + (V_L − V_C)² ] (V_L and V_C oppose by 180°!)

PHASE ANGLE φtanφ = X_L − X_CR (X_L > X_C → Inductive, I lags; X_C > X_L → Capacitive, I leads!)

🧠 Memory Trick: Draw R along +x, X_L straight UP (+y), X_C straight DOWN (−y) — Pythagoras immediately gives Z = √[R² + (X_L − X_C)²]!

💡 Simple Kid-Check: Because inductive voltage V_L leads current by 90° and capacitive voltage V_C lags current by 90°, they are exact opposites (180° apart)! So they fight each other first (|V_L − V_C|), and the leftover teams up at right angles with V_R!

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Phasor Addition: Take current phasor I along reference x-axis: V_R = I R (0°), V_L = I X_L (+90°), V_C = I X_C (−90°). Net vertical component is V_L − V_C, giving total source voltage V = √[V_R² + (V_L − V_C)²] = I √[R² + (X_L − X_C)²]!
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Impedance Triangle: Dividing voltage sides by I yields a right triangle with base R, vertical leg (X_L − X_C), hypotenuse Impedance Z = √[R² + (X_L − X_C)²], and angle φ where tanφ = (X_L − X_C)/R and cosφ = R/Z.
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Three Regimes: If X_L > X_C → Inductive (V leads I); if X_C > X_L → Capacitive (I leads V); if X_L = X_C → Purely Resistive (φ = 0°)!

🧒 Super-Easy 3-4-5 Triangle Example: In a series LCR circuit, voltmeters read V_R = 30 V, V_L = 80 V, and V_C = 40 V. Algebraic sum = 150 V! Yet true source voltage is V = √[30² + (80 − 40)²] = √[30² + 40²] = 50 V!

Top 2-mark board paradox: "Can the sum of RMS voltmeter readings V_R + V_L + V_C be GREATER than the AC source voltage V?" YES! Always explain that voltages are NOT in phase — V_L and V_C differ by 180° and cancel partly, so you MUST combine them vectorially as √[V_R² + (V_L − V_C)²]!

Topic 7.6.2: Electrical Resonance, Resonant Frequency & Q-Factor

Push a playground swing at its exact natural beat and every tiny push sends it flying high! Inside a radio or mobile receiver, turning the tuning capacitor adjusts C until X_L and X_C cancel each other dead — impedance crashes down to just R, letting your favourite FM station surge through crystal clear!

RESONANCE CONDITIONX_L = X_C ⇒ ω₀ = 1√(LC) • ν₀ = 12π√(LC)

AT RESONANCEZ_min = R • I_max = V / R • φ = 0° (cosφ = 1) • V_L = V_C!

SHARPNESS (Q-FACTOR)Q = ω₀ LR = 1ω₀ C R = 1R √(LC) (Smaller R → sharper peak!)

🧠 Memory Trick: At resonance, L and C cancel like twins (X_L = X_C), leaving ONLY R! Smaller R makes the Q-Factor taller & sharper so nearby radio stations don't mix!

💡 Simple Kid-Check: Imagine two wrestlers X_L (grows with frequency) and X_C (shrinks with frequency) pushing opposite directions. At one magic frequency ν₀, they tie in exact balance (X_L = X_C) — leaving zero reactance, minimum impedance R, and maximum current!

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Deriving Resonant Frequency: Set X_L = X_C ⇒ ω₀ L = 1/(ω₀ C) ⇒ ω₀² = 1/(LC) ⇒ ν₀ = 1 / (2π √(LC)). Notice ν₀ depends ONLY on L and C, NOT on R!
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Quality Factor (Q) & Selectivity: Bandwidth between half-power points (I = I_max/√2) is 2Δω = R/L. Quality factor Q = ω₀2Δω = ω₀ LR = 1R √(LC) (dimensionless!). At resonance, V_L = V_C = Q · V_source!
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Tuning Application: Series LCR accepts only signals right around ν₀ (used in radio/TV tuners — high Q = sharp station separation!).

🧒 NCERT Numerical Check: Series LCR has L = 1.0 mH, C = 1.0 nF, R = 100 Ω, V = 100 V.
• ω₀ = 1 / √(10⁻³ × 10⁻⁹) = 10⁶ rad/s • I_max = 100/100 = 1 A.
• Q = (10⁶ × 10⁻³)/100 = 10 ⇒ voltage across L or C magnifies to Q V = 1000 V!

Watch three classic traps: (1) Can resonance happen in a series RL or RC circuit? NEVER! You MUST have BOTH L and C so X_L and X_C cancel! (2) Changing resistance R changes peak current V/R and sharpness Q, but DOES NOT shift resonant frequency ν₀! (3) Even though V_L and V_C can each reach 1000 V (Q times source!), they stay equal and opposite so V_L − V_C = 0!

Topic 7.7: Power in AC Circuit, Wattless Current & Power Factor cos(φ)

Imagine pushing sideways on a toy train locked on straight tracks — you sweat and push hard, yet your sideways force does ZERO forward work! In AC circuits, current splits into two parts: an in-phase worker (Power Current) that heats appliances, and a sideways 90° component (Wattless Current) that just sloshes energy back and forth doing zero work!

AVERAGE AC POWERP = V_rms I_rms cosφ = I_rms² R = ½ v_m i_m cosφ!

POWER FACTOR (cos φ)cosφ = RZ = R√[R²+(X_L−X_C)²] (Range: 0 to 1!)

WATTLESS CURRENTI_q = I sinφ (90° out of phase with V → 0 power loss!)

🧠 Memory Trick: Only R ever turns electricity into heat (P = I² R)! Pure L or C has cos 90° = 0 (Wattless!), while Resonance or pure R hits cos 0° = 1!

💡 Simple Kid-Check: Why did old fluorescent tube lights use a heavy coil called a Choke Coil instead of a cheap resistor to limit AC current? Because a resistor burns huge electricity as wasted heat (cosφ = 1), whereas a choke coil has huge X_L and tiny R — choking current with almost wattless loss (cosφ ≈ 0)!

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Deriving Power Formula: Multiplying v = v_m sin(ωt) by i = i_m sin(ωt + φ) gives p = ½ v_m i_m [cosφ − cos(2ωt + φ)]. Averaged over a full cycle, cos(2ωt + φ) averages to zero, leaving P = V_rms I_rms cosφ!
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Power Factor Cases:
• Pure R or LCR at Resonance: φ = 0° ⇒ cosφ = 1 (Max power V I).
• Pure L or Pure C: φ = ±90° ⇒ cosφ = 0 (Average power = 0 — purely wattless!).
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Power Factor Improvement in Factories: Heavy motors lag with large wattless current I sinφ, heating supply cables (I²R_wire). Factories connect capacitors in parallel to cancel lagging wattless current and boost cosφ → 1!

🧒 NCERT Example 7.8: Series LCR has v_m = 283 V (V_rms = 200 V), R = 3 Ω, X_L = 8 Ω, X_C = 4 Ω.
• Z = √[3² + (8 − 4)²] = 5 Ω • Power factor cosφ = R/Z = 3/5 = 0.6.
• I_rms = 200/5 = 40 A ⇒ Power dissipated P = I² R = 40² × 3 = 4800 W (check: 200 × 40 × 0.6 = 4800 W!).

Watch three classic traps: (1) If an exam gives peak formulas like v = 100 sin(100t) and i = 10 sin(100t + π/3), average power is ½ v_m i_m cos(60°) = ½ × 100 × 10 × 0.5 = 250 W — don't forget the ½! (2) Total current I = I_p + I_q is a vector sum, so Wattless Current amplitude is I_rms sinφ! (3) Why capacitors in parallel for power factor correction? To neutralize lagging motor current with leading capacitor current!

Topic 7.8: Working Principle, Turns Ratio & Energy Losses in Transformers

From Idukki hydroelectric station to Kochi, power travels at gigantic 220,000 V so current stays small and cables don't melt from I²R heat — then your neighbourhood pole-mounted box steps it safely down to 230 V! With ZERO moving parts, the Transformer pulls off this voltage-swapping magic using pure Mutual Induction!

PRINCIPLE & RATIOMutual Induction (AC ONLY!). Ideal ratio: V_sV_p = N_sN_p = I_pI_s!

STEP-UP VS STEP-DOWNStep-Up: N_s > N_p ⇒ V_s > V_p, I_s < I_p • Step-Down: N_s < N_p ⇒ V_s < V_p, I_s > I_p!

4 ENERGY LOSSESFlux leakage • Copper I²R loss • Eddy currents (laminate core!) • Hysteresis (soft iron core!)

🧠 Memory Trick: A transformer trades Voltage for Current (V_p I_p = V_s I_s) — step voltage UP 10× and current drops 10×, keeping power & frequency unchanged!

💡 Simple Kid-Check: Two separate copper coils wrapped around the SAME soft-iron ring share the same alternating magnetic flux per turn dΦ/dt! Give the secondary coil 10× more turns than the primary, and every loop picks up equal voltage — multiplying total secondary voltage by 10×!

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Transformer Equations: Equal flux per turn gives V_p = −N_p (dΦ/dt) and V_s = −N_s (dΦ/dt) ⇒ V_sV_p = N_sN_p. For ideal ~100% efficiency (V_p I_p = V_s I_s): I_pI_s = V_sV_p = N_sN_p!
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High-Voltage Grid Transmission: To deliver power P = V I across long cables of resistance R_c, line heat loss is P_loss = I² R_c = (P/V)² R_c. Stepping voltage UP 100× slashes transmission losses 10,000×!
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Four Losses & Remedies (Top 3-Mark Table):

Energy Loss	Cause	Engineering Fix
1. Flux Leakage	Not all primary flux links secondary	Wind coils coaxially one over another on closed core
2. Copper (Joule) Loss	Wire resistance wastes `I²R` heat	Use thick low-resistance copper wire on high-current coil
3. Eddy Current Loss	Swirling currents induced in iron core	Build core from thin insulated laminated sheets
4. Hysteresis Loss	Repeated AC magnetisation of core	Use soft iron (narrow B–H loop, low coercivity)

Watch three classic DHSE traps: (1) Hook a transformer up to a DC battery and secondary output is ZERO (steady DC gives dΦ/dt = 0 — plus without inductive reactance X_L, primary coil can overheat and burn!). (2) A step-up transformer increases voltage, NOT power, energy, or frequency (f_out = f_in, η = P_out / P_in ≤ 1)! (3) In a step-up transformer, secondary carries lower current, so the PRIMARY coil needs thicker copper wire!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 2 Marks): Define RMS value of alternating current and derive the relation I_rms = iₘ / √2.

Definition: RMS current is that steady DC current which produces the same Joule heat in a resistor in one full cycle as the AC produces.
I² R T = ∫₀ᵀ iₘ² sin²(ωt) R dt = ½ iₘ² R T ⇒ I = iₘ / √2 ≈ 0.707 iₘ.

Q2 (Kerala HSE Say • 2 Marks): Show that the average power dissipated over one complete cycle in a pure inductor (or pure capacitor) connected to an AC source is zero.

In a pure inductor, v = vₘ sin(ωt) and i = iₘ sin(ωt − π/2) = −iₘ cos(ωt).
Average power P = V I cosφ = V I cos 90° = 0 (energy absorbed in one quarter-cycle is returned in the next).

Q3 (Kerala HSE March • 4 Marks): Using a phasor diagram, derive expressions for impedance Z and phase angle φ of a series LCR circuit driven by v = vₘ sin(ωt).

Take current I along reference axis: V_R = IR (0°), V_L = IX_L (+90°), V_C = IX_C (−90°).
Net reactive voltage V_L − V_C is perpendicular to V_R:
V = √[V_R² + (V_L − V_C)²] = I √[R² + (X_L − X_C)²] ⇒ Z = √[R² + (X_L − X_C)²] and tanφ = (X_L − X_C)/R.

Q4 (Kerala HSE March • 3 Marks): Obtain the expression for resonant frequency of a series LCR circuit. Define Quality factor (Q) and write its formula.

Resonance occurs when X_L = X_C ⇒ ω₀L = 1/(ω₀C) ⇒ ω₀ = 1/√(LC) ⇒ ν₀ = 1/(2π√(LC)) (Z_min = R, I_max = V/R).
Q-Factor: Measure of sharpness of resonance, Q = ω₀ / (2Δω) = (1/R)√(L/C).

Q5 (Kerala HSE Model • 2 Marks): A 200 Ω resistor and 15.0 μF capacitor are in series on a 220 V, 50 Hz source. Why can the algebraic sum V_R + V_C exceed 220 V?

X_C = 1/(2π × 50 × 15×10⁻⁶) ≈ 212.3 Ω ⇒ Z = √(200² + 212.3²) ≈ 291.7 Ω ⇒ I = 0.755 A.
V_R = 151 V, V_C = 160.3 V (sum 311.3 V > 220 V) because V_R and V_C are 90° out of phase, and √(151² + 160.3²) = 220 V!

Q6 (Kerala HSE Say • 2 Marks): What is power factor and wattless current in an AC circuit? How can power factor be improved?

Power factor: cosφ = R/Z in P = V I cosφ.
Wattless current: Component I sinφ perpendicular to V, consuming zero average power.
Improvement: Connect an appropriate capacitor in parallel with inductive loads to cancel lagging wattless current.

Q7 (Kerala HSE March • 3 Marks): State the working principle of a transformer. Mention any three energy losses in a transformer and how each is minimized.

Principle: Mutual induction (V_s/V_p = N_s/N_p).
1. Eddy current loss: Laminated soft-iron core sheets.
2. Hysteresis loss: Soft-iron core (narrow B–H loop).
3. Copper (I²R) loss: Thick copper wire.
4. Flux leakage: Coaxial winding on closed core.

Q8 (NCERT / DHSE Numerical • 2 Marks): An ideal step-down transformer steps 2200 V down to 220 V. If the secondary delivers 10 A to a load, find (i) turns ratio N_s/N_p and (ii) primary current I_p.

(i) N_s / N_p = V_s / V_p = 220 / 2200 = 1 : 10.
(ii) Ideal power V_p I_p = V_s I_s ⇒ I_p = (220 × 10) / 2200 = 1.0 A.