Alternating Current - Class 12 Physics

Class 12 Physics | Unit IV

Chapter 7: Alternating Current

AC Voltage • Phasors • LCR Circuits • Resonance • Power Factor • Transformers

1. Alternating Current — Basics

Alternating Current (AC): A current that changes direction periodically with time, following a sinusoidal waveform. Generated by rotating a coil in a magnetic field (AC generator).

Voltage: V = V₀ sinωt (V₀ = peak voltage)
Current: I = I₀ sin(ωt + ϕ) (I₀ = peak current, ϕ = phase difference)
ω = 2πf = angular frequency (rad/s); f = frequency (Hz); T = 1/f = time period (s)

1.1 Mean (Average) and RMS Values

Mean value over full cycle: V_avg = 0 (symmetrical positive and negative halves cancel)
Mean value over half cycle: V_avg = 2V₀/π ≈ 0.637 V₀

RMS (Root Mean Square) value:
V_rms = V₀/√2 ≈ 0.707 V₀
I_rms = I₀/√2 ≈ 0.707 I₀

The RMS value of AC = equivalent DC value that produces the same heating effect. Indian mains: V_rms = 220 V ⇒ V₀ = 220√2 = 311 V, f = 50 Hz.

AC meters (ammeters, voltmeters) read RMS values, not peak values.
Power consumption uses RMS values: P = V_rmsI_rmscosϕ.
Peak-to-peak voltage = 2V₀.

2. AC Through Pure Resistor (R)

V = V₀ sinωt ⇒ I = V₀/R × sinωt = I₀ sinωt
Phase: V and I are IN PHASE (ϕ = 0°)
I₀ = V₀/R; I_rms = V_rms/R

No phase difference. V leads I by 0° (neither leads nor lags).
Power consumed: P = V_rmsI_rms = I²_rmsR = V²_rms/R (fully dissipated as heat).
Power factor = cos0° = 1 (maximum power transfer).

3. AC Through Pure Inductor (L)

V = V₀ sinωt ⇒ I = V₀/(ωL) × sin(ωt − π/2) = I₀ sin(ωt − 90°)
Inductive reactance: X_L = ωL = 2πfL (unit: Ω)
Phase: Current LAGS voltage by 90° (π/2) or equivalently V leads I by 90°.

X_L ∝ f: at DC (f=0), X_L=0 (short circuit). At high f, X_L is large (blocks high-frequency AC).
Average power consumed = 0 (cos90° = 0). Energy is alternately stored in B field and returned.
Pure inductor: power factor = cos90° = 0 (wattless current).
Mnemonic: ELI — EMF (E) leads current (I) in inductor (L).

4. AC Through Pure Capacitor (C)

V = V₀ sinωt ⇒ I = V₀ωC × sin(ωt + π/2) = I₀ sin(ωt + 90°)
Capacitive reactance: X_C = 1/(ωC) = 1/(2πfC) (unit: Ω)
Phase: Current LEADS voltage by 90° (π/2) or equivalently I leads V by 90°.

X_C ∝ 1/f: at DC (f=0), X_C=∞ (open circuit — blocks DC). At high f, X_C≈0 (passes AC freely).
Average power consumed = 0 (cos90° = 0). Energy stored in E field and returned alternately.
Pure capacitor: power factor = cos90° = 0 (wattless current).
Mnemonic: ICE — Current (I) leads EMF (E) in capacitor (C).

⚠️ NEET: ELI the ICE man: E leads I in L (inductor); I leads E in C (capacitor). At DC: inductor = wire (X_L=0), capacitor = open break (X_C=∞). At very high f: inductor = break (X_L=∞), capacitor = wire (X_C=0). No power dissipation in pure L or pure C.

5. Series LCR Circuit

When L, C, and R are connected in series with an AC source, the voltages across L and C are 180° out of phase with each other (opposite in phasor diagram), while voltage across R is in phase with current.

Impedance: Z = √[R² + (X_L − X_C)²] (unit: Ω)
Where X_L = ωL and X_C = 1/(ωC)

Current: I₀ = V₀/Z or I_rms = V_rms/Z

Phase angle: tanϕ = (X_L − X_C)/R
• X_L > X_C: circuit is inductive; V leads I (ϕ positive)
• X_L < X_C: circuit is capacitive; I leads V (ϕ negative)
• X_L = X_C: circuit is at resonance; V and I in phase (ϕ = 0)

5.1 Phasor (Vector) Diagram

V_R is along current direction (in phase with I).
V_L is 90° ahead of I (V_L leads I).
V_C is 90° behind I (V_C lags I).
V_L and V_C are 180° apart → partially cancel → net reactive voltage = |V_L−V_C|.
V_total = √[V_R² + (V_L−V_C)²] (Pythagoras on phasor diagram).

5.2 Voltage Across Individual Components

V_R = IR | V_L = IX_L | V_C = IX_C
Note: V_R + V_L + V_C ≠ V_source (phasor addition, not scalar!)
V_source = √[V_R² + (V_L−V_C)²]

6. Resonance in Series LCR Circuit

Resonance: Occurs when X_L = X_C, i.e., ωL = 1/(ωC). At resonance, impedance Z is minimum (= R), current is maximum (= V/R), and the circuit behaves as purely resistive.

Resonant frequency:
ω₀ = 1/√(LC) ⇒ f₀ = 1/(2π√LC)

At resonance: Z = R (minimum); I = V/R (maximum); ϕ = 0°; cosϕ = 1 (max power).
V_L = V_C (equal and opposite; can be much larger than V_source!)
Quality factor: Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C)
Q = voltage amplification factor = V_L/V_source = V_C/V_source at resonance.

Sharp resonance: Low R → high Q → narrow bandwidth → sharper current peak → better frequency selection (used in radio tuning).
Flat resonance: High R → low Q → broad bandwidth → flatter curve.
Bandwidth = Δω ≈ R/L = ω₀/Q.
Resonance frequency is independent of R (depends only on L and C).

⚠️ NEET (2013, 2016, 2019, 2021): f₀=1/(2π√LC). At resonance: Z=R (min), I=V/R (max), ϕ=0, power factor=1. Q=ω₀L/R. V_L=V_C at resonance (can exceed source V!). Resonant f independent of R. Lower R → sharper resonance.

7. Power in AC Circuits

Instantaneous power: p = VI = V₀I₀ sinωt × sin(ωt+ϕ)
Average power: P_avg = V_rms I_rms cosϕ = I²_rms R
cosϕ = R/Z — the Power Factor

Circuit Type	Power Factor cosϕ	Average Power	Nature
Pure R	1 (ϕ=0°)	V_rmsI_rms (maximum)	Fully dissipative
Pure L	0 (ϕ=90°)	0	Wattless (reactive)
Pure C	0 (ϕ=90°)	0	Wattless (reactive)
Series LCR	R/Z (0 to 1)	V_rmsI_rmsR/Z	Partially dissipative
At resonance	1 (ϕ=0°)	V_rmsI_rms = V²/R	Max power transfer

Wattless current: The component of current I_rms sinϕ contributes zero power (90° out of phase with V). Only I_rms cosϕ produces power.
To improve power factor: add capacitor in inductive circuit (reduce net reactance → ϕ → 0).

8. Transformer

Transformer: A static device that transfers electrical energy from one coil (primary) to another (secondary) using the principle of mutual induction. Changes AC voltage level while keeping frequency unchanged.

Turns ratio: V_S/V_P = N_S/N_P = k (transformation ratio)
Power conservation (ideal): V_PI_P = V_SI_S ⇒ I_S/I_P = N_P/N_S = 1/k
Efficiency: η = (Output power / Input power) × 100%

8.1 Step-Up vs Step-Down

Feature	Step-Up	Step-Down
Turns	N_S > N_P	N_S < N_P
Voltage	V_S > V_P	V_S < V_P
Current	I_S < I_P	I_S > I_P
Usage	Power stations → transmission	Transmission → domestic use

8.2 Energy Losses in Transformer

Copper losses (I²R): Ohmic heating in primary and secondary windings. Minimised by using thick copper wires.
Iron/Eddy current losses: Eddy currents in core produce heat. Minimised by using laminated core (thin insulated iron sheets).
Hysteresis loss: Energy lost per cycle of magnetisation-demagnetisation. Minimised by using soft iron core (low hysteresis loss material).
Flux leakage: Not all flux from primary links secondary. Minimised by winding coils tightly over each other.

8.3 Long-Distance Power Transmission

Power loss in transmission lines: P_loss = I²R
For given power P: I = P/V ⇒ P_loss = P²R/V²
If V is doubled → I halves → loss reduces to 1/4th.
That is why high voltage transmission (11kV to 400kV) is used → very low losses → step down at destination.

⚠️ NEET (2014, 2017, 2020, 2022): V_S/V_P = N_S/N_P. Ideal: V_PI_P=V_SI_S. Transformer works on mutual induction; AC only (not DC!). Laminated core → reduces eddy current loss. High V → low I → low I²R loss. Step-up: V↑, I↓. Step-down: V↓, I↑.

9. LC Oscillations

When a charged capacitor is connected to an inductor (no resistance), energy oscillates between the electric field of C and the magnetic field of L. This is called LC oscillation (analogous to spring-mass SHM).

Angular frequency: ω = 1/√(LC)
Frequency: f = 1/(2π√LC)
Q(t) = Q₀ cosωt (if initially fully charged)
I(t) = I₀ sinωt = ωQ₀ sinωt

Energy conservation: ½LI² + Q²/(2C) = Q₀²/(2C) = constant

9.1 Analogy: LC Circuit ↔ Mechanical SHM

Electrical (LC)	Mechanical (spring-mass)
Charge Q	Displacement x
Current I = dQ/dt	Velocity v = dx/dt
Inductance L	Mass m
1/C (reciprocal capacitance)	Spring constant k
½LI² (magnetic energy)	½mv² (kinetic energy)
Q²/(2C) (electric energy)	½kx² (potential energy)
ω = 1/√(LC)	ω = √(k/m)

🎓 NEET Previous Year Questions

Q1. [NEET 2022] An AC source of 220 V, 50 Hz is connected to an inductor of 0.7 H and resistance 30 Ω in series. Find impedance and current.

Answer X_L = ωL = 2π×50×0.7 = 100π ≈ 220 Ω
Z = √(R²+X_L²) = √(900+48400) ≈ √49300 ≈ 222 Ω
I_rms = 220/222 ≈ 0.99 A

Q2. [NEET 2021] In a series LCR circuit, L = 10 mH, C = 10 µF. Find resonant frequency.

Answer f₀ = 1/(2π√LC) = 1/(2π√(10×10⁻³×10×10⁻⁶))
= 1/(2π√(10⁻⁷)) = 1/(2π×3.162×10⁻⁴) = 1/(1.987×10⁻³)
f₀ ≈ 503 Hz

Q3. [NEET 2020] A transformer has 500 turns in primary, 10 turns in secondary. If 220 V applied to primary, find secondary voltage and type.

Answer V_S = V_P×N_S/N_P = 220×10/500 = 4.4 V. Since V_S < V_P, it is a step-down transformer.

Q4. [NEET 2019] Power factor of a series LCR circuit at resonance is:

Answer At resonance: X_L=X_C ⇒ Z=R ⇒ cosϕ = R/Z = R/R = 1 (maximum power transfer).

Q5. [NEET 2017] The peak voltage of a 220 V AC supply is:

Answer V₀ = V_rms√2 = 220×1.414 = 311 V. Indian mains: 220 V is the RMS value.

💡 Rapid Revision — Key Formulas

V_rms = V₀/√2 | I_rms = I₀/√2 | Indian mains: 220V (rms), 50Hz
X_L = ωL | X_C = 1/ωC | Z = √[R²+(X_L−X_C)²]
tanϕ = (X_L−X_C)/R | cosϕ = R/Z (power factor)
Resonance: f₀=1/(2π√LC) | Z=R | I=V/R (max) | cosϕ=1
P_avg = V_rmsI_rmscosϕ | Pure L/C: P=0 | Pure R: P=V²/R
Transformer: V_S/V_P=N_S/N_P | V_PI_P=V_SI_S (ideal)
LC oscillations: ω=1/√LC | Q analogy with spring-mass SHM
ELI the ICE man: V leads I (inductor) | I leads V (capacitor)

Numericals - Alternating Current - Class 12

CLASS 12 PHYSICS | NCERT SOLUTIONS

Chapter 7 — Alternating Current

25 NCERT Exercise & Exemplar Questions — Step-by-Step Solutions

📝 NCERT Exercise Questions (7.1 – 7.15)

3 MarksQ1 (Ex 7.1). A 100 Ω resistor is connected to a 220 V, 50 Hz AC supply. Find (a) rms current, (b) peak current, (c) net power consumed over a full cycle.

✓ Solution
V_rms = 220 V, R = 100 Ω
(a) I_rms = V_rms/R = 220/100 = 2.2 A
(b) I₀ = I_rms√2 = 2.2×1.414 = 3.11 A
(c) P = V_rmsI_rms = 220×2.2 = 484 W
(Or P = I²_rmsR = (2.2)²×100 = 484 W ✓)

3 MarksQ2 (Ex 7.2). Pure inductor of 25 mH is connected to 220 V, 50 Hz. Find (a) inductive reactance, (b) rms current. What happens to current if frequency doubles?

✓ Solution
L = 25 mH = 0.025 H, f = 50 Hz
(a) X_L = 2πfL = 2π×50×0.025 = 2.5π ≈ 7.85 Ω
(b) I_rms = V_rms/X_L = 220/7.85 = 28.0 A
If f doubles to 100 Hz: X_L = 2π×100×0.025 = 5π = 15.7 Ω
I = 220/15.7 = 14.0 A → current halves (X_L ∝ f, so I ∝ 1/f for pure inductor).

3 MarksQ3 (Ex 7.3). A 15 µF capacitor is connected to 220 V, 50 Hz. Find (a) capacitive reactance, (b) rms current. Does current flow in the external circuit?

✓ Solution
C = 15 µF = 15×10⁻⁶ F, f = 50 Hz
(a) X_C = 1/(2πfC) = 1/(2π×50×15×10⁻⁶) = 1/(4.712×10⁻³) = 212 Ω
(b) I_rms = V_rms/X_C = 220/212 = 1.04 A
Yes, current flows in the external circuit as the capacitor charges/discharges 50 times per second. No charge physically crosses the gap between plates.

3 MarksQ4 (Ex 7.4). A series LCR circuit has L = 80 mH, C = 60 µF, R = 40 Ω, connected to 230 V, 50 Hz. Find impedance, current, and phase angle.

✓ Solution
X_L = 2π×50×0.08 = 8π = 25.13 Ω
X_C = 1/(2π×50×60×10⁻⁶) = 1/(0.01885) = 53.05 Ω
X_L − X_C = 25.13 − 53.05 = −27.92 Ω (capacitive)

Z = √(R² + (X_L−X_C)²) = √(1600+779.5) = √2379.5 = 48.8 Ω

I_rms = 230/48.8 = 4.71 A
tanϕ = (X_L−X_C)/R = −27.92/40 = −0.698 ⇒ ϕ = −34.9°
Negative ϕ → circuit is capacitive; current leads voltage by ~35°.

3 MarksQ5 (Ex 7.5). In the LCR circuit above, find V_R, V_L, V_C individually. Show that V_R+V_L+V_C ≠ 230 V but phasor sum = 230 V.

✓ Solution
V_R = IR = 4.71×40 = 188.4 V
V_L = IX_L = 4.71×25.13 = 118.4 V
V_C = IX_C = 4.71×53.05 = 249.9 V
Scalar sum: 188.4+118.4+249.9 = 556.7 ≠ 230 ✓ (expected: not scalar!)

Phasor: V = √[V_R²+(V_L−V_C)²] = √[(188.4)²+(118.4−249.9)²] = √[35495+17278] ≈ √52773 ≈ 230 V ✓

Note: V_C > V_source (249.9 > 230). This is possible because L and C exchange reactive energy.

3 MarksQ6 (Ex 7.6). Find the frequency at which the LCR circuit above resonates. Find current and power at resonance.

✓ Solution

f₀ = 1/(2π√LC) = 1/(2π√(0.08×60×10⁻⁶)) = 1/(2π√(4.8×10⁻⁶))

= 1/(2π×2.19×10⁻³) = 1/(0.01376) = 72.7 Hz
At resonance: Z = R = 40 Ω ⇒ I = 230/40 = 5.75 A
Power: P = V_rmsI_rmscosϕ = 230×5.75×1 = 1322.5 W (cosϕ=1 at resonance)

3 MarksQ7 (Ex 7.7). A charged 30 µF capacitor is connected to a 27 mH inductor. Find angular frequency and frequency of free LC oscillations.

✓ Solution

ω = 1/√(LC) = 1/√(27×10⁻³×30×10⁻⁶) = 1/√(8.1×10⁻⁷)

ω = 1/(9×10⁻⁴) = 1111 rad/s
f = ω/(2π) = 1111/6.283 = 176.8 Hz

3 MarksQ8 (Ex 7.8). A transformer converts 2400 V to 240 V. Primary has 4000 turns. Find (a) secondary turns, (b) secondary current if primary draws 10 A (assume ideal).

✓ Solution
(a) N_S/N_P = V_S/V_P ⇒ N_S = N_P×V_S/V_P = 4000×240/2400 = 400 turns
(b) Ideal: V_PI_P = V_SI_S
I_S = V_PI_P/V_S = 2400×10/240 = 100 A
Step-down transformer: voltage decreases, current increases.

3 MarksQ9 (Ex 7.9). A town receives 18 kW of power at 230 V. Find the line power loss if resistance of two-wire line = 0.5 Ω each (total 1 Ω). What if transmitted at 4600 V?

✓ Solution
At 230 V: I = P/V = 18000/230 = 78.3 A
P_loss = I²R = (78.3)²×1 = 6130 W ≈ 6.13 kW (34% loss!)

At 4600 V (step-up): I = 18000/4600 = 3.91 A

P_loss = (3.91)²×1 = 15.3 W (only 0.085% loss!)

Ratio: Loss at 230V / Loss at 4600V = 6130/15.3 = 400 = (4600/230)² ✓
Increasing V by factor 20 → loss decreases by factor 400. This is why high-voltage transmission is essential.

3 MarksQ10 (Ex 7.10). Find the time taken for current to reach peak in an AC circuit operating at 50 Hz.

✓ Solution
I = I₀ sinωt; peak when sinωt = 1 ⇒ ωt = π/2
t = π/(2ω) = π/(2×2π×50) = 1/(200) = 0.005 s = 5 ms
(T/4 = 1/(4×50) = 0.005 s — quarter time period to reach first peak.)

3 MarksQ11 (Ex 7.11). A series LCR circuit with R=20Ω, L=1.5H, C=35µF is connected to a variable-frequency 200V supply. Find the frequency at which power transferred is maximum. Also find Q-factor.

✓ Solution
Max power at resonance: f₀ = 1/(2π√LC)
√LC = √(1.5×35×10⁻⁶) = √(5.25×10⁻⁵) = 7.245×10⁻³

f₀ = 1/(2π×7.245×10⁻³) = 1/0.04552 = 21.97 Hz ≈ 22 Hz

Q = ω₀L/R = (2π×22×1.5)/20 = (207.35)/20 = 10.4
(Or Q = (1/R)√(L/C) = (1/20)√(1.5/35×10⁻⁶) = (1/20)√42857 = 0.05×207 ≈ 10.4 ✓)

3 MarksQ12 (Ex 7.12). An LC circuit has L=200mH, C=500pF. At what frequency is it tuned? What is the value of Q if R=40Ω?

✓ Solution

f₀ = 1/(2π√(200×10⁻³×500×10⁻¹²)) = 1/(2π√(10⁻¹⁰))

= 1/(2π×10⁻⁵) = 10⁵/(2π) = 15924 Hz ≈ 15.9 kHz
Q = (1/R)√(L/C) = (1/40)√(0.2/(500×10⁻¹²)) = (1/40)√(4×10⁸)
= (1/40)×20000 = 500 (very sharp; excellent frequency selectivity)

3 MarksQ13 (Ex 7.13). A power transmission line feeds 2 kW to a village at 5000 V. If resistance of line is 20 Ω, find power loss in transmission. What fraction of power lost?

✓ Solution
I = P/V = 2000/5000 = 0.4 A

P_loss = I²R = (0.4)²×20 = 0.16×20 = 3.2 W

Fraction lost = 3.2/2000 = 0.0016 = 0.16% (extremely small at high V!)

3 MarksQ14 (Ex 7.14). A 60 W lamp is connected to 220 V, 50 Hz supply through a capacitor. What capacitance is needed so that lamp gets full rated power without using a resistor?

✓ Solution
Lamp rated: P = 60 W at V_L = 220 V (assumed); I_L = P/V_L = 60/220 = 0.273 A
R_lamp = V²/P = (220)²/60 = 807 Ω
For lamp (resistive) to be at full brightness, I must match rated current.
In series RC: Z = V_source/I = 220/0.273 = 807 Ω (but this means X_C should be chosen such that Z is right for given source).
If source is at higher V (say 440V): Z = 440/0.273 = 1612; X_C = √(Z²−R²) = √(1612²−807²) = √(2598544−651249) = √1947295 = 1395 Ω
C = 1/(2πfX_C) = 1/(2π×50×1395) = 2.28 µF

3 MarksQ15 (Ex 7.15). The voltage across a capacitor in a series RC circuit is 5V rms, and voltage across resistor is 12V rms. Find supply voltage.

✓ Solution
V_R and V_C are 90° apart (R: in phase with I; C: lags I by 90°).

V_source = √(V_R² + V_C²) = √(144 + 25) = √169 = 13 V

This is NOT 12+5=17! Phasor addition applies (90° between V_R and V_C).

🌟 Additional / Exemplar Questions (Q16 – Q25)

3 MarksQ16. An AC source of voltage V = 200 sin(100πt) is connected to a series combination of R = 50 Ω and L = 100 mH. Find impedance, rms current, and power consumed.

✓ Solution
V₀ = 200 V, ω = 100π rad/s
X_L = ωL = 100π×0.1 = 10π ≈ 31.42 Ω
Z = √(R²+X_L²) = √(2500+987.5) = √3487.5 = 59.1 Ω
I_rms = V_rms/Z = (200/√2)/59.1 = 141.4/59.1 = 2.39 A
cosϕ = R/Z = 50/59.1 = 0.846
P = V_rmsI_rmscosϕ = 141.4×2.39×0.846 = 286 W

3 MarksQ17. In an LCR circuit, R=100Ω, L=0.5H, C=10µF. Supply is 200V at variable frequency. Find resonant ω₀, I at resonance, and V_L at resonance.

✓ Solution
ω₀ = 1/√(LC) = 1/√(0.5×10×10⁻⁶) = 1/√(5×10⁻⁶) = 1/(2.236×10⁻³) = 447 rad/s
At resonance: I = V/R = 200/100 = 2 A
V_L = I×X_L = I×ω₀L = 2×447×0.5 = 447 V (!)
Note: V_L = 447 V >> 200 V (source). Voltage amplification! Q = V_L/V = 447/200 ≈ 2.24.

3 MarksQ18. An inductor L, capacitor C, and resistor R have values 20 mH, 100 µF, and 50 Ω. They are connected in series. If applied V = 100 sin(314t), find whether circuit is at resonance.

✓ Solution
ω = 314 rad/s (from sin314t)
Resonant ω₀ = 1/√(LC) = 1/√(20×10⁻³×100×10⁻⁶) = 1/√(2×10⁻⁶)
= 1/(1.414×10⁻³) = 707 rad/s
Applied ω = 314 ≠ 707 ⇒ NOT at resonance.
X_L = 314×0.02 = 6.28 Ω; X_C = 1/(314×10⁻⁴) = 31.85 Ω
X_C > X_L → circuit is capacitive (I leads V).

3 MarksQ19. The power factor of a series LCR circuit is 0.5. If R = 100 Ω, find impedance Z and the net reactance (X_L−X_C).

✓ Solution
cosϕ = R/Z = 0.5 ⇒ Z = R/cosϕ = 100/0.5 = 200 Ω
Z² = R² + (X_L−X_C)²
(X_L−X_C)² = Z²−R² = 40000−10000 = 30000
|X_L−X_C| = √30000 = 173.2 Ω
Phase angle: ϕ = cos⁻¹(0.5) = 60°

3 MarksQ20. Explain why a capacitor blocks DC but allows AC to pass. Why does an inductor do the opposite?

✓ Solution
Capacitor: X_C = 1/(2πfC). At DC (f=0): X_C = ∞ (infinite reactance → open circuit → blocks DC). At AC (f>0): X_C is finite → current flows → passes AC. Higher f → lower X_C → more current → passes high-frequency AC easily.

Inductor: X_L = 2πfL. At DC (f=0): X_L = 0 (zero reactance → short circuit → passes DC freely → behaves as wire). At AC (f>0): X_L is finite → opposes AC → higher f → more opposition. At very high f: X_L → ∞ (blocks high-frequency AC).
This complementary behaviour makes L and C useful for frequency filtering (low-pass, high-pass, and band-pass filters).

3 MarksQ21. In a series LCR circuit at resonance, R = 10 Ω, L = 0.1 H, C = 25 µF. Find Q-factor and bandwidth.

✓ Solution
ω₀ = 1/√(LC) = 1/√(0.1×25×10⁻⁶) = 1/√(2.5×10⁻⁶) = 1/(1.581×10⁻³) = 632.5 rad/s
Q = ω₀L/R = 632.5×0.1/10 = 6.325
Bandwidth: Δω = ω₀/Q = 632.5/6.325 = 100 rad/s
(Or Δω = R/L = 10/0.1 = 100 rad/s ✓)

3 MarksQ22. A step-up transformer has primary coil 200 turns, secondary coil 3000 turns. Primary voltage is 240 V. (a) Find secondary voltage. (b) If primary draws 5A, find secondary current (assume 90% efficiency).

✓ Solution
(a) V_S = V_P×N_S/N_P = 240×3000/200 = 3600 V
(b) η = V_SI_S/(V_PI_P) ⇒ I_S = η×V_PI_P/V_S

I_S = 0.9×240×5/3600 = 1080/3600 = 0.3 A

(Ideal I_S would be 240×5/3600 = 0.333 A. With 90% efficiency, less output → 0.3 A.)

3 MarksQ23. In an AC circuit, V = 200 V (rms) and I = 5A (rms). If power consumed = 500 W, find impedance, resistance, and power factor.

✓ Solution
Z = V/I = 200/5 = 40 Ω
P = VI cosϕ ⇒ cosϕ = P/(VI) = 500/(200×5) = 0.5
R = Z cosϕ = 40×0.5 = 20 Ω
Phase angle: ϕ = 60°. Net reactance: |X_L−X_C| = Z sinϕ = 40×sin60° = 40×0.866 = 34.6 Ω.

3 MarksQ24. Explain wattless current. In a circuit with cosϕ = 0.8, find the wattless component of 10 A rms current.

✓ Solution
Wattless current: The component of AC current that is 90° out of phase with voltage. It does NOT contribute to power dissipation. Only the in-phase component (I cosϕ) produces power.
Wattless component: I_w = I sinϕ
cosϕ = 0.8 ⇒ sinϕ = 0.6 (using sin²+cos²=1)

I_wattless = 10×0.6 = 6 A

Power component: I_power = 10×0.8 = 8A. Check: √(8²+6²) = √100 = 10 A ✓

3 MarksQ25. Why can't a transformer work on DC? What happens if DC is applied to the primary of a transformer?

✓ Solution
Transformer works on the principle of mutual induction, which requires a changing magnetic flux in the primary coil (ε = −Md I/dt). For flux to change, the current must change with time → AC.

If DC is applied: After initial transient, current becomes constant → dI/dt = 0 → dϕ/dt = 0 → no EMF induced in secondary → transformer produces no output.
Additionally: the DC current encounters only the very low resistance of the copper winding (no reactance at f=0) → extremely large current → excessive heating → primary coil may burn out. This is why transformers must only be used with AC.

✍ Score Guide — 25 Questions
NCERT Ex 7.1–7.15: 3 marks each — 45 marks | Additional Q16–Q25: 3 marks each — 30 marks | Grand Total: 75 marks

Formula Capsule - Alternating Current - Class 12

CLASS 12 PHYSICS | FORMULA CAPSULE

Alternating Current

Chapter 7 — Complete Formula Sheet & High-Yield Facts for NEET/JEE

📅 Section A — AC Basics & RMS Values

Quantity	Formula	Key Note
AC voltage	V = V₀ sinωt	V₀ = peak; ω=2πf
AC current	I = I₀ sin(ωt ± ϕ)	ϕ = phase difference w.r.t. V
RMS voltage	V_rms = V₀/√2	Meters read RMS; Indian mains = 220V rms
RMS current	I_rms = I₀/√2	Heating effect = DC equivalent
Mean (half cycle)	V_avg = 2V₀/π	Full cycle avg = 0
Peak-to-peak	V_pp = 2V₀

📅 Section B — Reactance, Impedance & Phase

Circuit	Reactance/Impedance	Phase (V vs I)	Power
Pure R	Z = R	In phase (ϕ=0°)	P = V_rmsI_rms
Pure L	X_L = ωL = 2πfL	V leads I by 90° (ELI)	P = 0
Pure C	X_C = 1/(ωC) = 1/(2πfC)	I leads V by 90° (ICE)	P = 0
Series LCR	Z = √[R²+(X_L−X_C)²]	tanϕ = (X_L−X_C)/R	P = V_rmsI_rmscosϕ
At resonance	Z = R (minimum)	ϕ = 0° (in phase)	P = V²/R (max)

📅 Section C — Resonance & Power

Quantity	Formula	Key Note
Resonant frequency	f₀ = 1/(2π√LC)	Independent of R
Resonant ω	ω₀ = 1/√(LC)	X_L = X_C at resonance
Quality factor	Q = ω₀L/R = 1/(ω₀CR) = (1/R)√(L/C)	Higher Q → sharper resonance
Bandwidth	Δω = R/L = ω₀/Q	Low R → narrow band → sharp peak
Power factor	cosϕ = R/Z	1 at resonance; 0 for pure L or C
Average power	P = V_rmsI_rmscosϕ = I²_rmsR	Only R dissipates power
Wattless current	I sinϕ	90° out of phase; zero power

📅 Section D — Transformer & LC Oscillations

Quantity	Formula	Key Note
Voltage ratio	V_S/V_P = N_S/N_P	Step-up: N_S>N_P
Current ratio (ideal)	I_S/I_P = N_P/N_S	V↑ → I↓ (power conserved)
Power conservation	V_PI_P = V_SI_S (ideal)	Real: η = output/input < 100%
Transmission loss	P_loss = I²R = P²R/V²	High V → low I → low loss
LC oscillation freq	ω = 1/√(LC); f = 1/(2π√LC)	Same as resonant f of LCR
LC energy conservation	½LI² + Q²/(2C) = const	Analogous to KE + PE = const

🧠 Memory Tricks

“ELI the ICE man” In inductor (L): EMF (E) leads Current (I) by 90° → ELI. In capacitor (C): Current (I) leads EMF (E) by 90° → ICE. Remember: ELI for inductor, ICE for capacitor.

“At DC: L = Wire, C = Break” At f=0 (DC): X_L=0 (inductor is short circuit → wire), X_C=∞ (capacitor is open circuit → break). At very high f: opposite → L blocks, C passes. This is why capacitors block DC!

“Resonance: Z = R (min), I = max, cosϕ = 1” At resonance X_L=X_C, they cancel. Circuit is purely resistive. Maximum current. Maximum power. Power factor = 1. Resonant f = 1/(2π√LC) — depends only on L and C, NOT on R.

“Transformer: V↑ → I↓ (power constant)” Step-up: voltage increases but current decreases proportionally (P = VI = const). This is why high-V transmission works: less I → less I²R loss. Cannot work on DC (no dI/dt → no induced EMF).

“Phasor ≠ Scalar; V_R+V_L+V_C ≠ V_source” Individual voltages across R, L, C add as phasors (vector diagram), NOT as scalars. V = √[V_R²+(V_L−V_C)²]. This is why V_L or V_C can exceed V_source!

“Q = Voltage Amplification” Q-factor = ω₀L/R = V_L/V at resonance. High Q → large V across L and C even with small source V. Radio tuning uses high-Q circuits to select narrow frequency bands.

🔢 Critical Values

Indian mains: 220V rms, 50Hz V₀ = V_rms√2 = 311V Pure L/C: P = 0 (wattless) Resonance: Z=R, I=max, cosϕ=1 f₀ = 1/(2π√LC) Q = ω₀L/R = (1/R)√(L/C) Bandwidth = R/L P_loss = P²R/V²

❌ Common Mistakes to Avoid

V_R+V_L+V_C ≠ V_source: These are phasor sums. V = √[V_R²+(V_L−V_C)²]. Scalar addition is wrong.
RMS vs Peak: Meters read RMS. Peak = RMS×√2. Confusing V₀ and V_rms gives factor-of-√2 errors.
Capacitor blocks DC, not AC: At f=0, X_C=∞. Inductor blocks high-frequency AC (X_L=ωL). Don't swap these.
Power in pure L or C = 0: Even though current flows, average power dissipated is zero. Only R dissipates power.
Resonant frequency independent of R: f₀=1/(2π√LC). R affects sharpness (Q, bandwidth), NOT the resonant frequency itself.
Transformer works on AC only: DC gives constant flux, dI/dt = 0 → no induced EMF. Also, DC can burn the primary (no reactance, only tiny R).

📱 Practice MCQs for this topic inside our App