Alternating Current — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Alternating Current" — 6 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Alternating Current" — 6 important questions with detailed answers for CBSE board exam…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 12 Physics — Alternating Current. Step-by-step answers to all 6 textbook questions from this chapter, written for CBSE board preparation and free to use.
Alternating Current — All 6 Questions Solved
Q1. Define AC current and voltage. Explain peak value, RMS value, and average value.
AC (Alternating Current): Electric current that reverses direction periodically with time.
I = I₀ sin(ωt + φ) or I = I₀ cos(ωt + φ)
where I₀ = peak (maximum) current
ω = angular frequency (rad/s)
t = time
φ = phase constant
AC Voltage: V = V₀ sin(ωt + φ)
where V₀ = peak voltage
Three Important Values:
1. Peak (Maximum) Value:
I_peak = I₀ (maximum value of current)
V_peak = V₀ (maximum value of voltage)
- Used in calculations of power dissipation in resistor
- Important for insulation design
2. RMS (Root Mean Square) Value:
I_rms = I₀/√2 = 0.707 I₀
V_rms = V₀/√2 = 0.707 V₀
Definition: √[(1/T)∫₀^T I² dt] for complete cycle
Calculation:
For I = I₀ sin(ωt)
I_rms² = (1/T)∫₀^T I₀² sin²(ωt) dt = I₀²/2
I_rms = I₀/√2
SI Unit: Ampere (A), Volt (V)
Significance:
- AC meter readings show RMS values
- Equivalent to DC current producing same heating effect
- Household voltage 230 V is RMS value
- Most practical calculations use RMS values
3. Average Value (over half cycle):
I_avg = (2/π)I₀ = 0.637 I₀
V_avg = (2/π)V₀ = 0.637 V₀
Calculation:
I_avg = (1/T/2)∫₀^(T/2) I₀ sin(ωt) dt = 2I₀/π
Used in rectified AC (diode circuits)
Relationships:
I_rms : I_avg : I_peak = 1 : π/2 : √2 = 1 : 1.57 : 1.414
Power calculation (most important):
P = I_rms² R = V_rms² / R = V_rms I_rms cos φ
Note: Using peak values directly gives incorrect power calculations
Q2. Explain AC circuit with pure resistance. Draw phasor diagram and power equation.
AC Circuit with Pure Resistance (R):
Circuit: AC source connected to resistor R only (no L or C)
Voltage applied: V = V₀ sin(ωt)
Current through resistor (Ohm's law):
I = V/R = (V₀/R) sin(ωt) = I₀ sin(ωt)
where I₀ = V₀/R (peak current)
Key Observations:
1. Current and voltage in phase (φ = 0°)
2. No phase difference between I and V
3. Peak values related by: V₀ = I₀R
4. RMS values: V_rms = I_rms × R
Phasor Diagram (at t = 0):
- Draw V and I vectors both along same direction (horizontal)
- Both at angle 0° (in phase)
- Magnitude of V vector: V_rms
- Magnitude of I vector: I_rms
- Angle between them: 0°
Power:
Instantaneous power:
P(t) = V(t) × I(t) = V₀ sin(ωt) × I₀ sin(ωt)
P(t) = V₀I₀ sin²(ωt)
P(t) = (V₀I₀/2)[1 - cos(2ωt)]
Average power (over complete cycle):
P_avg = (1/T)∫₀^T P(t) dt = V₀I₀/2 = V_rms × I_rms
Alternatively:
P = I_rms² R = V_rms²/R = V_rms I_rms cos φ
where cos φ = power factor
For resistance, φ = 0°, so cos φ = 1
P = V_rms I_rms (pure resistance)
Power factor:
PF = cos φ = 1 (for pure resistance)
Energy dissipated as heat in resistance, no reactive power
SI Units: Power in Watts (W)
Characteristics:
- Voltage and current in phase
- Power always positive (no reactive component)
- Average power equals VI (unlike L or C circuits)
- Complete energy absorption by resistance
Q3. Explain AC circuit with pure inductance (L). Draw impedance and phasor diagram.
AC Circuit with Pure Inductance (L):
Circuit: AC source connected to pure inductor only (no R or C)
Voltage applied: V = V₀ sin(ωt)
Induced emf in inductor:
ε = -L(dI/dt) = -V (opposes applied voltage)
From Faraday's law:
V = L(dI/dt)
V₀ sin(ωt) = L(dI/dt)
Integrating:
I = -(V₀/ωL) cos(ωt) = (V₀/ωL) sin(ωt - π/2)
I = I₀ sin(ωt - π/2)
where I₀ = V₀/ωL
Inductive Reactance:
X_L = ωL = 2πfL (Ohms, Ω)
where ω = 2πf (angular frequency)
f = frequency (Hz)
L = inductance (H)
Key Observations:
1. Current lags voltage by 90° (φ = -90°)
2. When V = maximum, I = 0
3. When V = 0, I = maximum
4. Peak current: I₀ = V₀/X_L
5. X_L increases with frequency (ω)
Phasor Diagram:
- V vector: horizontal (0°)
- I vector: vertically downward (90° behind V)
- Angle between V and I: 90°
- Magnitude: V_rms, I_rms = V_rms/X_L
Impedance (Z):
Z = √(X_L²) = X_L = ωL (pure inductance)
Phase angle:
tan φ = -X_L/0 → φ = -90°
cos φ = 0
Power:
Instantaneous power:
P(t) = V(t) × I(t) = V₀ sin(ωt) × I₀ sin(ωt - π/2)
P(t) = V₀I₀ sin(ωt) × (-cos(ωt))
P(t) = -(V₀I₀/2) sin(2ωt)
Average power:
P_avg = 0 (no net energy dissipation)
Reactive power:
Q_L = V_rms I_rms = V_rms²/X_L = I_rms² X_L (VAR)
Power factor:
PF = cos φ = cos(-90°) = 0
Characteristics:
- Current lags voltage by 90°
- No real power dissipation (average power = 0)
- Energy stored and returned each cycle (reactive power)
- Inductor acts as short circuit at DC (X_L = 0)
- Inductor acts as open circuit at infinite frequency
Q4. Explain AC circuit with pure capacitance (C). Compare with inductance.
AC Circuit with Pure Capacitance (C):
Circuit: AC source connected to pure capacitor only (no R or L)
Voltage applied: V = V₀ sin(ωt)
Charge on capacitor:
Q = CV = CV₀ sin(ωt)
Current (rate of charge flow):
I = dQ/dt = CV₀ω cos(ωt)
I = CV₀ω sin(ωt + π/2)
I = I₀ sin(ωt + π/2)
where I₀ = CV₀ω = V₀/(1/ωC)
Capacitive Reactance:
X_C = 1/(ωC) = 1/(2πfC) (Ohms, Ω)
where ω = 2πf
C = capacitance (F)
Key Observations:
1. Current leads voltage by 90° (φ = +90°)
2. When V = maximum, I = 0
3. When V = 0, I = maximum
4. Peak current: I₀ = V₀/X_C
5. X_C decreases with frequency
Phasor Diagram:
- V vector: horizontal (0°)
- I vector: vertically upward (90° ahead of V)
- Angle between V and I: 90°
- I leads V by 90°
Comparison: Inductance vs Capacitance
| Property | Inductance | Capacitance |
|----------|-----------|-------------|
| Reactance | X_L = ωL | X_C = 1/(ωC) |
| Frequency effect | Increases | Decreases |
| Phase angle | φ = -90° | φ = +90° |
| Current relation | Lags by 90° | Leads by 90° |
| At DC | X_L = 0 (short) | X_C = ∞ (open) |
| At high f | X_L = ∞ (open) | X_C = 0 (short) |
| Average power | P = 0 | P = 0 |
| Reactive power | Q_L = V_rms I_rms | Q_C = -V_rms I_rms |
| Power factor | cos(-90°) = 0 | cos(90°) = 0 |
| Energy storage | Magnetic field | Electric field |
| Effect on voltage | Opposes rise | Allows rise |
Power:
Instantaneous power:
P(t) = V₀ sin(ωt) × I₀ sin(ωt + π/2)
P(t) = V₀I₀ sin(ωt) cos(ωt)
P(t) = (V₀I₀/2) sin(2ωt)
Average power:
P_avg = 0 (no dissipation)
Reactive power:
Q_C = -V_rms I_rms = -V_rms²/X_C (capacitive)
Characteristics:
- Current leads voltage by 90°
- No real power dissipation
- Energy stored and returned each cycle
- Acts as open circuit at DC
- Acts as short circuit at high frequency
Q5. Analyze AC series RLC circuit. Derive impedance and resonance condition.
AC Series RLC Circuit: Resistor, Inductor, Capacitor in series with AC source
Voltage distribution:
V = V_R + V_L + V_C = V₀ sin(ωt)
Voltage across elements:
V_R = IR (in phase with I)
V_L = IX_L (leads I by 90°)
V_C = IX_C (lags I by 90°)
Phasor Analysis:
- All currents same (series circuit)
- Net voltage from phasors
- V_L and V_C opposite direction (cancel partially)
Net reactance:
X = X_L - X_C = ωL - 1/(ωC)
Impedance (Z):
Z = √[R² + (X_L - X_C)²] = √[R² + (ωL - 1/ωC)²]
Alternatively:
Z = √[R² + X²]
Peak current:
I₀ = V₀/Z = V₀/√[R² + (X_L - X_C)²]
Phase angle φ:
tan φ = X/R = (X_L - X_C)/R = (ωL - 1/ωC)/R
cos φ = R/Z (power factor)
Power:
P_avg = V_rms I_rms cos φ = I_rms² R
Resonance Condition:
At resonance, X_L = X_C (net reactance = 0)
ωL = 1/(ωC)
ω² = 1/(LC)
ω₀ = 1/√(LC) (resonant angular frequency)
Resonant frequency:
f₀ = 1/(2π√(LC)) (Hertz)
At Resonance:
1. Impedance minimum: Z = R (only resistance)
2. Current maximum: I_max = V₀/R
3. Phase angle: φ = 0° (voltage and current in phase)
4. Power factor: PF = cos 0° = 1 (maximum)
5. Power dissipation maximum: P_max = V_rms²/R
6. X_L = X_C (voltages cancel)
Quality Factor (Q):
Q = ω₀L/R = 1/(ω₀RC) = f₀/Δf
where Δf = bandwidth (frequency range for P > P_max/2)
Bandwidth at half-power points:
Δf = R/(2πL) = R × C (in radians)
Characteristics near resonance:
- Narrow Q (high Q): Sharp resonance, narrow bandwidth
- Wide Q (low Q): Broad resonance, wide bandwidth
Applications:
- Radio tuning (selecting specific frequency)
- Filtering circuits
- Power factor correction
Q6. Explain transformer principle. Derive voltage and current transformation ratios.
Transformer: Device that converts AC voltage and current using electromagnetic induction and mutual inductance.
Principle:
1. Primary coil (input): Connected to AC source
2. Iron core: Carries alternating magnetic field
3. Secondary coil (output): Magnetic field induces emf
4. Mutual inductance couples the coils
Theory (Ideal Transformer):
Primary coil:
- N₁ turns
- Voltage V₁ = V₁₀ sin(ωt)
- Current I₁
- Induced emf by changing flux: ε₁ = -N₁(dΦ/dt)
Secondary coil:
- N₂ turns
- Voltage V₂ (induced emf)
- Current I₂
- Induced emf: ε₂ = -N₂(dΦ/dt)
Both experience same flux change (ideal core):
ε₁/ε₂ = N₁/N₂
Voltage Transformation Ratio:
V₁/V₂ = N₁/N₂
Alternatively:
V₂ = V₁ × (N₂/N₁)
where V₁ = primary voltage
V₂ = secondary voltage
N₁/N₂ = turns ratio
Current Transformation Ratio:
From conservation of power (ideal transformer):
P₁ = P₂
V₁I₁ = V₂I₂
I₁/I₂ = V₂/V₁ = N₂/N₁
Alternatively:
I₂ = I₁ × (N₁/N₂)
Impedance Transformation:
Z₂' = Z₂ × (N₁/N₂)² (impedance reflected to primary)
Types of Transformers:
1. Step-up transformer: N₂ > N₁
- V₂ > V₁ (voltage increases)
- I₂ < I₁ (current decreases)
2. Step-down transformer: N₂ < N₁
- V₂ < V₁ (voltage decreases)
- I₂ > I₁ (current increases)
3. Isolation transformer: N₁ = N₂
- V₂ = V₁
- I₂ = I₁
Ideal Transformer Equations:
1. Voltage ratio: V₁/V₂ = N₁/N₂
2. Current ratio: I₁/I₂ = N₂/N₁
3. Power conservation: V₁I₁ = V₂I₂
4. Impedance ratio: Z₁/Z₂' = (N₁/N₂)²
Losses in Real Transformer:
1. Copper loss: I²R heating in coils
2. Iron loss: Hysteresis and eddy currents in core
3. Flux leakage: Not all flux links both coils
Efficiency:
η = (P_out/P_in) × 100% = (V₂I₂/V₁I₁) × 100%
Typical efficiency: 95-99% for large transformers
Applications:
- Power transmission (step-up to reduce loss)
- Distribution (step-down to safe voltages)
- Isolation of circuits
- Impedance matching
- Voltage regulation
Alternating Current — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. The peak value of an alternating voltage is 220 V. Its RMS value is:
- A. 220 V
- B. 110 V
- C. 220/√2 V
- D. 220√2 V
Answer: C. 220/√2 V — For a sinusoidal AC voltage, the RMS (Root Mean Square) value is the peak value divided by √2. So, V_RMS = 220/√2 V.
MCQ 2. In an AC circuit containing only a pure resistor, the voltage across the resistor and the current flowing through it are:
- A. In phase
- B. Voltage leads current by π/2
- C. Current leads voltage by π/2
- D. Voltage lags current by π/4
Answer: A. In phase — For a purely resistive AC circuit, the voltage and current are always in phase, meaning their peaks and zeros occur simultaneously.
MCQ 3. The inductive reactance of an inductor connected to an AC source increases with:
- A. Increase in frequency
- B. Decrease in frequency
- C. Decrease in inductance
- D. Increase in resistance
Answer: A. Increase in frequency — Inductive reactance, X_L = ωL = 2πfL. Thus, X_L is directly proportional to the frequency (f) of the AC source.
MCQ 4. A capacitor acts as an infinite resistance for:
- A. DC
- B. AC
- C. Both DC and AC
- D. Neither DC nor AC
Answer: A. DC — The capacitive reactance is X_C = 1/(ωC) = 1/(2πfC). For DC, the frequency f = 0, so X_C becomes infinite, effectively blocking the DC current.
MCQ 5. In a series LCR circuit, if X_L > X_C, then the circuit behaves as a:
- A. Purely resistive circuit
- B. Inductive circuit
- C. Capacitive circuit
- D. Resonant circuit
Answer: B. Inductive circuit — When the inductive reactance (X_L) is greater than the capacitive reactance (X_C), the net reactance of the circuit is inductive, causing the voltage to lead the current.
MCQ 6. A series LCR circuit has R = 400 Ω, L = 0.8 H, C = 2.5 μF, and an AC source of angular frequency ω = 1000 rad/s. The phase difference between the current and the voltage is approximately:
- A. 0 rad
- B. π/2 rad
- C. π/4 rad
- D. -π/4 rad
Answer: C. π/4 rad — First, calculate X_L = ωL = 1000 * 0.8 = 800 Ω. Then, calculate X_C = 1/(ωC) = 1/(1000 * 2.5 * 10^-6) = 400 Ω. The phase angle φ is given by tan φ = (X_L - X_C) / R = (800 - 400) / 400 = 1. Therefore, φ = π/4 rad.
MCQ 7. The resonant frequency of a series LCR circuit is given by:
- A. 1/(LC)
- B. 1/(2π√LC)
- C. 2π√LC
- D. √(LC)
Answer: B. 1/(2π√LC) — The resonant frequency (f_0) for a series LCR circuit occurs when the inductive and capacitive reactances are equal (X_L = X_C), which gives f_0 = 1/(2π√LC).
MCQ 8. The average power dissipated in an AC circuit is given by:
- A. V_rms * I_rms
- B. V_rms * I_rms * sin φ
- C. V_rms * I_rms * cos φ
- D. I_rms^2 * Z
Answer: C. V_rms * I_rms * cos φ — The average power dissipated in an AC circuit is P_avg = V_rms * I_rms * cos φ, where cos φ is the power factor.
MCQ 9. For a purely inductive AC circuit, the power factor is:
- A. 0
- B. 1
- C. 0.5
- D. -1
Answer: A. 0 — In a purely inductive circuit, the phase difference (φ) between voltage and current is π/2. The power factor is cos φ = cos(π/2) = 0.
MCQ 10. A step-up transformer converts 220 V AC to 2200 V AC. If the primary coil has 100 turns, how many turns does the secondary coil have?
- A. 10 turns
- B. 100 turns
- C. 1000 turns
- D. 10000 turns
Answer: C. 1000 turns — For an ideal transformer, the ratio of voltages is equal to the ratio of turns: V_s/V_p = N_s/N_p. So, N_s = N_p * (V_s/V_p) = 100 * (2200/220) = 100 * 10 = 1000 turns.
MCQ 11. A series LCR circuit has L = 1 H, C = 1 μF, R = 40 Ω. The quality factor (Q-factor) of the circuit is:
- A. 10
- B. 20
- C. 25
- D. 50
Answer: C. 25 — The Q-factor is given by Q = (1/R)√(L/C). Substituting the values, Q = (1/40)√(1 / (1 * 10^-6)) = (1/40)√(10^6) = (1/40) * 1000 = 25.
MCQ 12. The angular frequency of free oscillations in an LC circuit is:
- A. 1/LC
- B. 1/√(LC)
- C. √(LC)
- D. 2π√(LC)
Answer: B. 1/√(LC) — The angular frequency (ω) of LC oscillations is given by ω = 1/√(LC), assuming ideal components with no damping.
MCQ 13. A choke coil is preferred over a resistance for controlling current in an AC circuit because:
- A. It consumes less power
- B. It has a high resistance
- C. It can vary current more easily
- D. It is cheaper
Answer: A. It consumes less power — A choke coil has low resistance and high inductance. Due to low resistance, the power dissipated (P = I^2 R) is minimal, making it energy-efficient compared to a resistor which dissipates power as heat.
MCQ 14. In an AC circuit, the wattless component of current is:
- A. I sin φ
- B. I cos φ
- C. I
- D. I tan φ
Answer: A. I sin φ — The wattless current (or reactive current) is the component of current that is 90 degrees out of phase with the voltage, contributing zero to the average power. It is given by I sin φ.
MCQ 15. An LCR series circuit has a voltage source of 200 V and a current of 2 A. The power dissipated in the circuit is 200 W. The power factor of the circuit is:
- A. 0.5
- B. 0.25
- C. 0.75
- D. 1
Answer: A. 0.5 — The average power P = V_rms * I_rms * cos φ. Given P = 200 W, V_rms = 200 V, I_rms = 2 A. So, 200 = 200 * 2 * cos φ, which gives cos φ = 200 / 400 = 0.5.
MCQ 16. A transformer works on the principle of:
- A. Self-induction
- B. Mutual induction
- C. Eddy currents
- D. Lenz's law
Answer: B. Mutual induction — A transformer consists of two coils (primary and secondary) magnetically coupled. When AC flows through the primary, it produces a varying magnetic flux, which induces an EMF in the secondary coil due to mutual induction.
MCQ 17. An AC generator converts:
- A. Electrical energy into mechanical energy
- B. Mechanical energy into electrical energy
- C. Chemical energy into electrical energy
- D. Heat energy into electrical energy
Answer: B. Mechanical energy into electrical energy — An AC generator (alternator) operates on the principle of electromagnetic induction, converting mechanical energy (e.g., from rotation) into electrical energy.
MCQ 18. The average value of AC current over a full cycle is:
- A. I_peak
- B. I_rms
- C. 0
- D. (2/π)I_peak
Answer: C. 0 — Over a complete cycle, the alternating current is positive for half the cycle and negative for the other half, with symmetrical values. Thus, the net average value over a full cycle is zero.
MCQ 19. A series LCR circuit consists of a coil with inductance 5 H and resistance 20 Ω, and a capacitor of capacitance 20 μF. The resonant angular frequency of the circuit is:
- A. 50 rad/s
- B. 100 rad/s
- C. 200 rad/s
- D. 250 rad/s
Answer: B. 100 rad/s — The resonant angular frequency ω_0 = 1/√(LC). Given L = 5 H, C = 20 μF = 20 × 10^-6 F. So, ω_0 = 1/√(5 * 20 * 10^-6) = 1/√(100 * 10^-6) = 1/√(10^-4) = 1/(10^-2) = 100 rad/s.
MCQ 20. In a series LCR circuit, the RMS voltage across the inductor (V_L) is 50 V, across the capacitor (V_C) is 20 V, and across the resistor (V_R) is 40 V. The RMS voltage of the AC source is:
- A. 110 V
- B. 70 V
- C. 50 V
- D. 30 V
Answer: C. 50 V — The total RMS voltage V_source = √[V_R^2 + (V_L - V_C)^2]. Substituting the values, V_source = √[40^2 + (50 - 20)^2] = √[1600 + 30^2] = √[1600 + 900] = √2500 = 50 V.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.