NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current provides detailed topic summaries, essential formulas, phasor analysis, and step-by-step solutions for AC circuits containing pure R, L, and C components, series LCR circuits, resonance, power factor, LC oscillations, and transformers.
Overview of Topics & Subtopics (Click to Jump)
1. AC Voltage Applied to Resistor, Inductor, and Capacitor
Short Note:
• AC Current & Voltage: Alternating voltage V = V₀ sin(ω t), where V₀ is peak voltage and ω = 2π f is angular frequency.
• RMS Value (Root Mean Square): V_rms = V₀ / √2 ≈ 0.707 V₀ and I_rms = I₀ / √2 ≈ 0.707 I₀.
• Pure Resistive Circuit: Current and voltage are in same phase (φ = 0°).
• Pure Inductive Circuit: Current lags voltage by 90° (π/2 rad). Inductive Reactance X_L = ω L = 2π f L (increases linearly with frequency).
• Pure Capacitive Circuit: Current leads voltage by 90° (π/2 rad). Capacitive Reactance X_C = 1 / (ω C) = 1 / (2π f C) (decreases with frequency; blocks DC where f = 0).
2. LCR Series Circuit, Resonance & Power Factor
Short Note:
• Impedance (Z): Total effective opposition in an AC circuit containing R, L, and C in series: Z = √[R² + (X_L - X_C)²].
• Phase Angle (φ): Angle by which voltage leads current: tan φ = (X_L - X_C) / R.
• Electrical Resonance: Occurs when X_L = X_C (i.e., ω L = 1 / (ω C)).
- Resonant Frequency: f₀ = 1 / [2π √(L C)] or ω₀ = 1 / √(L C).
- At resonance, impedance is minimum (Z = R) and current amplitude is maximum (I_max = V₀ / R).
• Quality Factor (Q-factor): Measure of sharpness of resonance: Q = (ω₀ L) / R = 1 / (ω₀ C R) = (1 / R) × √(L / C).
• Average Power in AC Circuit: P_avg = V_rms × I_rms × cos φ, where cos φ = R / Z is the Power Factor.
- Pure R circuit: cos φ = 1 (Maximum power dissipated).
- Pure L or C circuit: cos φ = 0 (Zero average power dissipation; current is Wattless Current).
3. LC Oscillations & Transformers
Short Note:
• LC Oscillations: When a charged capacitor C is connected across an inductor L, energy oscillates back and forth between electric field of capacitor and magnetic field of inductor with frequency ω = 1 / √(L C).
• Transformer: Electrical device used to step-up or step-down alternating voltages based on mutual induction.
- Transformation Ratio Equation: V_s / V_p = N_s / N_p = I_p / I_s = k.
- Step-up Transformer: N_s > N_p => V_s > V_p and I_s < I_p.
- Step-down Transformer: N_s < N_p => V_s < V_p and I_s > I_p.
• Energy Losses in Transformers: Flux leakage, Resistance of windings (Joule heat loss), Eddy currents (minimized by laminated core), Hysteresis loss, and Magnetostriction (humming noise).
Intext & Conceptual Questions
• Definition: The component of alternating current that consumes zero average electrical power over a full cycle is called wattless current.
• Condition: It flows in a purely inductive or purely capacitive AC circuit where the phase difference between current and voltage is φ = 90° (π/2 rad).
• Proof: Average power is given by P_avg = V_rms × I_rms × cos φ. Since cos(90°) = 0, the power dissipated is P_avg = 0, making the current purely wattless.
| Feature | Step-Up Transformer | Step-Down Transformer |
|---|---|---|
| Output Voltage (V_s) | Increases voltage (V_s > V_p). | Decreases voltage (V_s < V_p). |
| Number of Turns | N_s > N_p (Secondary has more turns). | N_s < N_p (Primary has more turns). |
| Output Current (I_s) | Decreases current (I_s < I_p). | Increases current (I_s > I_p). |
| Transformation Ratio (k) | k > 1 | k < 1 |
| Primary Application | Power generating stations for long-distance power transmission. | Substations, mobile chargers, welding transformers. |
Textbook Chapter End Exercise Questions
Given:
• Resistance (R) = 100 Ω
• RMS Voltage (V_rms) = 220 V
• Frequency (f) = 50 Hz
Part (a): RMS Value of Current
I_rms = V_rms / R = 220 / 100 = 2.2 A
Part (b): Net Power Consumed
For a purely resistive circuit, power factor cos φ = 1.
P_avg = V_rms × I_rms = 220 × 2.2 = 484 W
• The rms value of current is 2.2 A.
• The net power consumed over a full cycle is 484 W.
Given:
• Inductance (L) = 44 mH = 44 × 10⁻³ H
• RMS Voltage (V_rms) = 220 V
• Frequency (f) = 50 Hz
Step 1: Calculate Inductive Reactance (X_L)
X_L = 2π f L = 2 × 3.1416 × 50 × (44 × 10⁻³)
X_L = 314.16 × 0.044 ≈ 13.82 Ω
Step 2: Calculate RMS Current (I_rms)
I_rms = V_rms / X_L = 220 / 13.82 ≈ 15.92 A
The rms value of the current in the circuit is 15.92 A.
Read More & Chapter Resources
These short notes are designed for quick revision. Read the main points once, then use the related material below to practise the same chapter. Revising a concept and checking it with questions can make your preparation more focused.