class 12 notes

Class 12 Physics Notes Chapter 7 (Alternating current) – Physics Part-I Book

Philoid

Philoid

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Physics Part-I
Detailed Notes with MCQs of Chapter 7: Alternating Current. This is a crucial chapter, not just for your board exams but also extensively featured in various government entrance examinations like NEET, JEE, and others for technical posts. Pay close attention to the concepts, formulas, and their applications.

Chapter 7: Alternating Current (AC)

1. Introduction to Alternating Current (AC) and Voltage

  • Direct Current (DC): Current whose magnitude remains constant and direction is unidirectional.
  • Alternating Current (AC): Current whose magnitude changes continuously with time and direction reverses periodically. Similarly, alternating voltage varies sinusoidally.
  • Representation: AC voltage and current are typically represented by sinusoidal functions:
    • V = V₀ sin(ωt)
    • I = I₀ sin(ωt + φ)
      where:
      • V, I: Instantaneous values of voltage and current.
      • V₀, I₀: Peak or maximum values (amplitude) of voltage and current.
      • ω: Angular frequency (ω = 2πf = 2π/T).
      • f: Frequency (number of cycles per second, Unit: Hertz (Hz)). In India, standard AC frequency is 50 Hz.
      • T: Time period (time for one complete cycle, T = 1/f).
      • φ: Phase difference between voltage and current.

2. Average and RMS Values

  • Average (Mean) Value:
    • The average value of AC over one complete cycle (0 to T) is zero because the positive and negative half-cycles cancel out.
    • Average value over a positive half-cycle (0 to T/2):
      I_avg = (2/π) I₀ ≈ 0.637 I₀
      V_avg = (2/π) V₀ ≈ 0.637 V₀
  • Root Mean Square (RMS) Value (Effective or Virtual Value):
    • Defined as the value of steady DC that would produce the same amount of heat in a given resistor in a given time as produced by the AC.
    • It's the square root of the mean of the squares of the instantaneous values over one full cycle.
    • I_rms = I₀ / √2 ≈ 0.707 I₀
    • V_rms = V₀ / √2 ≈ 0.707 V₀
    • Significance: AC measuring instruments (ammeters, voltmeters) usually measure the RMS value. The voltage rating of household AC supply (e.g., 220 V in India) refers to the RMS value.

3. AC Circuits with Single Components

  • AC Applied to a Resistor (R):

    • Voltage: V = V₀ sin(ωt)
    • Current: I = (V₀/R) sin(ωt) = I₀ sin(ωt) where I₀ = V₀/R.
    • Phase: Voltage and current are in phase (φ = 0).
    • Phasor Diagram: V and I phasors are parallel.
    • Resistance (R): Opposes current, independent of frequency.
    • Power: P_avg = V_rms * I_rms = I_rms² * R = V_rms² / R

  • AC Applied to an Inductor (L):

    • Voltage: V = V₀ sin(ωt)
    • Current: I = (V₀/ωL) sin(ωt - π/2) = I₀ sin(ωt - π/2) where I₀ = V₀/X_L.
    • Phase: Voltage leads the current by π/2 (or 90°), or current lags voltage by π/2. (φ = +π/2)
    • Inductive Reactance (X_L): Opposition offered by the inductor to AC flow.
      X_L = ωL = 2πfL (Unit: Ohm (Ω)). X_L is directly proportional to frequency.
    • Phasor Diagram: V phasor leads I phasor by 90°.
    • Power: P_avg = V_rms * I_rms * cos(π/2) = 0. Average power consumed by a pure inductor over a full cycle is zero. Energy is stored in the magnetic field during one quarter cycle and returned in the next.

  • AC Applied to a Capacitor (C):

    • Voltage: V = V₀ sin(ωt)
    • Current: I = (V₀ / (1/ωC)) sin(ωt + π/2) = I₀ sin(ωt + π/2) where I₀ = V₀/X_C.
    • Phase: Current leads the voltage by π/2 (or 90°). (φ = -π/2)
    • Capacitive Reactance (X_C): Opposition offered by the capacitor to AC flow.
      X_C = 1 / (ωC) = 1 / (2πfC) (Unit: Ohm (Ω)). X_C is inversely proportional to frequency. For DC (f=0), X_C is infinite (capacitor blocks DC).
    • Phasor Diagram: I phasor leads V phasor by 90°.
    • Power: P_avg = V_rms * I_rms * cos(-π/2) = 0. Average power consumed by a pure capacitor over a full cycle is zero. Energy is stored in the electric field during one quarter cycle and returned in the next.

4. Series LCR Circuit

  • Consider a resistor (R), inductor (L), and capacitor (C) connected in series to an AC source V = V₀ sin(ωt).
  • Let the current be I = I₀ sin(ωt + φ).
  • Voltage across R: V_R = I R (in phase with I)
  • Voltage across L: V_L = I X_L (leads I by π/2)
  • Voltage across C: V_C = I X_C (lags I by π/2)
  • Phasor Diagram Analysis: V_L and V_C are 180° out of phase. The net reactive voltage is (V_L - V_C). The source voltage V is the vector sum of V_R and (V_L - V_C).
  • Impedance (Z): Total effective opposition offered by the LCR circuit to AC flow.
    V² = V_R² + (V_L - V_C)²
    (I Z)² = (I R)² + (I X_L - I X_C)²
    Z² = R² + (X_L - X_C)²
    Z = √[R² + (ωL - 1/ωC)²] (Unit: Ohm (Ω))
    Impedance Z acts like the effective resistance of the AC circuit. I₀ = V₀ / Z and I_rms = V_rms / Z.
  • Impedance Triangle: A right-angled triangle with base R, perpendicular (X_L - X_C), and hypotenuse Z.
  • Phase Difference (φ): Angle between source voltage V and current I.
    tan φ = (V_L - V_C) / V_R = (I X_L - I X_C) / (I R)
    tan φ = (X_L - X_C) / R = (ωL - 1/ωC) / R
    • If X_L > X_C (ωL > 1/ωC): φ is positive, circuit is inductive, voltage leads current.
    • If X_L < X_C (ωL < 1/ωC): φ is negative, circuit is capacitive, current leads voltage.
    • If X_L = X_C (ωL = 1/ωC): φ is zero, circuit is purely resistive (Resonance).

5. Resonance in Series LCR Circuit

  • Condition: The frequency at which inductive reactance equals capacitive reactance (X_L = X_C).
  • Resonant Angular Frequency (ω_r):
    ω_r L = 1 / (ω_r C)
    ω_r² = 1 / (LC)
    ω_r = 1 / √(LC)
  • Resonant Frequency (f_r):
    2πf_r = 1 / √(LC)
    f_r = 1 / (2π√(LC))
  • Characteristics at Resonance:
    • X_L = X_C
    • Impedance Z = √[R² + (X_L - X_C)²] = √R² = R. Impedance is minimum.
    • Current I_rms = V_rms / Z = V_rms / R. Current is maximum.
    • Phase difference φ = 0. Voltage and current are in phase. The circuit behaves like a purely resistive circuit.
    • Power factor cos φ = cos 0 = 1 (Maximum power dissipation).
  • Quality Factor (Q-factor): Measures the sharpness of resonance or voltage magnification.
    Q = ω_r L / R = 1 / (ω_r C R) = (1/R) * √(L/C)
    • A higher Q-factor means a sharper resonance curve (current falls off rapidly as frequency deviates from f_r). Used in tuning circuits (radio/TV receivers).
  • Bandwidth (Δω): The difference between the two frequencies (ω₁ and ω₂) on either side of ω_r where the current amplitude is 1/√2 times the maximum current (or power is half the maximum power). Δω = ω₂ - ω₁ = R/L. Sharpness of resonance = ω_r / Δω = Q.

6. Power in AC Circuits

  • Instantaneous Power: p(t) = V(t) * I(t) = [V₀ sin(ωt)] * [I₀ sin(ωt + φ)]
  • Average Power (P_avg) over a full cycle:
    P_avg = V_rms * I_rms * cos φ
  • Power Factor (cos φ):
    cos φ = R / Z = R / √[R² + (X_L - X_C)²]
    • It represents the fraction of the apparent power (V_rms * I_rms) that is actually dissipated as true power.
    • 0 ≤ cos φ ≤ 1
    • Purely resistive circuit: φ = 0, cos φ = 1, P_avg = V_rms * I_rms.
    • Purely inductive or capacitive circuit: φ = ±π/2, cos φ = 0, P_avg = 0.
  • Wattless Current: The component of current that is perpendicular (90° out of phase) to the voltage (I_rms sin φ). This component does not contribute to the average power dissipation. The component I_rms cos φ is in phase with voltage and contributes to power.

7. LC Oscillations

  • An ideal LC circuit (no resistance) can produce electrical oscillations when a charged capacitor is connected across an inductor.
  • Energy oscillates between the capacitor's electric field (U_E = ½ Q²/C = ½ C V²) and the inductor's magnetic field (U_B = ½ L I²).
  • The total energy U = U_E + U_B remains constant (in an ideal circuit).
  • These oscillations are analogous to the mechanical oscillations of a mass-spring system.
  • Natural Frequency of Oscillation: f = 1 / (2π√(LC)) (Same as the resonant frequency of a series LCR circuit).
  • In practical circuits, resistance causes damping, and the oscillations eventually die out unless external energy is supplied (like in an oscillator circuit).

8. Transformers

  • Principle: Mutual Induction - An EMF is induced in a secondary coil when the magnetic flux linked with it changes due to a changing current in the primary coil. Works only with AC.
  • Use: To change (step-up or step-down) AC voltage levels with a corresponding change in current, keeping power (ideally) constant.
  • Construction: Two coils (primary and secondary) wound on a common laminated soft iron core. Lamination reduces eddy currents. Soft iron minimizes hysteresis loss and increases magnetic flux linkage.
  • Working & Theory (Ideal Transformer - 100% efficient):
    • Voltage ratio: V_s / V_p = N_s / N_p
    • Current ratio: I_p / I_s = N_s / N_p (Since Input Power = Output Power => V_p I_p = V_s I_s)
    • Transformation Ratio (k): k = N_s / N_p
      • Step-up Transformer: N_s > N_p (k > 1) => V_s > V_p and I_s < I_p. Increases voltage, decreases current.
      • Step-down Transformer: N_s < N_p (k < 1) => V_s < V_p and I_s > I_p. Decreases voltage, increases current.
  • Efficiency (η):
    η = (Output Power) / (Input Power) = (V_s I_s) / (V_p I_p)
    Practical transformers have efficiencies around 90-99%.
  • Energy Losses in Transformers:
    1. Copper Loss: Heat produced (I²R) in primary and secondary windings. Minimized by using thick copper wires.
    2. Flux Leakage: Not all flux from the primary links with the secondary. Minimized by winding coils over each other or using a shell-type core.
    3. Eddy Currents: Currents induced in the bulk of the iron core due to changing magnetic flux. Cause heating (I²R). Minimized by using a laminated core (thin sheets insulated from each other).
    4. Hysteresis Loss: Energy lost in magnetizing and demagnetizing the iron core during each AC cycle. Minimized by using soft iron with a thin hysteresis loop.

Key Formulas Summary:

  • V = V₀ sin(ωt), I = I₀ sin(ωt + φ)
  • ω = 2πf = 2π/T
  • I_rms = I₀ / √2, V_rms = V₀ / √2
  • X_L = ωL, X_C = 1 / (ωC)
  • Z = √[R² + (X_L - X_C)²]
  • tan φ = (X_L - X_C) / R
  • Resonance: X_L = X_C, f_r = 1 / (2π√(LC))
  • Q-factor: Q = ω_r L / R = (1/R) * √(L/C)
  • Average Power: P_avg = V_rms * I_rms * cos φ
  • Power Factor: cos φ = R / Z
  • Transformer: V_s / V_p = N_s / N_p = I_p / I_s (Ideal)
  • Efficiency: η = (V_s I_s) / (V_p I_p)

This covers the core concepts of Alternating Current. Remember to practice numerical problems based on these formulas and concepts, especially related to LCR circuits, resonance, and transformers, as they are frequently tested.

Multiple Choice Questions (MCQs)

  1. The RMS value of an AC voltage given by V = 200√2 sin(100πt) V is:
    a) 100 V
    b) 200 V
    c) 200√2 V
    d) 100√2 V

  2. In a pure inductive circuit, the current:
    a) Leads the voltage by π/2
    b) Lags the voltage by π/2
    c) Is in phase with the voltage
    d) Lags the voltage by π

  3. The opposition offered by a capacitor to the flow of AC is called:
    a) Resistance
    b) Inductive Reactance
    c) Capacitive Reactance
    d) Impedance

  4. At resonance frequency in a series LCR circuit, the impedance is:
    a) Maximum
    b) Minimum and equal to R
    c) Equal to X_L
    d) Equal to X_C

  5. The power factor of a purely resistive AC circuit is:
    a) 0
    b) 1
    c) 0.5
    d) -1

  6. A transformer works on the principle of:
    a) Self-induction
    b) Mutual induction
    c) Electromagnetic induction (general)
    d) Static electricity

  7. In a step-up transformer:
    a) N_s < N_p
    b) V_s < V_p
    c) I_s > I_p
    d) N_s > N_p

  8. Laminating the core of a transformer reduces energy loss due to:
    a) Hysteresis
    b) Eddy currents
    c) Copper loss
    d) Flux leakage

  9. The quality factor (Q-factor) of a series LCR circuit is given by:
    a) R / ω_r L
    b) ω_r L / R
    c) ω_r C / R
    d) R / Z

  10. An AC source is connected to a series LCR circuit. If the capacitive reactance is greater than the inductive reactance, the phase difference between voltage and current is:
    a) Positive (Voltage leads current)
    b) Negative (Current leads voltage)
    c) Zero
    d) π

Answers to MCQs:

  1. b) 200 V (V_rms = V₀ / √2 = 200√2 / √2 = 200 V)
  2. b) Lags the voltage by π/2
  3. c) Capacitive Reactance
  4. b) Minimum and equal to R
  5. b) 1
  6. b) Mutual induction
  7. d) N_s > N_p
  8. b) Eddy currents
  9. b) ω_r L / R
  10. b) Negative (Current leads voltage) (tan φ = (X_L - X_C) / R. If X_C > X_L, tan φ is negative, so φ is negative).

Study these notes thoroughly. Let me know if any specific part needs more clarification.

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