EMI & Alternating Current (AC)
Practice JEE Main EMI & AC chapter-wise PYQs online. Faraday’s law, series LCR resonance, transformer turn ratio, power factor & 4-step solutions.
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4-Step Solution & Tricks
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🎯 High-Yield Syllabus Breakdown
These high-frequency concepts account for over 90% of all questions asked in EMI & Alternating Current (AC):
- Faraday’s Law of Induction |ε| = |dΦ/dt| & Lenz’s Law (Energy Conservation)
- Motional EMF ε = B v L and Rotating Rod EMF ε = ½ B ω L²
- Self-Inductance of Solenoid L = μ₀ N² A / l & Stored Energy U = ½ L I²
- Series LCR Circuit, Resonance (ω₀ = 1/√(LC)), Impedance Z_min = R
- Power in AC Circuits P = V_rms I_rms cos φ and Transformers (V_s/V_p = N_s/N_p)
⚡ Essential Formulas & Speed Shortcuts
Formulas and time-saving shortcuts to solve EMI & AC questions in under 45 seconds:
Series Resonance: Impedance Z = R (minimum), Current is maximum, power factor cos φ = 1.
Quality Factor: Q = (1/R) √(L/C).
Pure Inductor or Capacitor has phase difference 90° => Power Factor cos 90° = 0 (Wattless Current).
📖 Sample Solved Exam Questions
Preview authentic questions from this chapter with full step-by-step solutions:
💡 View Step-by-Step Solution & Answer
Correct Answer: B v L
- Step 1: Each free charge q inside the moving rod experiences a magnetic Lorentz force F = q (v × B).
- Step 2: Since v and B are mutually perpendicular, F = q v B along the length of rod.
- Step 3: This force produces an electric field E = F / q = v B across the ends.
- Step 4: Induced potential difference (motional EMF) is ε = E × L = B v L.
⚡ Speed Tip: Motional EMF for translation: ε = B v L; for rotation about one end: ε = ½ B ω L².
💡 View Step-by-Step Solution & Answer
Correct Answer: 19 V
- Step 1: According to Faraday's Law of Electromagnetic Induction, |ε| = |dΦ/dt|.
- Step 2: Given flux function: Φ(t) = 6t² - 5t + 1.
- Step 3: Differentiate with respect to time t: dΦ/dt = 12t - 5.
- Step 4: Evaluate at t = 2 s: |ε| = |12(2) - 5| = |24 - 5| = 19 V.
⚡ Speed Tip: Faraday's Law: ε = -dΦ/dt. At t = 2: |d/dt(6t² - 5t + 1)| = 12(2) - 5 = 19 V.
💡 View Step-by-Step Solution & Answer
Correct Answer: R
- Step 1: Impedance of a series LCR circuit is Z = √[ R² + (X_L - X_C)² ].
- Step 2: At electrical resonance, inductive reactance equals capacitive reactance: X_L = X_C (i.e. ωL = 1/ωC).
- Step 3: The reactive term becomes zero: (X_L - X_C)² = 0.
- Step 4: Therefore, impedance is purely resistive and reaches minimum value Z_min = R.
⚡ Speed Tip: At series resonance: Z = R (minimum), I = V/R (maximum), and power factor cos φ = 1.
❓ Frequently Asked Questions (FAQs)
Why does power dissipated in a pure capacitor or inductor equal zero?
Because current and voltage are 90° out of phase (φ = 90°), making the power factor cos 90° = 0.