Alternating Current (AC) – Complete Formula Framework (JEE Mains)

Phasors | Impedance | Resonance | Power | PYQ Mapping

1️⃣ AC Basics

Alternating current:

\[ i = I_0 \sin \omega t \]

Peak current = \( I_0 \)
Angular frequency = \( \omega = 2\pi f \)
Time period = \( T = \frac{2\pi}{\omega} \)

RMS value:

\[ I_{rms} = \frac{I_0}{\sqrt{2}} \] \[ V_{rms} = \frac{V_0}{\sqrt{2}} \]

⚠ Mains Trap: Household voltage is RMS, not peak.

2️⃣ AC through Pure Resistor

Voltage & current in phase

\[ V = IR \] Power: \[ P = V_{rms} I_{rms} \] Phase angle: \[ \phi = 0 \]

3️⃣ AC through Pure Inductor

Inductive reactance: \[ X_L = \omega L \] Current: \[ I = \frac{V}{X_L} \] Phase: \[ \phi = +90^\circ \]

Voltage leads current.

4️⃣ AC through Pure Capacitor

Capacitive reactance: \[ X_C = \frac{1}{\omega C} \] Current: \[ I = \frac{V}{X_C} \] Phase: \[ \phi = -90^\circ \]

Current leads voltage.

5️⃣ RLC Series Circuit

Impedance: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] Current: \[ I = \frac{V}{Z} \] Phase angle: \[ \tan\phi = \frac{X_L - X_C}{R} \]

6️⃣ Resonance

Condition: \[ X_L = X_C \] Resonant frequency: \[ \omega_0 = \frac{1}{\sqrt{LC}} \] \[ f_0 = \frac{1}{2\pi \sqrt{LC}} \] At resonance:

Z = R
Current maximum
Phase angle = 0

7️⃣ Power in AC Circuit

Average power: \[ P = V_{rms} I_{rms} \cos\phi \] Power factor: \[ \cos\phi = \frac{R}{Z} \]

⚠ Pure inductor/capacitor → power = 0

8️⃣ Quality Factor (Q)

\[ Q = \frac{\omega_0 L}{R} \] Also: \[ Q = \frac{1}{R} \sqrt{\frac{L}{C}} \] Higher Q → Sharper resonance.

9️⃣ Transformer

Turns ratio: \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \] Current ratio: \[ \frac{I_s}{I_p} = \frac{N_p}{N_s} \] Ideal transformer: \[ V_p I_p = V_s I_s \]

🎯 JEE Mains Question Profiling

RMS & peak conversion
Reactance comparison
Phase angle calculation
Resonance frequency
Power factor based numericals
Transformer ratio problems

StudyBeacon

Search This Blog