Alternating Current & Resonant Circuits: Master Notes

Chapter Overview

Alternating Current (AC) analysis transforms linear differential equations into simple algebraic complex phasor equations. In IIT-JEE (Advanced), top problems focus on series/parallel RLC resonance, Q-factor sharpness and bandwidth, power factor correction ($\cos \varphi$), and LC electromagnetic oscillations.

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1. Phasor Formalism & Complex Impedance

$$v(t)=V_0\sin (\omega t)⟹\tilde{V}=V_{\text{rms}}\mathrm{\angle }{0}^{\circ }$$$$i(t)=I_0\sin (\omega t-\varphi )⟹\tilde{I}=\frac{\tilde{V}}{\tilde{Z}}$$

Circuit Element	Reactance / Impedance $\tilde{Z}$	Phase Relationship
Resistor ($R$)	$Z_R=R$	Current and Voltage are strictly in phase ($\varphi =0^{\circ }$)
Inductor ($L$)	$X_L=\omega L,Z_L=+i\omega L$	Voltage leads Current by ${90}^{\circ }$ ($\varphi =+{90}^{\circ }$)
Capacitor ($C$)	$X_C=\frac{1}{\omega C},Z_C=-i\frac{1}{\omega C}$	Current leads Voltage by ${90}^{\circ }$ ($\varphi =-{90}^{\circ }$)

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2. Series RLC Circuit Dynamics

Multi-Mode DiagramAC Circuits: Series RLC Circuit & Resonance Impedance

Option 1: Publication-Grade Scientific Vector SVG

RLC series circuit phasor triangle and resonance curve omega_0 = 1 / sqrt(LC) with quality factor Q.

Impedance Z: $Z = \sqrt{R^2 + (\omega L - 1/\omega C)^2}$

Resonance Frequency: $\omega_0 = \frac{1}{\sqrt{LC}}$

Option 2: Interactive 60 FPS Canvas Physics Engine

Simulate RLC frequency response and resonance peak.

Frequency ω: 50 rad/s

Impedance ($Z$):10.0 Ω

Option 3: Interactive 3D Spatial Viewport

3D impedance phasor cylinder (R, XL, XC).

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Impedance Triangle:$Z = \sqrt{R^2 + X^2}$

Option 5: Reactive Parametric Physics with Live Calculus

Resonance frequency calculator.

Inductance L: 10 mH

Capacitance C: 10 μF

Resonance f_0 ($f_0$):$503 Hz$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

$$\boxed{{\displaystyle Z=\sqrt{R^2+(X_L-X_C{)}^2}=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}}}$$$$\boxed{{\displaystyle \tan \varphi =\frac{X_L-X_C}{R}=\frac{\omega L-\frac{1}{\omega C}}{R}}}$$

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2.1 Series Resonance & The Quality Factor ($Q$)

At resonance frequency ${\omega }_0$:

$$X_L=X_C⟹{\omega }_{0}L=\frac{1}{{\omega }_{0}C}⟹\boxed{{\displaystyle {\omega }_0=\frac{1}{\sqrt{LC}}}}$$

• Impedance is Minimum: $Z_{min}=R$ (purely resistive).
• Current is Maximum: $I_{max}=\frac{V_0}{R}$.
• Quality Factor ($Q$): Measures sharpness of resonance and voltage magnification:$$\boxed{{\displaystyle Q=\frac{{\omega }_{0}L}{R}=\frac{1}{{\omega }_{0}CR}=\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{{\omega }_0}{\mathrm{\Delta }{\omega }_{\text{bandwidth}}}}}$$
• Voltage Magnification across Inductor / Capacitor:$$V_L=V_C=Q\cdot V_{\text{source}}$$

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3. AC Power & Wattless Current

$$\boxed{{\displaystyle P_{\text{avg}}=V_{\text{rms}}{I}_{\text{rms}}\cos \varphi }}$$

Where $\cos \varphi =\frac{R}{Z}$ is the Power Factor of the circuit.

• Apparent Power: $S=V_{\text{rms}}{I}_{\text{rms}}$ (Volt-Amperes).
• Active / True Power: $P=V_{\text{rms}}{I}_{\text{rms}}\cos \varphi$ (Watts, dissipated purely in resistors).
• Wattless (Reactive) Current:$$I_{\text{wattless}}=I_{\text{rms}}\sin \varphi$$(Current that draws zero average power over a complete cycle!)

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4. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2021 (Paper 2) — Series RLC Quality Factor & Half-Power Frequencies

Question:
In a series RLC circuit, $R=10\mathrm{\Omega }$, $L=0.1\text{ H}$, and $C=10\mu \text{F}$.

1. Find the resonant frequency ${\omega }_0$.
2. Calculate the Quality factor $Q$.
3. Determine the half-power bandwidth $\mathrm{\Delta }\omega$.

Step-by-Step Solution:

• 1. Resonant Frequency:

  $${\omega }_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.1\times 10\times {10}^{-6}}}=\frac{1}{\sqrt{{10}^{-6}}}=\boxed{{\displaystyle 1000\text{ rad/s}}}$$

• 2. Quality Factor:

  $$Q=\frac{{\omega }_{0}L}{R}=\frac{1000\times 0.1}{10}=\boxed{{\displaystyle 10}}$$

• 3. Bandwidth:

  $$\mathrm{\Delta }\omega =\frac{{\omega }_0}{Q}=\frac{1000}{10}=\boxed{{\displaystyle 100\text{ rad/s}}}$$$$\text{Half-power frequencies: }{\omega }_1={\omega }_0-\frac{\mathrm{\Delta }\omega }{2}=950\text{ rad/s},{\omega }_2={\omega }_0+\frac{\mathrm{\Delta }\omega }{2}=1050\text{ rad/s}$$

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PYQ 2: JEE Advanced 2019 — Parallel AC Branch Current Phasors

Question:
An alternating voltage source $V(t)=100\sqrt{2}\sin (100t)\text{ V}$ is connected across two parallel branches. Branch 1 contains a pure resistor $R=100\mathrm{\Omega }$. Branch 2 contains a pure inductor $L=1\text{ H}$. Find the total RMS current drawn from the source.

Step-by-Step Solution:

1. $V_{\text{rms}}=100\text{ V}$, $\omega =100\text{ rad/s}$.
2. Current in branch 1 (in phase with $V$):$$I_{1,\text{rms}}=\frac{V_{\text{rms}}}{R}=\frac{100}{100}=1.0\text{ A}$$
3. Inductive reactance of branch 2:$$X_L=\omega L=100\times 1=100\mathrm{\Omega }$$Current in branch 2 (lags voltage by ${90}^{\circ }$):$$I_{2,\text{rms}}=\frac{V_{\text{rms}}}{X_L}=\frac{100}{100}=1.0\text{ A}$$
4. Vector addition of orthogonal phasor currents:$$I_{\text{total, rms}}=\sqrt{I_1^2+I_2^2}=\sqrt{(1.0{)}^2+(1.0{)}^2}=\boxed{{\displaystyle \sqrt{2}\text{ A}\approx 1.414\text{ A}}}$$

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5. High-Yield Formula Sheet

Concept	Formula	Notes
Resonant Frequency	${\omega }_0=\frac{1}{\sqrt{LC}}$	$Z=R$
Quality Factor	$Q=\frac{1}{R}\sqrt{\frac{L}{C}}$	Sharpness of resonance
Average AC Power	$P_{\text{avg}}=V_{\text{rms}}{I}_{\text{rms}}\cos \varphi$	$\cos \varphi =R/Z$
LC Free Oscillation	$q(t)=Q_0\cos ({\omega }_{0}t)$	Energy sloshes between $C$ and $L$