Axiom Physics | Class 12 - AC through Pure R, L, and C

⚡ 01. The Spark (Analogy)

Socho ek jhule (swing) ko push karne ke process ko:
1. Pure Resistor (Mechanical Friction): Jab aap push karte hain, jhoola turant usi waqt aage badhta hai—force aur velocity bilkul sath-sath peak aur zero reach karte hain ($\phi = 0^\circ$).
2. Pure Inductor (Heavy Mass / Inertia): Jhoole ka mass achanak speed pakadne se inkaar karta hai. Pehle aapko poora zor (Voltage) lagana padta hai, tab jakar kuch der baad movement (Current) aati hai—isliye Voltage leads Current by $90^\circ$ ($\pi/2$)!
3. Pure Capacitor (Elastic Spring): Shuru mein spring dheeli hoti hai toh tezi se aage bhagti hai (Current reaches peak first), aur jab tak spring poori stretch hokar opposing tension (Voltage) banati hai, movement ruk chuki hoti hai—isliye Current leads Voltage by $90^\circ$ ($\pi/2$)!

📐 02. The Core Law & Formulae

Phase Angles & In-Phase Resistor

Mathematical formulations for applied voltage $v = V_0\sin\omega t$:

$$i_R = I_0 \sin\omega t \quad (\phi = 0^\circ)$$ $$i_L = I_0 \sin\left(\omega t - \frac{\pi}{2}\right) \quad (\text{V leads I by } \pi/2)$$ $$i_C = I_0 \sin\left(\omega t + \frac{\pi}{2}\right) \quad (\text{I leads V by } \pi/2)$$

Power dissipation across pure resistor is maximum: $P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}}$. In pure $L$ and $C$, $P_{\text{avg}} = 0$.

Reactances ($X_L$ and $X_C$)

Frequency-dependent effective oppositions to AC current:

$$X_L = \omega L = 2\pi f L \quad (\text{SI Unit: }\Omega)$$ $$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \quad (\text{SI Unit: }\Omega)$$

DC Circuit Boundary ($f = 0$): $X_L = 0$ (Inductor acts as short circuit wire); $X_C = \infty$ (Capacitor blocks DC completely).

03. Interactive Visualizer: Phasor Wheel & Time-Domain Waveforms

AC Frequency ($f$) 50.0 Hz

Peak Voltage ($V_0$) 100.0 V

Inductance ($L$) 100 mH

Opposition ($R, X_L, \text{or } X_C$):

31.42 Ω

Peak Current ($I_0$):

3.18 A

Phase Difference ($\phi$):

+90° (V leads I)

Average Power ($P_{\text{avg}}$):

0.00 W (Wattless)

📝 04. Comprehensive Stepwise Derivations Blueprint

Case 01

AC through Pure Resistor ($R$): Derivation, Phasor & Waveforms

Fig 7.1A: Pure resistive circuit; voltage and current phasors are strictly collinear ($\phi = 0^\circ$).

1. Applied alternating voltage across resistor $R$: $v = V_0\sin\omega t$.

2. By Ohm's law, instantaneous current $i$ is given by:

$$i = \frac{v}{R} = \frac{V_0\sin\omega t}{R} = \left(\frac{V_0}{R}\right)\sin\omega t$$

$$i = I_0 \sin\omega t \quad \text{where } I_0 = \frac{V_0}{R}$$

Phase Comparison: Phase difference $\phi = \omega t - \omega t = 0^\circ$. Voltage and current oscillate in phase.

Case 02

AC through Pure Inductor ($L$): Derivation, Inductive Reactance & Phasor

Fig 7.1B: Pure inductor circuit; voltage phasor leads current phasor by $90^\circ$ ($\pi/2$).

1. Applied alternating voltage across pure inductor: $v = V_0\sin\omega t$.

2. Opposing back-EMF by Faraday's Law is $\mathcal{E} = -L\frac{di}{dt}$. Applying Kirchhoff's Loop Rule ($v + \mathcal{E} = 0$):

$$V_0\sin\omega t - L\frac{di}{dt} = 0 \implies L\frac{di}{dt} = V_0\sin\omega t \implies di = \frac{V_0}{L}\sin\omega t \, dt$$

3. Integrating both sides with respect to time:

$$i = \frac{V_0}{L}\int \sin\omega t \, dt = -\frac{V_0}{\omega L}\cos\omega t$$

4. Using identity $-\cos\theta = \sin(\theta - \frac{\pi}{2})$:

$$i = I_0 \sin\left(\omega t - \frac{\pi}{2}\right) \quad \text{where } I_0 = \frac{V_0}{\omega L} = \frac{V_0}{X_L} \quad (X_L = \omega L = 2\pi f L)$$

Deduction: Current lags voltage by phase angle $\frac{\pi}{2}$ ($90^\circ$). Inductive reactance $X_L \propto f$.

Case 03

AC through Pure Capacitor ($C$): Derivation, Capacitive Reactance & Phasor

Fig 7.1C: Pure capacitor circuit; current phasor leads voltage phasor by $90^\circ$ ($\pi/2$).

1. Let applied voltage across capacitor $C$ be $v = V_0\sin\omega t$. Instantaneous charge $q = Cv = C V_0\sin\omega t$.

2. Instantaneous current $i$ entering capacitor plates:

$$i = \frac{dq}{dt} = \frac{d}{dt}(C V_0\sin\omega t) = C V_0 \omega \cos\omega t = \frac{V_0}{1/\omega C} \cos\omega t$$

3. Using identity $\cos\theta = \sin(\theta + \frac{\pi}{2})$:

$$i = I_0 \sin\left(\omega t + \frac{\pi}{2}\right) \quad \text{where } I_0 = \frac{V_0}{X_C} \quad \left(X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}\right)$$

Deduction: Current leads voltage by phase angle $\frac{\pi}{2}$ ($90^\circ$). Capacitive reactance $X_C \propto 1/f$.

Energy Proof

Proof of Zero Average Power Dissipation in Pure L and C (Wattless Current)

1. Instantaneous power delivered to circuit: $p = v \cdot i$.

2. For pure inductor or capacitor, phase difference is $\phi = \pm\frac{\pi}{2}$:

$$p = (V_0\sin\omega t) \times \left[I_0\sin\left(\omega t \pm \frac{\pi}{2}\right)\right] = \pm V_0 I_0 \sin\omega t\cos\omega t = \pm \frac{V_0 I_0}{2} \sin(2\omega t)$$

3. Average power over complete cycle of time period $T$:

$$P_{\text{avg}} = \frac{1}{T}\int_0^T p \, dt = \pm \frac{V_0 I_0}{2T} \int_0^T \sin(2\omega t) dt$$

4. Since the average of $\sin(2\omega t)$ over full cycle is identically zero ($\int_0^T \sin(2\omega t)dt = 0$):

$$P_{\text{avg}} = 0 \quad (\text{Wattless Current: Zero continuous energy is consumed in pure } L \text{ or } C)$$

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: The "220 V AC vs 220 V DC" Shock Hazard

*Board Question:* Why is $220\text{ V AC}$ significantly more dangerous than $220\text{ V DC}$?
• $220\text{ V}$ rated AC is its RMS value ($V_{\text{rms}}$).
• Its peak voltage reaches: $$V_0 = \sqrt{2} \times V_{\text{rms}} = 1.414 \times 220\text{ V} \approx \mathbf{311\text{ Volts}}!$$ AC fluctuates between $+311\text{ V}$ and $-311\text{ V}$ (a peak-to-peak swing of $622\text{ V}$), whereas DC remains fixed at $220\text{ V}$.

Trap 2: DC Response of Inductors and Capacitors

For direct current (DC), frequency is zero ($f = 0, \omega = 0$):
• Inductor: $X_L = 2\pi(0)L = \mathbf{0}$. Acts as a pure zero-resistance wire (short circuit).
• Capacitor: $X_C = \frac{1}{2\pi(0)C} \to \mathbf{\infty}$. Acts as a complete insulator/open circuit (blocks steady DC completely).

Trap 3: Choke Coil vs Variable Resistor (Rheostat)

To control AC current without wasting energy in heat, a Choke Coil (pure high $L$, negligible $R$) is used instead of a rheostat. A resistor dissipates continuous energy ($I^2R$), whereas an ideal choke coil controls current via inductive reactance $X_L$ while consuming zero average power ($P_{\text{avg}} = 0$).

Trap 4: Average of AC Current over Full Cycle vs Half Cycle

• Average current over a FULL cycle: $\mathbf{I_{\text{avg}} = 0}$ (positive half cancels negative half). Moving coil DC meters read 0 in AC!
• Mean current over a POSITIVE HALF cycle: $$I_{\text{half}} = \frac{2}{\pi} I_0 \approx 0.637 I_0$$ Hot-wire meters measure RMS current based on heating effect ($I^2_{\text{rms}}$).

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: A $44\text{ mH}$ inductor is connected to a $220\text{ V}, 50\text{ Hz}$ AC supply. Determine (i) the inductive reactance of the circuit, (ii) the RMS value of the current in the circuit, and (iii) the average power absorbed by the circuit over a complete cycle.

Solution:

1. Given: $L = 44 \times 10^{-3}\text{ H}$, $V_{\text{rms}} = 220\text{ V}$, $f = 50\text{ Hz}$.

2. Inductive reactance ($X_L$):

$$X_L = 2\pi f L = 2 \times \frac{22}{7} \times 50 \times (44 \times 10^{-3}) = \frac{4400 \times 44 \times 10^{-3}}{7} \approx 13.83\text{ }\Omega$$

3. RMS current ($I_{\text{rms}}$):

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{X_L} = \frac{220\text{ V}}{13.83\text{ }\Omega} \approx 15.91\text{ Amperes}$$

4. Average power absorbed:

$$P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos(90^\circ) = 0.00\text{ Watts}$$

JEE/NEET Competitive Trap

Problem: A $15.0\text{ }\mu\text{F}$ capacitor is connected to a $220\text{ V}, 50\text{ Hz}$ source. Find the capacitive reactance and the current (RMS and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current?

Solution:

1. $C = 15.0 \times 10^{-6}\text{ F}, V_{\text{rms}} = 220\text{ V}, f = 50\text{ Hz}$:

$$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times (15.0 \times 10^{-6})} = \frac{10^6}{1500\pi} \approx 212.2\text{ }\Omega$$

2. Currents:

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = \frac{220\text{ V}}{212.2\text{ }\Omega} \approx 1.037\text{ A}$$ $$I_0 = \sqrt{2} \times I_{\text{rms}} = 1.414 \times 1.037\text{ A} \approx 1.47\text{ A}$$

3. When frequency is doubled ($f' = 2f$):

$$X_C \propto \frac{1}{f} \implies X_C' = \frac{X_C}{2} = 106.1\text{ }\Omega \quad (\text{Reactance is Halved})$$ $$I_{\text{rms}} \propto f \implies I_{\text{rms}}' = 2 \times I_{\text{rms}} = 2.074\text{ A} \quad (\text{Current is Doubled!})$$

🗂️ 07. Quick Revision Flashcard Core Formulae

Pure Inductor (L)

$V$ leads $I$ by $90^\circ$ ($X_L = \omega L$)

Pure Capacitor (C)

$I$ leads $V$ by $90^\circ$ ($X_C = \frac{1}{\omega C}$)

Pure Resistor (R)

In phase ($\phi = 0^\circ$)

Wattless Power

$P_{\text{avg}} = 0$ in pure $L$ & $C$

AC through Pure Resistor, Inductor & Capacitor