CBSE Class XII • Unit 04 (Alternating Currents)

Chapter 07: Alternating Current

Master Revision Dossier • RMS Foundations, Pure R/L/C Phasors, Series LCR Resonance, Power Factor & Transformers

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Unit IV Combined Weightage

1. Definitive Formula & AC Parameter Matrix

Concept	Mathematical Formulation	Phase & Units	Boundary Conditions & Notes
RMS / Effective Value	$$I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \approx 0.707 I_0, \quad V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$$	$\text{Ampere (A)}$ $\text{Volt (V)}$	DC heating equivalence ($H = I_{\text{rms}}^2 R t$). $220\text{ V}$ domestic mains has peak $V_0 \approx 311\text{ V}$.
Half-Cycle Mean Value	$$I_{\text{avg}} = \frac{2}{\pi}I_0 \approx 0.637 I_0$$	$\text{Ampere (A)}$	Evaluated over $T/2$. Over full cycle, $I_{\text{avg}} = 0$. DC meters read zero in AC circuits.
Pure Inductor ($L$)	$$X_L = \omega L = 2\pi f L, \quad i = I_0\sin\left(\omega t - \frac{\pi}{2}\right)$$	$X_L \text{ in }\Omega$ $\phi = +90^\circ$	Voltage leads current by $\pi/2$. DC limit ($f=0$): $X_L = 0$ (short circuit wire). Average power $= 0$.
Pure Capacitor ($C$)	$$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}, \quad i = I_0\sin\left(\omega t + \frac{\pi}{2}\right)$$	$X_C \text{ in }\Omega$ $\phi = -90^\circ$	Current leads voltage by $\pi/2$. DC limit ($f=0$): $X_C \to \infty$ (blocks steady DC). Average power $= 0$.
Series LCR Impedance	$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$	$\text{Ohm (}\Omega\text{)}$	Phasor hypotenuse. $\tan\phi = \frac{X_L - X_C}{R}$. Inductive if $X_L > X_C$; Capacitive if $X_C > X_L$.
Series Resonance	$$\omega_0 = \frac{1}{\sqrt{LC}} \implies f_0 = \frac{1}{2\pi\sqrt{LC}}$$	$\text{rad/s}$ $\text{Hertz (Hz)}$	$X_L = X_C \implies Z_{\min} = R \implies I_{\max} = V_0/R$. Power factor $\cos\phi = 1$. Cannot occur with DC.
Quality Factor ($Q$)	$$Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 CR} = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{\omega_0}{2\Delta\omega}$$	Dimensionless	Sharpness of tuning. Bandwidth $2\Delta\omega = R/L$. Voltage magnification across $L$ or $C$ is $Q \times V_{\text{source}}$.
AC Real Power	$$P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos\phi$$	$\text{Watt (W)}$	$\cos\phi = R/Z$ is the Power Factor. Active consumed power over full cycle in any reactive circuit.
Wattless Current	$$I_{\text{wattless}} = I_{\text{rms}}\sin\phi$$	$\text{Ampere (A)}$	Quadrature component at $90^\circ$ to voltage. Consumes zero net power: $P = V_{\text{rms}}(I_{\text{rms}}\sin\phi)\cos 90^\circ = 0$.
Transformer Ratio	$$\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} = k$$	Dimensionless	Step-up: $k > 1$. Step-down: $k < 1$. Frequency remains strictly invariant ($f_s = f_p$). Fails on pure DC.

2. Core Derivations Blueprint (Step-by-Step Board Proofs)

Blueprint 7.1: Derivation of RMS Value of Alternating Current ($I_{\text{rms}} = I_0/\sqrt{2}$)

1. Let $i = I_0\sin\omega t$. Heat produced in resistor $R$ during small time $dt$: $dH = i^2 R \, dt = (I_0^2\sin^2\omega t) R \, dt$.

2. Total heat produced over full cycle of period $T = 2\pi/\omega$:

$$H = \int_0^T I_0^2 R \sin^2\omega t \, dt = I_0^2 R \int_0^T \left(\frac{1 - \cos 2\omega t}{2}\right) dt = \frac{I_0^2 R}{2} \left[ \int_0^T dt - \int_0^T \cos 2\omega t \, dt \right]$$

3. Since $\int_0^T \cos 2\omega t \, dt = 0$: $H = \frac{I_0^2 R T}{2}$.

4. By definition, equivalent DC current producing identical heat over time $T$ is $H = I_{\text{rms}}^2 R T$:

$$I_{\text{rms}}^2 R T = \frac{I_0^2 R T}{2} \implies I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \approx 0.707 I_0, \quad V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \approx 0.707 V_0$$

Blueprint 7.2: Phase Relations for Pure Inductor ($L$) and Pure Capacitor ($C$)

1. Pure Inductor: Applied $v = V_0\sin\omega t$. Loop rule: $v - L\frac{di}{dt} = 0 \implies di = \frac{V_0}{L}\sin\omega t \, dt$.

$$i = \frac{V_0}{L} \int \sin\omega t \, dt = -\frac{V_0}{\omega L}\cos\omega t = I_0\sin\left(\omega t - \frac{\pi}{2}\right) \quad \left(X_L = \omega L, \text{ Voltage leads Current by } \frac{\pi}{2}\right)$$

2. Pure Capacitor: Applied $v = V_0\sin\omega t$. Charge $q = C v = C V_0\sin\omega t$. Current $i = \frac{dq}{dt}$:

$$i = \frac{d}{dt}(C V_0\sin\omega t) = \omega C V_0\cos\omega t = I_0\sin\left(\omega t + \frac{\pi}{2}\right) \quad \left(X_C = \frac{1}{\omega C}, \text{ Current leads Voltage by } \frac{\pi}{2}\right)$$

Blueprint 7.3: Series LCR Circuit Impedance & Resonance Formulation

1. Current $i = I_0\sin\omega t$ is common. Individual component voltages: $V_R = I_0 R$ (in-phase), $V_L = I_0 X_L$ (leads by $90^\circ$), and $V_C = I_0 X_C$ (lags by $90^\circ$).

2. Combining opposite reactive voltages: net reactive vector is $(V_L - V_C)$ perpendicular to $V_R$. Applying Pythagoras theorem on phasor triangle:

$$V_0^2 = V_R^2 + (V_L - V_C)^2 = I_0^2 [R^2 + (X_L - X_C)^2] \implies Z = \frac{V_0}{I_0} = \sqrt{R^2 + (X_L - X_C)^2}$$ $$\tan\phi = \frac{V_L - V_C}{V_R} = \frac{X_L - X_C}{R}$$

3. Resonance Condition: Current is maximum when $Z$ is minimum, which occurs when $X_L = X_C$:

$$\omega_0 L = \frac{1}{\omega_0 C} \implies \omega_0 = \frac{1}{\sqrt{LC}} \implies f_0 = \frac{1}{2\pi\sqrt{LC}} \quad (Z_{\min} = R, \cos\phi = 1)$$

Blueprint 7.4: Average Power in AC Circuit & Transformer Transformation Ratio

1. Instantaneous power $p = v \times i = [V_0\sin\omega t][I_0\sin(\omega t - \phi)] = \frac{V_0 I_0}{2}[(1 - \cos 2\omega t)\cos\phi - \sin 2\omega t\sin\phi]$.

2. Full-cycle average: $\langle \cos 2\omega t \rangle = 0$ and $\langle \sin 2\omega t \rangle = 0$:

$$P_{\text{avg}} = \frac{V_0 I_0}{2}\cos\phi = \left(\frac{V_0}{\sqrt{2}}\right)\left(\frac{I_0}{\sqrt{2}}\right)\cos\phi = V_{\text{rms}} I_{\text{rms}}\cos\phi \quad \left(\text{Power Factor } \cos\phi = \frac{R}{Z}\right)$$

3. Transformer Equations: Primary induced EMF $\mathcal{E}_p = -N_p \frac{d\Phi}{dt}$, Secondary induced EMF $\mathcal{E}_s = -N_s \frac{d\Phi}{dt}$.

$$\frac{V_s}{V_p} = \frac{N_s}{N_p} = k \quad \text{and for 100% ideal efficiency } (V_p I_p = V_s I_s) \implies \frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} = k$$

3. High-Yield "Red Alert" Exam Trap Matrix

Master Trap 1: Comparative Response across AC Circuit Elements

Element	Opposition Formula	Phase Angle ($\phi$)	Power Factor ($\cos\phi$)	DC Behavior ($f = 0$)
Pure Resistor ($R$)	Opposition $= R$	$\phi = 0^\circ$ (In-phase)	$\cos 0^\circ = 1$ (Max $P_{\text{avg}}$)	Resists identically ($R$).
Pure Inductor ($L$)	$X_L = \omega L = 2\pi f L$	+$\pi/2$ (V leads I)	$\cos 90^\circ = 0$ ($P_{\text{avg}} = 0$)	$X_L = 0$ (Acts as short-circuit wire).
Pure Capacitor ($C$)	$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$	-$\pi/2$ (I leads V)	$\cos 90^\circ = 0$ ($P_{\text{avg}} = 0$)	$X_C \to \infty$ (Blocks DC completely).
Series LCR (Resonant)	$Z_{\min} = R$	$\phi = 0^\circ$	$\cos\phi = 1$ (Purely resistive)	Cannot resonate with DC ($X_C \to \infty$).

⚠️ Trap 2: Transformer DC Burnout Mechanism

Connecting a DC battery to a transformer primary produces constant flux ($d\Phi/dt = 0$).
• Secondary output is strictly ZERO.
• Primary inductive back-EMF is zero ($X_L = 0$). Current is limited only by tiny wire resistance: $I_p = V_{\text{DC}}/R_p \to \text{Massive}$! The primary coil catches fire or burns out.

⚠️ Trap 3: Choke Coil vs Rheostat

To reduce AC current without power loss, use a Choke Coil (high $L$, negligible $R$).
• Rheostat reduces current by ohmic dissipation: $P = I^2 R$ (wasted as heat).
• Choke coil reduces current by inductive reactance $X_L$, with $\phi \approx 90^\circ \implies \cos\phi \approx 0 \implies \mathbf{P_{\text{avg}} \approx 0}$.

4. Handcrafted Exemplars (Analytical Numerical Solutions)

Problem 1: Series LCR Circuit Complete Analysis Exemplar 01

A series LCR circuit with $R = 10\text{ }\Omega$, $L = 100\text{ mH}$, and $C = 25\text{ }\mu\text{F}$ is connected to a $200\text{ V}, 50\text{ Hz}$ AC source. Calculate: (i) Reactances $X_L, X_C$ and total impedance $Z$, (ii) RMS current, and (iii) Power factor and average power dissipated.

Solution: $\omega = 2\pi(50) = 100\pi \approx 314.16\text{ rad/s}$.

$$X_L = \omega L = 314.16 \times 0.10 = 31.42\text{ }\Omega, \quad X_C = \frac{1}{\omega C} = \frac{1}{314.16 \times (25 \times 10^{-6})} \approx 127.32\text{ }\Omega$$ $$Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{10^2 + (127.32 - 31.42)^2} = \sqrt{100 + 9196.8} \approx 96.42\text{ }\Omega$$

Current and Power parameters:

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200\text{ V}}{96.42\text{ }\Omega} \approx 2.07\text{ A}$$ $$\text{Power Factor } \cos\phi = \frac{R}{Z} = \frac{10}{96.42} \approx 0.1037 \quad (\text{Capacitive Leading})$$ $$P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos\phi = 200 \times 2.07 \times 0.1037 \approx 42.93\text{ Watts}$$

Problem 2: Transformer Load and Current Transmission Exemplar 02

A step-down transformer operates on a $2.2\text{ kV}$ line and delivers $220\text{ V}$ to a load consuming $11\text{ kW}$. If transformer efficiency is $90\%$, calculate (i) turns ratio $N_p/N_s$, (ii) secondary current $I_s$, and (iii) primary current $I_p$.

Solution: $V_p = 2200\text{ V}, V_s = 220\text{ V}, P_{\text{out}} = 11{,}000\text{ W}, \eta = 0.90$.

$$\text{Turns ratio } \frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{2200}{220} = 10 : 1$$ $$I_s = \frac{P_{\text{out}}}{V_s} = \frac{11{,}000\text{ W}}{220\text{ V}} = 50.0\text{ Amperes}$$

From efficiency $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{P_{\text{out}}}{V_p I_p}$:

$$I_p = \frac{P_{\text{out}}}{\eta V_p} = \frac{11{,}000}{0.90 \times 2200} = \frac{11{,}000}{1980} \approx 5.56\text{ Amperes}$$

5. Printable Pocket Flashcard Grid

RMS Formula

$I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$

Series Resonance

$\omega_0 = \frac{1}{\sqrt{LC}}$ ($Z=R$)

Power Factor

$\cos\phi = \frac{R}{Z}$

Quality Factor

$Q = \frac{1}{R}\sqrt{\frac{L}{C}}$