Class 12 Physics Chapter 7: Alternating Current (AC) | Complete Notes & Numericals

📅 Tuesday, 20 January 2026 📖 3-5 min read
Class 12 Physics • NCERT

Chapter 7: Alternating Current

Complete Textbook Edition (Phasors, LCR & Transformers)

Vol 1: AC Fundamentals & RMS Value

Alternating Current (AC): The electric current whose magnitude changes continuously with time and direction reverses periodically.

Why AC? Alternating current is widely preferred over Direct Current (DC) because it can be easily transformed to higher or lower voltages using transformers, minimizing energy loss during long-distance transmission.
$$I = I_0 \sin(\omega t)$$
$$V = V_0 \sin(\omega t)$$

1. Average and RMS Value

Mean (Average) Value ($I_{avg}$): Over one complete cycle, the average value of AC is Zero. Hence, we calculate it over a half-cycle.

  • $I_{avg} = \frac{2I_0}{\pi} \approx 0.637 I_0$
Root Mean Square (RMS) Value ($I_{rms}$):
It is that steady current (DC) which produces the same amount of heat in a resistor as produced by the AC in the same time.

$$I_{rms} = \frac{I_0}{\sqrt{2}} \approx 0.707 I_0$$
$$V_{rms} = \frac{V_0}{\sqrt{2}}$$

Note: Household supply (220V) represents the RMS value. The peak voltage ($V_0$) is actually $220\sqrt{2} \approx 311V$.

Vol 2: AC Through R, L, and C (Phasors)

Phasors: Rotating vectors used to represent sinusoidally varying quantities (Voltage and Current). They rotate counter-clockwise with frequency $\omega$.

Circuit Element Phase Relation Opposition (Reactance)
Resistor (R) V and I are in Same Phase. Resistance ($R$)
Inductor (L) Voltage leads Current by $\pi/2$. Inductive Reactance
$$X_L = \omega L = 2\pi \nu L$$
Capacitor (C) Current leads Voltage by $\pi/2$. Capacitive Reactance
$$X_C = \frac{1}{\omega C} = \frac{1}{2\pi \nu C}$$

Note: For DC frequency $\nu = 0$, so $X_L = 0$ (Inductor allows DC) and $X_C = \infty$ (Capacitor blocks DC).

Vol 3: Series LCR Circuit & Resonance

Consider a circuit containing Inductor (L), Capacitor (C), and Resistor (R) connected in series across an AC source $V = V_0 \sin \omega t$.

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1. Impedance (Z)

The total effective opposition offered by the LCR circuit to the flow of AC.

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$

2. Electrical Resonance

A condition where the current in the LCR circuit becomes Maximum. This happens when Inductive Reactance equals Capacitive Reactance ($X_L = X_C$).

Resonant Frequency ($\omega_r$):
$\omega L = \frac{1}{\omega C} \Rightarrow \omega^2 = \frac{1}{LC}$

$$\nu_r = \frac{1}{2\pi \sqrt{LC}}$$
At resonance, Impedance is minimum ($Z = R$) and Current is maximum.

3. Power Factor ($\cos \phi$)

The average power dissipated is $P = V_{rms} I_{rms} \cos \phi$.
Power Factor: $\cos \phi = \frac{R}{Z}$.
• Pure Resistive: $\phi = 0, \cos \phi = 1$ (Maximum Power).
• Pure Inductive/Capacitive: $\phi = 90^\circ, \cos \phi = 0$ (Wattless Current).

Vol 4: The Transformer

A device used to change (step-up or step-down) alternating voltage. It works on the principle of Mutual Induction.

Transformation Ratio:

$$\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} = k$$
  • Step-Up Transformer: $N_s > N_p$ (Voltage increases, Current decreases).
  • Step-Down Transformer: $N_s < N_p$ (Voltage decreases, Current increases).

Energy Losses in Transformer

Ideally, input power equals output power. However, practical transformers have small losses.
Note: The efficiency of a well-designed practical transformer is very high (approx. 95–99%).

  1. Flux Leakage: Not all flux from primary passes to secondary. Minimized by soft iron core.
  2. Copper Loss: Heat loss in windings ($I^2R$). Minimized by using thick wires.
  3. Eddy Currents: Heat loss in the core. Minimized by using Laminated Cores.
  4. Hysteresis Loss: Energy lost in magnetization/demagnetization. Minimized by using Soft Iron.

Vol 5: Mission 100 Question Bank

Section A: MCQs

Q1. In a pure inductive circuit, the current:
(a) Leads voltage by $\pi/2$   (b) Lags voltage by $\pi/2$   (c) Is in phase   (d) None
Ans: (b) Lags voltage by $\pi/2$.

Q2. The core of a transformer is laminated to reduce:
(a) Flux leakage   (b) Hysteresis loss   (c) Copper loss   (d) Eddy currents
Ans: (d) Eddy currents.

Section B: Numericals (High Probability)

Q3. A 100 $\Omega$ resistor is connected to a 220 V, 50 Hz AC supply. (a) What is the RMS value of current? (b) What is the net power consumed over a full cycle?

Solution:
(a) $I_{rms} = V_{rms} / R = 220 / 100 = 2.2$ A.
(b) $P = V_{rms} I_{rms} = 220 \times 2.2 = 484$ Watts.
Ans: 2.2 A, 484 W.

Q4. In a series LCR circuit, $R = 20 \Omega$, $X_L = 50 \Omega$, and $X_C = 30 \Omega$. Calculate the Impedance ($Z$).

Solution:
Formula: $Z = \sqrt{R^2 + (X_L - X_C)^2}$
$Z = \sqrt{20^2 + (50 - 30)^2} = \sqrt{400 + 400}$
$Z = \sqrt{800} = 20\sqrt{2} \approx 28.28 \Omega$.
Ans: $28.28 \Omega$.

Section C: Conceptual

Q5. Why is AC preferred over DC for long-distance transmission?
Ans: AC voltage can be stepped up using transformers. Transmission at high voltage and low current ($P=VI$) significantly reduces heat loss ($I^2R$) in transmission wires.

Mission 100 Physics Series

Next Chapter: Electromagnetic Waves (EM Waves) - The Connection!

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